Circles

∠ADB = ∠ACB = 50° (∵ angles in same segment are equal.)

We know that sum of angles in a triangle is 180°.

Considering △ADB,

⇒ ∠ADB + ∠DAB + ∠ABD = 180°
⇒ 50° + 42° + x° = 180°
⇒ x° + 92° = 180°
⇒ x° = 180° - 92°
⇒ x° = 88°

Hence, the value of x = 88°.

From figure,

∠ACB = ∠ADB = 45° (∵ Angles in same segment are equal)

∠DCB = ∠DCA + ∠ACB = 32° + 45° = 77°

We know that,

The opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠DCB + ∠DAB = 180°

⇒ 77° + x° = 180°

⇒ x° = 180° - 77°

⇒ x° = 103°

⇒ x = 103

Hence, the value of x = 103.

Considering △ABC and △ADC,

∠ABC = ∠ADC = 20° (∵ angles in same segment are equal.)

∠DOC = ∠DOB = 90° (As DA is perpendicular to BC)

We know that sum of angles in a triangle is 180°.

Considering △DOC,

⇒ ∠ODC + ∠DOC + ∠OCD = 180°
⇒ 20° + 90° + x° = 180°
⇒ x° + 110° = 180°
⇒ x° = 180° - 110°
⇒ x° = 70°

Hence, the value of x = 70°.

Considering △ABC and △DBC,

∠BAC = ∠BDC = x° (∵ angles in same segment are equal.)

We know that sum of angles in a triangle is 180°.

Considering △ABC,

⇒ ∠BAC + ∠ABC + ∠BCA = 180°
⇒ x° + 69° + 31° = 180°
⇒ x° + 100° = 180°
⇒ x° = 180° - 100°
⇒ x° = 80°

Hence, the value of x = 80°.

Considering △ACB and △CDB,

∠CAB = ∠CDB = x° (∵ angles in same segment are equal.)

Considering △ACP,

∠CPB = ∠APD = 120° (∵ vertically opposite angles are equal.)

Since exterior angle in a triangle is equal to the sum of the opposite interior angles,

⇒ ∠CAP + ∠ACP = ∠APD
⇒ x° + 70° = 120°
⇒ x° = 120° - 70°
⇒ x° = 50°

Hence, the value of x = 50°.

From figure,

∠DAB = ∠BCD = 25° (∵ angles in same segment are equal)

In △DAP,

Exterior angle ∠CDA = ∠DAP + ∠DPA ....(i)

From figure,

∠DAP = ∠DAB = 25°.

Putting values in equation (i),

⇒ x° = 25° + 35°
⇒ x° = 60°.

Hence, the value of x = 60°.

Considering △ACB and △ADB,

∠ACB = ∠ADB = x° (∵ angles in same segment are equal.)

Considering △ACB,

∠ABC = 90° (∵ angle in semicircle is 90°.)

Since sum of angles in a triangle is equal to 180°.

⇒ ∠BAC + ∠ACB + ∠ABC = 180°.
⇒ 40° + x° + 90° = 180°
⇒ x° + 130° = 180°
⇒ x° = 50°.

Hence, the value of x = 50°.

∠ADB = ∠ACB = x° (∵ angles in same segment are equal.)

Considering △AOD,

∠ODA = ∠OAD = 62° (∵ OD = OA, and angles of equal sides are equal.)

∴ x = 62°.

Hence, the value of x = 62°.

Angle of a circle is 360°.

⇒ ∠AOB + ∠AOC + ∠BOC = 360°
⇒ ∠AOB + 80° + 130° = 360°
⇒ ∠AOB + 210° = 360°
⇒ ∠AOB = 360° - 210°
⇒ ∠AOB = 150°.

From figure,

⇒ ∠AOB = 2∠ACB (∵ angle subtended by an arc at the center = double the angle subtended by it any point on the remaining part of the circle.)

⇒ 150° = 2∠ACB
⇒ ∠ACB = $\frac{150°}{2}$
⇒ ∠ACB = 75°
⇒ x° = 75°.

Hence, the value of x = 75°.

AC subtends reflex angle ∠AOC at center and ∠ABC at point B. Since angle subtended at center is double the angle subtended at any other part of the circle.

⇒ Reflex ∠AOC = 2∠ABC .....(i)

From figure,

∠ABC + ∠CBD = 180° (As they are linear pair.)

∠ABC + 75° = 180°
∠ABC = 180° - 75°
∠ABC = 105°.

Putting value of ∠ABC = 105° in (i)

x° = 2(105°)
x° = 210°

Hence, the value of x = 210.

From figure,

⇒ ∠AOC + ∠COB = 180°
⇒ ∠COB = 180° - ∠AOC
⇒ ∠COB = 180° - 135°
⇒ ∠COB = 45°.

BC subtends ∠COB = 45° at center and ∠CDB = x° at point D. Since angle subtended at center is double the angle subtended at any other part of the circle.

⇒ ∠COB = 2∠CDB

⇒ 45° = 2x°

⇒ x° = $\frac{45^{\circ}}{2} = 22\frac{1^{\circ}}{2}.$

Hence, the value of x = $22\frac{1^{\circ}}{2}$.

Label the points as shown in the figure below:

Arc AD subtends ∠AOD = 70° at center.

⇒ ∠AOD = 2∠ABD (∵ angle subtended by an arc at the center = double the angle subtended by it any point on the remaining part of the circle.)

⇒ 70° = 2∠ABD
⇒ ∠ABD = $\frac{70°}{2}$
⇒ ∠ABD = 35°

∠ABM = ∠ABD.

Considering △ABM,

Since sum of angles in a triangle is equal to 180°.

⇒ ∠AMB + ∠ABM + ∠BAM = 180°.
⇒ 90° + 35° + x° = 180°
⇒ x° + 125° = 180°
⇒ x° = 55°.

Hence, the value of x = 55.

From figure,

∠DAC = ∠ACB = 35° (∵ alternate angles are equal.)

∠DAC = ∠DBC = 35° (∵ angles in same segment are equal.)

Hence, the value of ∠DBC = 35°.

From figure,

⇒ ∠AOC + Reflex ∠AOC = 360°
⇒ 130° + Reflex ∠AOC = 360°
⇒ Reflex ∠AOC = 360° - 130°
⇒ Reflex ∠AOC = 230°.

Arc AC subtends Reflex ∠AOC at center and ∠ABC at another point of circle.

⇒ Reflex ∠AOC = 2 ∠ABC

⇒ 2∠ABC = 230°

⇒ ∠ABC = $\frac{230°}{2}$

⇒ ∠ABC = 115°.

Hence, the value of ∠ABC = 115°.

In the figure,

ABCD is a cyclic quadrilateral.

⇒ ∠B + ∠D = 180°
⇒ 40° + 45° + y = 180°
⇒ 85° + y = 180°
⇒ y = 180° - 85°
⇒ y = 95°.

Considering △ABD and △ACD,

∠ABD = ∠ACD = 40° (∵ angles in same segment are equal.)

x = 40°.

Hence, the value of x = 40° and y = 95°.

From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ 120° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - 120°
⇒ Reflex ∠AOB = 240°.

Arc AB subtends Reflex ∠AOB at center and ∠ADB at point D of circle.

⇒ Reflex ∠AOB = 2 ∠ADB
⇒ 2∠ADB = 240°
⇒ ∠ADB = $\frac{240°}{2}$
⇒ ∠ADB = 120°
⇒ y° = 120°

Arc AB subtends ∠AOB at center and ∠ACB at point C of circle.

⇒ ∠AOB = 2 ∠ACB
⇒ 2∠ACB = 120°
⇒ ∠ACB = $\frac{120°}{2}$
⇒ ∠ACB = 60°
⇒ x° = 60°

Hence, the value of x = 60 and y = 120.

Considering △YBN,

Since sum of angles in a triangle is equal to 180°.

⇒ ∠NYB + ∠YNB + ∠YBN = 180°.
⇒ 50° + 20° + ∠YBN = 180°
⇒ 70° + ∠YBN = 180°
⇒ ∠YBN = 180° - 70°
⇒ ∠YBN = 110°.

From figure,

∠MBN = ∠YBN = 110°.

Considering △MAN and △MBN,

∠MAN = ∠MBN = 110° (∵ angles in same segment are equal.)

MN subtends Reflex ∠MON at center and ∠MAN at point A of circle.

⇒ Reflex ∠MON = 2 ∠MAN = 2 × 110° = 220°.

Hence, ∠MAN = 110° and Reflex ∠MON = 220°.

(i) From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ 140° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - 140°
⇒ Reflex ∠AOB = 220°.

Arc AB subtends Reflex ∠AOB at center and ∠ACB at point C of circle.

⇒ Reflex ∠AOB = 2 ∠ACB
⇒ 2∠ACB = 220°
⇒ ∠ACB = $\frac{220°}{2}$
⇒ ∠ACB = 110°.

Hence, the value of ∠ACB = 110°.

(ii) In Quadrilateral OABC,

⇒ ∠OAC + ∠ACB + ∠BOA + ∠OBC = 360°
⇒ 50° + 110° + 140° + ∠OBC = 360°
⇒ 300° + ∠OBC = 360°
⇒ ∠OBC = 360° - 300°
⇒ ∠OBC = 60°

Hence, the value of ∠OBC = 60°.

(iii) In △OAB,

OA = OB (Radius of the circle)

∠OAB = ∠OBA = x (∵ angles of equal sides in isosceles triangle are equal.)

Sum of angles in a triangle are equal,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°.
⇒ 140° + x + x = 180°
⇒ 140° + 2x = 180°
⇒ 2x = 40°
⇒ x = 20°.

Hence, the value of ∠OAB = 20°.

(iv) ∠CBA = ∠OBC - ∠OBA

⇒ ∠CBA = 60° - 20° ⇒ ∠CBA = 40°

Hence, the value of ∠CBA = 40°.

(i) From figure,

⇒ ∠ADC = ∠ABC (Angles in same segment are equal)

⇒ ∠ADC = 62°.

Hence, ∠ADC = 62°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ACB = 90°

Using angle sum property,

⇒ ∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 90° + 62° = 180°

⇒ ∠CAB + 152° = 180°

⇒ ∠CAB = 180° - 152°

⇒ ∠CAB = 28°.

Hence, ∠CAB = 28°.

(i) In △EDC,

∠ECD = ∠EDC = 32° (Given)

Since sum of angles of triangle = 180°.

⇒ ∠DEC + ∠ECD + ∠EDC = 180°
⇒ ∠DEC + 32° + 32° = 180°
⇒ ∠DEC = 180° - 64°
⇒ ∠DEC = 116°.

Since, ∠CEF and ∠DEC are linear pair,

∴ ∠CEF + ∠DEC = 180°
⇒ ∠CEF + 116° = 180°
⇒ ∠CEF = 180° - 116°
⇒ ∠CEF = 64°

Hence, ∠CEF = 64°.

(ii) ∠FDC = ∠EDC = 32°. (From figure)

Arc FC subtends ∠COF at center and ∠FDC at point D of circle so,

⇒ ∠COF = 2 ∠FDC
⇒ ∠COF = 2 × 32°
⇒ ∠COF = 64°

Hence, the value of ∠COF = 64°.

(i) ∠PRB = ∠PAB = 35° (∵ angles in same segment are equal.)

Hence, the value of ∠PRB = 35°

(ii) From figure,

∠APB = 90° (∵ angle in semicircle is 90°.)

⇒ ∠APB + ∠BPQ = 180° (∵ angles form a linear pair).
⇒ 90° + ∠BPQ = 180°
⇒ ∠BPQ = 90°.

Exterior angle in a triangle is equal to the sum of opposite two interior angles.

In △PBQ,

Ext. ∠PBR = ∠PQB + ∠BPQ = 25° + 90° = 115°.

Hence, the value of ∠PBR = 115°

(iii) In △PRQ,

Ext. ∠APR = ∠PRQ + ∠PQR = ∠PRB + ∠PQR = 35° + 25° = 60°.

∠APB = 90° (∵ angle in semicircle is 90°.)

From figure,

∠BPR = ∠APB - ∠APR = 90° - 60° = 30°.

Hence, the value of ∠BPR = 30°

Consider △ABC and △ADC,

∠ABC = ∠ADC = 40° (∵ angles in same segment are equal.)

In △ADC,

∠DCA = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠DAC + ∠ADC + ∠DCA = 180°.
⇒ ∠DAC + 40° + 90° = 180°
⇒ ∠DAC + 130° = 180°
⇒ ∠DAC = 180° - 130°
⇒ ∠DAC = 50°.

Hence, the value of ∠DAC = 50°.

Arc AB subtends ∠APB at center and ∠ACB at the point C on the circle.

∴ ∠APB = 2∠ACB

⇒ ∠APB = 2∠ACB
⇒ 130° = 2∠ACB
⇒ ∠ACB = 65°.

From figure,

∠ACB + ∠BCD = 180°. (∵ both angles form a linear pair)

⇒ 65° + ∠BCD = 180°
⇒ ∠BCD = 180° - 65°
⇒ ∠BCD = 115°.

In circle with center Q,

⇒ ∠BQD + Reflex ∠BQD = 360°
⇒ x° + Reflex ∠BQD = 360°
⇒ Reflex ∠BQD = 360° - x°.

Arc BD subtends reflex ∠BQD at center and ∠BCD at the point C on the circle.

∴ Reflex ∠BQD = 2∠BCD

⇒ 360° - x° = 2 × 115°
⇒ 360° - x° = 230°
⇒ x° = 360° - 230°
⇒ x° = 130°.

Hence, the value of x = 130.

(i) From figure,

AC = BC so,

∠CBA = ∠CAB (As angles of equal sides are equal)

In △ABC,

∠CAB + ∠CBA + ∠ACB = 180°
2∠CAB + 56° = 180°
2∠CAB = 180° - 56°
2∠CAB = 124°
∠CAB = 62°.

Hence, ∠CAB = 62°.

(ii) OC is the radius of the circle. OC bisects ∠ACB.

∠OCA = $\frac{1}{2}$∠ACB = $\frac{1}{2} \times$ 56° = 28°.

Now in △OCA,

OA = OC (Radius of the same circle)

∠OAC = ∠OCA = 28°.

Hence, ∠OAC = 28°.

Consider △AEC and △EBC,

∠EAC = ∠EBC = 65° (∵ angles in same segment are equal.)

In △AEC,

∠AEC = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠AEC + ∠EAC + ∠ACE = 180°.
⇒ 90° + 65° + ∠ACE = 180°
⇒ ∠ACE + 155° = 180°
⇒ ∠ACE = 180° - 155°
⇒ ∠ACE = 25°.

We know that,

Alternate angles are equal.

∴ ∠DEC = ∠ACE = 25°.

Hence, the value of ∠DEC = 25°.

From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ q° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - q°.

Arc AB subtends reflex ∠AOB at center and ∠ACB at the point C on the circle.

∴ Reflex ∠AOB = 2∠ACB

⇒ 360° - q° = 2 × p°
⇒ 360° - q° = 2p°
⇒ 2p° + q° = 360°
⇒ q° = 360° - 2p°
⇒ q° = 2(180° - p°)
⇒ q = 2(180 - p).

Given, OABC is a parallelogram, then

Opposite angles are equal.

∴ ∠AOB = ∠ACB

⇒ p° = q°
⇒ p° = 360° - 2p°
⇒ 3p° = 360°
⇒ p° = 120°.

Hence, q = 2(180 - p) and the value of p = 120.

(i) In △CED,

∠CED = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠CED + ∠DCE + ∠CDE = 180°.
⇒ 90° + 40° + ∠CDE = 180°
⇒ ∠CDE + 130° = 180°
⇒ ∠CDE = 180° - 130°
⇒ ∠CDE = 50°.

Hence, the number of degrees in ∠CDE = 50.

(ii) From figure,

∠AOD + ∠DOB = 180° (∵ they form linear pair)
⇒ 75° + ∠DOB = 180°
⇒ ∠DOB = 180° - 75°
⇒ ∠DOB = 105°.

In △DOB,

∠ODB = ∠CDE = 50°

We know that sum of angles of a triangle is 180°.

⇒ ∠DOB + ∠ODB + ∠DBO = 180°.
⇒ 105° + 50° + ∠DBO = 180°
⇒ ∠DBO + 155° = 180°
⇒ ∠DBO = 180° - 155°
⇒ ∠DBO = 25°.

From figure,

∠OBE = ∠DBO

∴ ∠OBE = 25°.

Hence, the number of degrees in ∠OBE = 25.

(i) Join BI and CI as shown in the figure below:

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° (∵ sum of angles = 180°.)
⇒ ∠BAC + 55° + 65° = 180°
⇒ ∠BAC + 120° = 180°
⇒ ∠BAC = 180° - 120°
⇒ ∠BAC = 60°.

I is the incentre,

∴ I lies on the bisectors of angle of the △ABC,

∴ ∠BAD = ∠CAD = $\frac{60°}{2}$ = 30°.

∠BCD = ∠BAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠BCD = 30°

(ii) Similarly,

∠CBD = ∠CAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠CBD = 30°

(iii) The line CI bisects ∠C (∵ I lies on the bisectors of angle of the △ABC).

∴ ∠BCI = $\frac{65°}{2} = 32\frac{1}{2}°$.

From figure,

∠DCI = ∠BCD + ∠BCI = $30° + 32\frac{1}{2}° = 62\frac{1}{2}°$.

Hence, the value of ∠DCI = $62\frac{1}{2}°$.

(iv) ∠IBC = $\frac{55°}{2} = 27\frac{1}{2}°$

∠ICB = $\frac{65°}{2} = 32\frac{1}{2}°$

∠BIC = 180° - (∠IBC + ∠ICB)

$$\begin{aligned}= 180° - \Big(\frac{55°}{2} + \frac{65°}{2}\Big) \\[1em] = 180° - \Big(\frac{120°}{2}\Big) \\[1em] = 180° - 60° \\[1em] = 120°.\end{aligned}$$

Hence, the value of ∠BIC = 120°.

From the below figure:

Arc BC subtends ∠BOC at center and ∠BAC at the point A on the circle.

∴ ∠BOC = 2∠A

In △OBD and △ODC,

OD = OD (Common side)

BD = CD (As D is the mid-point of BC)

OB = OC (Radius of the same circle)

∴ △OBD ≅ △ODC (SSS rule of congruency).

∴ ∠BOD = ∠COD (As corresponding part of congruent triangles are congruent.)

Since, ∠BOD = ∠COD so,

∠BOD = $\frac{1}{2}$∠BOC ....(i)

∠BOC = 2∠A
∠A = $\frac{1}{2}$ ∠BOC .....(ii)

From (i) and (ii) we get,

∠BOD = ∠A

Hence, proved that ∠BOD = ∠A.

In △AEB and △CED,

∠A = ∠C (∵ angles in same segment of a circle are equal.)

∠B = ∠D (∵ angles in same segment of a circle are equal.)

AB = CD (Given)

∴ △AEB ≅ △CED (By ASA axiom)

As corresponding part of congruent triangles are congruent hence,

AE = CE and BE = DE.

Hence, proved that AE = CE and BE = DE.

Join AE.

∠AEB = 90° (∵ angle in semicircle is 90°.)

∠AED = 90° (∵ ∠AEB and ∠AED form a linear pair.)

Hence, we can say that,

DE is also perpendicular to AE, since DE is also perpendicular to PQ hence,

AE || PQ.

Since, CA and DE both are perpendicular to PQ hence,

CA || DE.

Hence, proved that ACDE is a rectangle.

In rectangle opposite sides are equal so,

AC = DE.

Hence, proved that AC = DE.

Join OC and OD as shown in the figure below:

AC subtends angle AOC at centre and ∠ABC at point B.

∴ ∠AOC = 2∠ABC = 2 × 25° = 50°.

From figure,

∠OCD = ∠AOC (Alternate angles)

Hence, ∠OCD = 50°.

In △OCD,

OC = OD (Both are radius of the circle)

so, ∠ODC = ∠OCD.

Since, sum of angles of a triangle is 180°.

⇒ ∠COD + ∠OCD + ∠ODC = 180°
⇒ ∠COD + 50° + 50° = 180°
⇒ ∠COD + 100° = 180°
⇒ ∠COD = 80°.

CD subtends ∠COD at center and ∠CED at point E of the circle.

∴ ∠COD = 2∠CED
⇒ 80° = 2∠CED
⇒ ∠CED = 40°.

Hence, ∠CED = 40°.

Join AD.

Arc AC subtends ∠AOC at the center and ∠ADC at the point D of the circle.

∴ ∠AOC = 2∠ADC (As angle at center = double the angle at the remaining part of the circle)

∠OAD = ∠ADC (∵ alternate angles are equal.)

∴ ∠AOC = 2∠OAD .....(i)

Since, opposite angles are equal in parallelogram,

∴ ∠ABC = ∠AOC

Putting values of ∠AOC in eqn (i) we get,

∠ABC = 2∠OAD.

Hence, proved that ∠ABC = 2∠OAD.

Given, two chords AQ and BC intersect each other at P inside the circle. AB and CQ are joined and AB = AP.

To prove : CP = CQ

Construction : Join AC.

Proof :

In △ABP and △CQP

∠B = ∠Q (∵ angles in same segment are equal)

∠BAP = ∠PCQ (∵ angles in same segment are equal)

∠BPA = ∠CPQ (∵ vertically opposite angles are equal.)

∴ △ABP ~ △CQP (By AAA axiom of similarity.)

Since, triangles are similar hence, the ratio of the corresponding sides are equal.

$\frac{AB}{CQ} = \frac{AP}{CP}$

We know, AB = AP,

$$\begin{aligned}\therefore \frac{AP}{CQ} = \frac{AP}{CP} \\[1em] CQ = CP.\end{aligned}$$

Hence, proved that CQ = CP.

(i) In △ACD,

AC = CD

∴ ∠CAD = ∠ADC = 38° (∵ angles of equal sides in triangle are equal)

∠ACB = ∠CAD + ∠ADC = 38° + 38° = 76° (∵ exterior angle = sum of two opposite interior angles.)

In △ABC,

AB = AC

∴ ∠ABC = ∠ACB = 76° (As angles of equal sides in triangle are equal)

Hence, the value of ∠ABC = 76°.

(ii) We know that sum of angles in a triangle = 180°.

⇒ ∠BAC + ∠ABC + ∠ACB = 180°
⇒ ∠BAC + 76° + 76° = 180°
⇒ ∠BAC + 152° = 180°
⇒ ∠BAC = 180° - 152°
⇒ ∠BAC = 28°.

∠BEC = ∠BAC (∵ angles in same segment are equal.)

∠BEC = 28°.

Hence, the value of ∠BEC = 28°.

From figure,

∠ACB = ∠ADB (∵ angles in same segment are equal.) ....(i)

∠ACP = ∠ADP (∵ angles in same segment are equal.) ....(ii)

We know,

∠ACP = $\frac{1}{2}$∠ACB (∵ CP bisects ∠ACB.) .....(iii)

Using values from eq (i) and eq (ii) and putting in eq (iii) we get,

∠ADP = $\frac{1}{2}$∠ADB

Hence, proved that DP bisects ∠ADB.

Join CD as shown in the figure below:

In △ABE and △BCD,

∠A = ∠D (∵ angles in same segment are equal.)

∠ABE = ∠DBC (As BD is bisector of ∠ABC)

△ABE ~ △BCD (AA rule of similarity).

Since, ratio of corresponding sides of similar triangles are equal,

∴ $\frac{AB}{BD} = \frac{BE}{BC}.$

Hence, proved.

(i) In △CBE and △ADE,

∠B = ∠D (∵ angles in same segment are equal.)

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

△CBE ~ △ADE. (By AA axiom)

Hence, proved that △CBE ~ △ADE.

(ii) Given, DC = 12 cm.

From figure,

⇒ DC = DE + EC
⇒ 12 = 4 + EC
⇒ EC = 12 - 4
⇒ EC = 8 cm.

Chords AB and CD intersect each other at E.

Considering △BEC and △AED,

∠BEC = ∠DEA (∵ vertically opposite angles are equal.)

∠CBE = ∠EDA (∵ both angles are subtended on circle by arc AC and angles in same segment are equal.)

Hence, △BEC ~ △AED.

Since triangles are similar hence the ratio of the corresponding sides are equal.

$$\begin{aligned}\therefore \frac{BE}{DE} = \frac{EC}{EA} \\[1em] \Rightarrow BE \times EA = EC \times DE \\[1em] BE \times 16 = 8 \times 4 \\[1em] BE = \frac{32}{16} \\[1em] BE = 2.\end{aligned}$$

Hence, the length of BE = 2 cm.

In △APD and △CPB,

∠DAB = ∠DCB (∵ angles in same segment are equal.)

∠APD = ∠CPB (∵ vertically opposite angles are equal.)

△APD ~ △CPB. (By AA axiom)

Hence, proved that △APD ~ △CPB.

Chords AB and CD intersect each other at P.

Since △APD ~ △CPB, Hence, the ratio of corresponding sides are equal.

$\therefore \frac{AP}{CP} = \frac{PD}{PB}$

∴ AP × PB = CP × PD .....(i)

From figure,

CD = CP + PD

Let CP = x cm.

⇒ 14 = x + PD
⇒ PD = (14 - x) cm.

Putting values in eq (i)

⇒ 6 × 4 = x × (14 - x)

⇒ 24 = 14x - x²

⇒ x² - 14x + 24 = 0

⇒ x² - 12x - 2x + 24 = 0

⇒ x(x - 12) - 2(x - 12) = 0

⇒ (x - 2)(x - 12) = 0

⇒ x - 2 = 0 or x - 12 = 0

⇒ x = 2 or x = 12.

Since, given PC > PD so, CP = 12 cm.

Hence, the length of CP = 12 cm.

Join A and B as shown in the figure below:

From figure,

Since, ∠CDE = 90° so, ∠ADB = 90° (∵ vertically opposite angles are equal.)

In right angle triangle △ADB, by pythagoras theorem,

$$\begin{aligned}\Rightarrow AB^2 = AD^2 + BD^2 \\[1em] \Rightarrow 5^2 = AD^2 + 4^2 \\[1em] \Rightarrow 25 = AD^2 + 16 \\[1em] \Rightarrow AD^2 = 25 - 16 \\[1em] \Rightarrow AD^2 = 9 \\[1em] \Rightarrow AD = 3 \text{ cm}.\end{aligned}$$

Chords AE and CB intersect each other at D.

In △ADB and △CDE,

∠BAD = ∠DCE (∵ angles in same segment are equal.)

∠ADB = ∠CDE (∵ vertically opposite angles are equal.)

△ADB ~ △CDE. (By AA axiom)

Since △ADB ~ △CDE, Hence, the ratio of corresponding sides are equal.

$\therefore \frac{AD}{CD} = \frac{BD}{DE}$

∴ AD × DE = CD × BD

⇒ 3 × DE = 9 × 4

⇒ DE = $\frac{36}{3}$

⇒ DE = 12 cm.

Hence, the length of DE = 12 cm.

We know that ∠PQR = 90° as angle in semicircle is equal to 90°.

So, △PQR is a right angled triangle.

By pythagoras theorem,

$$\begin{aligned}PR^2 = PQ^2 + QR^2 \\[1em] PR^2 = 7^2 + 6^2 \\[1em] PR^2 = 49 + 36 \\[1em] PR^2 = 85.\end{aligned}$$

In △PRS,

∠PSR = 90° as angle in semicircle is equal to 90°.

So, △PRS is a right angled triangle.

By pythagoras theorem,

$$\begin{aligned}PR^2 = PS^2 + RS^2 \\[1em] 85 = PS^2 + 2^2 \\[1em] PS^2 = 85 - 4 \\[1em] PS^2 = 81 \\[1em] PS = 9.\end{aligned}$$

Perimeter of PQRS = PQ + QR + RS + SP = 7 + 6 + 2 + 9 = 24 cm.

Hence, the perimeter of cyclic quadrilateral is 24 cm.

In △ABP and △DPC,

∠APB = ∠DPC (∵ vertically opposite angles are equal.)

∠ABP = ∠DCP (∵ angles in same segment are equal.)

△APB ~ △DPC. (By AA axiom)

We know that ratio of the area of similar triangles is the ratio of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APB}}{\text{Area of △DPC}} = \frac{AB^2}{CD^2} \\[1em] \Rightarrow \frac{24}{\text{Area of △DPC}} = \frac{64}{25} \\[1em] \Rightarrow \text{Area of △DPC} = \frac{24 \times 25}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \frac{600}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \frac{75}{8} \\[1em] \Rightarrow \text{Area of △DPC} = 9\frac{3}{8}. \\[1em]\end{aligned}$$

Hence, the area of △DPC = $9\frac{3}{8}$ cm².

(i) As we know that,

If two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

∴ PA.PD = PB.PC

$\Rightarrow \frac{PA}{PC} = \frac{PB}{PD}$

⇒ ∠APB = ∠CPD (Vertically opposite angles)

If the corresponding sides of two triangles are proportional and one angle are equal, then the two triangles are similar.

Hence, proved that Δ PAB ∼ Δ PCD (By SAS rule of similarity).

(ii) Since, Δ PAB ∼ Δ PCD

$\therefore \frac{PA}{PC} = \frac{AB}{CD}$

Substituting the values, we get :

$$\begin{aligned}\Rightarrow \frac{7.5}{5} = \frac{9}{CD} \\[1em] \Rightarrow 1.5 = \frac{9}{CD} \\[1em] \Rightarrow CD = \frac{9}{1.5} \\[1em] \Rightarrow CD = 6.\end{aligned}$$

Hence, the length of CD = 6 cm.

(iii) As we know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of ΔPAB}}{\text{Area of ΔPCD}} = \frac{PA^2}{PC^2} \\[1em] = \frac{(7.5)^2}{5^2} \\[1em] = \frac{56.25}{25} \\[1em] = \frac{9}{4}\end{aligned}$$

Hence, the area of ΔPAB : area of ΔPCD = 9 : 4.

In △XYQ and △XPZ,

∠Q = ∠Z (∵ angles in same segment are equal as arc XY subtends both the angles at the circle.)

∠YXQ = ∠PXZ (∵ QPX is the bisector of ∠YXZ hence it divides the angle in two equal halves.)

△XYQ ~ △XPZ. (By AA axiom)

Since triangles are similar hence, the ratio of corresponding sides are similar,

$$\begin{aligned}\therefore \frac{XY}{XP} = \frac{XQ}{XZ} \\[1em] \Rightarrow \frac{XY}{XQ} = \frac{XP}{XZ} \\[1em] \Rightarrow XY : XQ = XP : XZ \\[1em]\end{aligned}$$

Hence, proved that XY : XQ = XP : XZ.

(i) From figure,

∠PAD + ∠DAB = ∠PCB + ∠BCD (∵ both are equal to 180°)....(i)

∠DAB = ∠BCD (∵ angles in same segment are equal.)

Putting this value of ∠DAB in (i) we get,

⇒ ∠PAD + ∠BCD = ∠PCB + ∠BCD
⇒ ∠PAD = ∠PCB + ∠BCD - ∠BCD
⇒ ∠PAD = ∠PCB

Hence, proved that ∠PAD = ∠PCB.

(ii) In △PBC and △PAD,

∠PAD = ∠PCB (Proved above.)

∠P = ∠P (Common angle.)

△PBC ~ △PAD. (By AA axiom)

Since triangles are similar hence, the ratio of corresponding sides are similar,

$$\begin{aligned}\therefore \frac{PC}{PA} = \frac{PB}{PD} \\[1em] \Rightarrow PA \times PB = PC \times PD.\end{aligned}$$

Hence proved that PA × PB = PC × PD.

From figure,

ABCD is a cyclic quadrilateral.

By property of cyclic quadrilateral — An exterior angle of cyclic quadrilateral is equal to opposite interior angle.

We get,

∠BAD = ∠DCE = x°

Arc BD subtends ∠BOD on center and ∠BAD at point A of the circle.

∴ ∠BOD = 2∠BAD (∵ as angle subtended on center of circle is double the angle subtended on any other part.)

⇒ 150° = 2x°
⇒ x° = $\frac{150°}{2}$
⇒ x° = 75°.

Hence, the value of x = 75.

From figure,

ABCD is a cyclic quadrilateral.

By property of cyclic quadrilateral — An exterior angle of cyclic quadrilateral is equal to opposite interior angle.

We get,

∠BAD = ∠DCE = 80°

Arc BD subtends ∠BOD on center and ∠BAD at point A of the circle.

From figure,

∠BOD = 360° - x°

∴ ∠BOD = 2∠BAD (∵ as angle subtended on center of circle is double the angle subtended on any other part.)

⇒ 360° - x° = 2 × 80°
⇒ x° = 360° - 160°
⇒ x° = 200°.

Hence, the value of x = 200.

From figure,

ABCD is a cyclic quadrilateral.

∠ACB = 90°. (∵ angle in semi circle is 90°)

Since sum of angles in triangle is 180°.

∴ In △ABC

⇒ ∠CAB + ∠ACB + ∠ABC = 180°
⇒ 25° + 90° + ∠ABC = 180°
⇒ 115° + ∠ABC = 180°
⇒ ∠ABC = 65°.

Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠ABC + ∠ADC = 180°
⇒ 65° + x° = 180°
⇒ x° = 180° - 65°
⇒ x° = 115°.

Hence, the value of x = 115.

(i) Arc AC subtends ∠AOC on center and ∠ABC at point B of the circle.

∴ ∠AOC = 2∠ABC (∵ as angle subtended on center of circle is double the angle subtended on any other part.)

⇒ 150° = 2∠ABC

⇒ ∠ABC = $\frac{150°}{2}$

⇒ ∠ABC = 75°.

Hence, the value of ∠ABC = 75°.

(ii) From figure,

ABCD is a cyclic quadrilateral.

Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠ABC + ∠ADC = 180°
⇒ 75° + x° = 180°
⇒ x° = 180° - 75°
⇒ x° = 105°.

Hence, the value of ∠ADC = 105°.

(i) From figure,

∠ABC = 90° (∵ angle in semicircle is equal to 90°)

Hence, the value of ∠ABC = 90°.

(ii) Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠BAD + ∠BCD = 180°
⇒ ∠BAD + 75° = 180°
⇒ ∠BAD = 180° - 75°
⇒ ∠BAD = 105°.

From figure,

∠EAF = ∠BAD = 105°.

Hence, the value of ∠EAF = 105°.

(i) Given,

∠DBC = 58°

From figure,

∠BCD = 90° (∵ angle in semicircle is equal to 90°.)

Since sum of angles in triangle is 180°.

∴ In △BCD

⇒ ∠DBC + ∠BCD + ∠BDC = 180°
⇒ 58° + 90° + ∠BDC = 180°
⇒ 148° + ∠BDC = 180°
⇒ ∠BDC = 180° - 148°
⇒ ∠BDC = 32°.

Hence, the value of ∠BDC = 32°.

(ii) Considering quadrilateral BDCE.

From figure,

BDCE is a cyclic quadrilateral.

Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠BDC + ∠BEC = 180°
⇒ 32° + ∠BEC = 180°
⇒ ∠BEC = 180° - 32°
⇒ ∠BEC = 148°.

Hence, the value of ∠BEC = 148°.

(iii) From figure,

∠BAC = ∠BDC (∵ angles in same segment are equal.)

∴ ∠BAC = 32°.

Hence, the value of ∠BAC = 32°.

(i) ABCD is a cyclic quadrilateral.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∴ ∠BCE = ∠DAB

⇒ ∠BCE = ∠CAD + ∠BAC
⇒ 80° = ∠CAD + 25°
⇒ ∠CAD = 80° - 25°
⇒ ∠CAD = 55°.

Hence, the value of ∠CAD = 55°.

(ii) From figure,

∠CBD = ∠CAD (∵ angles in same segment are equal.)

∠CBD = 55°.

Hence, the value of ∠CBD = 55°.

(iii) ∠BAC = ∠BDC (∵ angles in same segment are equal.)

∠BDC = 25°.

Now AB || DC and BD is the transversal

∠ABD = ∠BDC (∵ ∵ alternate angles are equal.)

⇒ ∠ABD = 25°

From figure,

∠ABC = ∠ABD + ∠CBD = 25° + 55° = 80°.

Since, sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠ABC + ∠ADC = 180°
⇒ 80° + ∠ADC = 180°
⇒ ∠ADC = 180° - 80°
⇒ ∠ADC = 100°.

Hence, the value of ∠ADC = 100°.

In the given figure,

ABCD is a cyclic quadrilateral.

Since, sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠ABC + ∠ADC = 180°
⇒ ∠ABC + 80° = 180°
⇒ ∠ABC = 180° - 80°
⇒ ∠ABC = 100°.

From figure,

∠DBA = ∠DCA = 52°. (∵ angles in same segment are equal.)

⇒ ∠ABC = ∠DBA + ∠CBD
⇒ 100° = 52° + ∠CBD
⇒ ∠CBD = 100° - 52°
⇒ ∠CBD = 48°.

Hence, ∠ABC = 100° and ∠CBD = 48°.

From figure,

∠DAB = ∠DAO = 51°.

ABED is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

∠BEC = ∠DAB = 51° (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)

Reflex ∠AOE = 360° - ∠AOE = 360° - 150° = 210°.

Arc AE subtends ∠ADE at point D and Reflex ∠AOE at center.

Reflex ∠AOE = 2∠ADE (∵ angle subtended by an arc at center is double the angle subtended at any other point of the circle.)

210° = 2∠ADE
∠ADE = $\frac{210°}{2}$
∠ADE = 105°.

∠EBC = ∠ADE = 105° (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)

Hence, ∠BEC = 51° and ∠EBC = 105°.

ADFE is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

∠ADF = ∠FEB = 80° (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)

As opposite angles of a parallelogram are equal.

∠ABC = ∠ADC

From figure,

∠ADC = ∠ADF = 80°.

∴ ∠ABC = 80°.

Hence, the value of ∠ABC = 80°.

(i) In trapezium sum of angles on same side = 180°.

⇒ ∠ABC + ∠BAD = 180°
⇒ 70° + ∠BAD = 180°
⇒ ∠BAD = 180° - 70°
⇒ ∠BAD = 110°

Hence, the value of ∠BAD = 110°.

(ii) ABCD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠BAD + ∠BCD = 180°
⇒ 110° + ∠BCD = 180°
⇒ ∠BCD = 180° - 110°
⇒ ∠BCD = 70°.

Hence, the value of ∠BCD = 70°.

From figure,

ABCD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠A + ∠C = 180°
⇒ 30° + p = 180°
⇒ p = 180° - 30°
⇒ p = 150°.

∠BAD = ∠BED (∵ angles in same segment are equal.)

⇒ r = 30°.

Arc BD subtends ∠BAD at point A and ∠BOD at center.

∠BOD = 2∠BAD (∵ angle subtended by an arc at center is double the angle subtended at any other point of the circle.)

q = 2 × 30°
q = 60°.

Hence, the value of p = 150°, q = 60° and r = 30°.

Join PQ as shown in the figure below:

PQAD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠DAQ + ∠DPQ = 180°
⇒ 80° + ∠DPQ = 180°
⇒ ∠DPQ = 180° - 80°
⇒ ∠DPQ = 100°.

Also,

⇒ ∠PDA + ∠PQA = 180°
⇒ 84° + ∠PQA = 180°
⇒ ∠PQA = 180° - 84°
⇒ ∠PQA = 96°.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠QBC = ∠DPQ = 100°

∠BCP = ∠PQA = 96°.

Hence, the value of ∠QBC = 100° and ∠BCP = 96°.

(i) From figure,

∠PRQ = 90° (∵ angle in semicircle is equal to 90°.)

Since sum of angles in triangle is 180°.

⇒ ∠PRQ + ∠RQP + ∠RPQ = 180°
⇒ 90° + 58° + ∠RPQ = 180°
⇒ ∠RPQ + 148° = 180°
⇒ ∠RPQ = 180° - 148
⇒ ∠RPQ = 32°.

Hence, the value of ∠RPQ = 32°.

(ii) From figure,

∠SRP = ∠RPQ = 32° (∵ alternate angles are equal.)

Since, PTSR is a cyclic quadrilateral so sum of its opposite angles is equal to 180°.

⇒ ∠SRP + ∠STP = 180°
⇒ 32° + ∠STP = 180°
⇒ ∠STP = 180° - 32°
⇒ ∠STP = 148°.

Hence, the value of ∠STP = 148°.

From figure,

∠CAE = ∠CAF = 62°

Since sum of angles in triangle is 180°.

In △ACE,

⇒ ∠CAE + ∠ACE + ∠CEA = 180°
⇒ 62° + 43° + ∠CEA = 180°
⇒ ∠CEA + 105° = 180°
⇒ ∠CEA = 180° - 105°
⇒ ∠CEA = 75°.

From figure,

⇒ ∠CEA + ∠DEF = 180° (As they are linear pair.)
⇒ 75° + ∠DEF = 180°
⇒ ∠DEF = 180° - 75°
⇒ ∠DEF = 105°.

ABDE is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

⇒ ∠ABD = ∠DEF = 105°
⇒ a = 105°.

Since sum of angles in triangle is 180°.

In △ABF,

⇒ ∠BAF + ∠ABF + ∠BFA = 180°
⇒ 62° + a + b = 180°
⇒ 62° + 105° + b = 180°
⇒ b + 167° = 180°
⇒ b = 180° - 167°
⇒ b = 13°

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

⇒ ∠EDF = ∠BAE = 62°
⇒ c = 62°.

Hence, the value of a = 105°, b = 13° and c = 62°.

Join CB as shown in the figure below:

So, ABCD becomes a cyclic quadrilateral.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠CBA + ∠ADC = 180°
⇒ ∠CBA + 120° = 180°
⇒ ∠CBA = 180° - 120°
⇒ ∠CBA = 60°.

Join AC.

In △ABC

∠ACB = 90° (∵ angle in semicircle is equal to 90°.)

Sum of angles of triangle = 180°

⇒ ∠CAB + ∠ACB + ∠CBA = 180°
⇒ ∠CAB + 90° + 60° = 180°
⇒ ∠CAB + 150° = 180°
⇒ ∠CAB = 180° - 150°
⇒ ∠CAB = 30°.

Hence, the value of ∠CAB = 30°.

In figure,

ABCD is a cyclic quadrilateral.

∠DAB = ∠BCE = x (Property of cyclic quadrilateral by which an exterior angle = opposite interior angle.)

In △BCE

∠CBE = 180° - (x° + y°)

From figure,

∠CBE and ∠CBA are linear pairs.

So,

⇒ ∠CBE + ∠CBA = 180°
⇒ 180° - (x° + y°) + ∠CBA = 180°
⇒ ∠CBA = 180° - 180° + (x° + y°)
⇒ ∠CBA = x° + y°

In △ABF,

So,

∠BAF + ∠ABF + ∠AFB = 180°
x + (x + y) + z = 180° (∵ From figure, ∠BAF = ∠DAB and ∠ABF = ∠CBA)

Given x : y : z = 3 : 4 : 5 , so x = 3k, y = 4k and z = 5k

3k + (3k + 4k) + 5k = 180°
15k = 180°
k = 12°

Hence, x = 3k = 3 × 12° = 36°, y = 4k = 4 × 12° = 48° and z = 5k = 5 × 12° = 60°.

Hence, the value of x = 36°, y = 48° and z = 60°.

Join OA, OB, OC and AC as shown in the figure below:

(i) ABCD is a cyclic quadrilateral as all the vertices lie on the circumference.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

⇒ ∠ADE = ∠ABC = 70°.

Arc AC subtends ∠AOC at center and ∠ABC at point B,

∠AOC = 2∠ABC (∵ angle subtended by an arc at centre is double the angle subtended at any other point of the circle.)

⇒ ∠AOC = 2 × 70° = 140°.

From figure,

OA = OC = Radius of the circle.

So, ∠OCA = ∠OAC = x.

Since sum of angles of triangle = 180°

In △OCA,

⇒ ∠AOC + ∠OCA + ∠OAC = 180°
⇒ 140° + x + x = 180°
⇒ 140° + 2x = 180°
⇒ 2x = 180° - 140°
⇒ 2x = 40°
⇒ x = 20°.

Hence, the value of ∠OCA = 20°.

(ii) From above solution,

∠ABC = 70°

From figure,

∠ABC = ∠OBA + ∠OBC
70° = 45° + ∠OBC
∠OBC = 70° - 45°
∠OBC = 25°.

As, OB = OC = radius of the circle.

∴ ∠OCB = ∠OBC = 25°.

From figure,

⇒ ∠ACB = ∠OCB + OCA
⇒ ∠ACB = 25° + 20° = 45°

Since sum of angles of triangle = 180°

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 70° + 45° + ∠BAC = 180°
⇒ 115° + ∠BAC = 180°
⇒ ∠BAC = 180° - 115°
⇒ ∠BAC = 65°

Hence, the value of ∠BAC = 65°.

(i) Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠DAB + ∠BCD = 180°
⇒ 92° + ∠BCD = 180°
⇒ ∠BCD = 180° - 92°
⇒ ∠BCD = 88°.

Hence, the value of ∠BCD = 88°.

(ii) ∠CBE = ∠BCD = 88° (∵ ∠CBE and ∠BCD are alternate angles)

∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

⇒ ∠ADC = ∠CBF
⇒ ∠ADC = ∠CBE + ∠EBF
⇒ ∠ADC = 88° + 20°
⇒ ∠ADC = 108°.

Hence, the value of ∠ADC = 108°.

Given, ∠QPR = 52°.

Since PQ = PR so, ∠QRP = ∠QPR = 52°.

Since sum of angles of triangle = 180°

In △PQR

⇒ ∠QPR + ∠QRP + ∠PQR = 180°
⇒ 52° + 52° + ∠PQR = 180°
⇒ 104° + ∠PQR = 180°
⇒ ∠PQR = 180° - 104°
⇒ ∠PQR = 76°.

∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠PST = ∠PQR = 76°

Hence, the value of ∠PST = 76°

From figure,

ABCD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠BCD + ∠BAD = 180°
⇒ x + 50° = 180°
⇒ x = 180° - 50°
⇒ x = 130°.

OA = OD = radius of the circle.

So, in △ODA,

∠ODA = ∠OAD = 50°.

In triangle exterior angle is equal to the sum of the opposite two interior angle.

y = ∠ODA + ∠OAD = 50° + 50° = 100°.

Hence, the value of x = 130° and y = 100°.

(i) ABCD is a cyclic quadrilateral.

∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠ADC = ∠CBE = 100°.

Hence, the value of ∠ADC = 100°

(ii) Arc DC subtends ∠DOC at center and ∠DAC at point A.

⇒ ∠DOC = 2∠DAC (∵ angle subtended on center by an arc is double the angle subtended on the remaining part of circle.)

⇒ 40° = 2∠DAC
⇒ ∠DAC = 20°

Hence, the value of ∠DAC = 20°.

(iii) In △COD, OC = OD = radius of the same circle.

∠CDO = ∠DCO = x

Since sum of angles of triangle = 180°

In △COD

⇒ ∠CDO + ∠DCO + ∠COD = 180°
⇒ x + x + 40° = 180°
⇒ 40° + 2x = 180°
⇒ 2x = 180° - 40°
⇒ 2x = 140°
⇒ x = 70°.

From figure,

⇒ ∠ADC = ∠ODA + ∠CDO
⇒ 100° = ∠ODA + 70°
⇒ ∠ODA = 100° - 70° = 30°.

Hence, the value of ∠ODA = 30°.

(iv) Since sum of angles of triangle = 180°

In △ADC

⇒ ∠ADC + ∠DAC + ∠ACD = 180°
⇒ 100° + 20° + ∠ACD = 180°
⇒ 120° + ∠ACD = 180°
⇒ ∠ACD = 180° - 120°
⇒ ∠ACD = 60°.

From figure,

∠OCA = ∠DCO - ∠ACD = 70° - 60° = 10°.

Hence, the value of ∠OCA = 10°.

Join OC and BD as shown in the figure below:

(i) ∠BOD = 2 × ∠BAD (∵ angle subtended by an arc at center is double the angle subtended at any point on the remaining part of the circle.)

∠BOD = 2 × 75° = 150°.

Hence, the value of ∠BOD = 150°.

(ii) ABCD is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠BCD + ∠BAD = 180°
⇒ ∠BCD + 75° = 180°
⇒ ∠BCD = 180° - 75°
⇒ ∠BCD = 105°.

Hence, the value of ∠BCD = 105°.

(iii) Join OC.

As equal chords of a circle subtend equal angles at the center and chord BC = chord CD, so ∠BOC = ∠COD.

∠BOC = $\frac{1}{2}$∠BOD = $\frac{1}{2} \times 150°$ = 75°.

Hence, the value of ∠BOC = 75°.

(iv) Join BD.

Since, OB = OD

∴ ∠OBD = ∠ODB = x

Since sum of angles of triangle = 180°

In △OBD

⇒ ∠BOD + ∠OBD + ∠ODB = 180°
⇒ 150° + x + x = 180°
⇒ 150° + 2x = 180°
⇒ 2x = 180° - 150°
⇒ 2x = 30°
⇒ x = 15°.

Hence, the value of ∠OBD = 15°.

(i) AECB is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

From figure,

∠ABE = 90° (∵ angle in semicircle is 90°.)

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠AEC + ∠ABC = 180°
⇒ ∠AEC + ∠ABE + ∠CBE = 180°
⇒ 50° + 90° + ∠CBE = 180°
⇒ ∠CBE + 140° = 180°
⇒ ∠CBE = 180° - 140° = 40°.

Hence, the value of ∠CBE = 40°.

(ii) BEDC is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠CBE + ∠CDE = 180°
⇒ 40° + ∠CDE = 180°
⇒ ∠CDE = 180° - 40° = 140°.

Hence, the value of ∠CDE = 140°.

(iii) Given,

AB = BC

∴ ∠AEB = ∠BEC = $\frac{1}{2}$∠AEC = $\frac{1}{2} \times 50° = 25°.$ (∵ equal chords subtend equal angle at circumference.)

In △OBE,

OB = OE = radius of the same circle

∴ ∠OBE = ∠OEB = 25°.

In triangle exterior angle is equal to the sum of opposite two interior angles.

∠AOB = ∠OBE + ∠OEB = 25° + 25° = 50°.

Hence, the value of ∠AOB = 50°.

∠AOB = ∠OEC (∵ both are equal to 50°)

Since these angles are corresponding angles and are equal which is property of parallel lines.

Hence proved that OB || EC.

BEDC is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠ABC = ∠D ....(i)

From figure, as ED is parallel to BC, ∠ABC and ∠E are corresponding angles,

∴ ∠ABC = ∠E ....(ii)

From (i) and (ii)

∠D = ∠E

In △AED,

∠D = ∠E,

∴ AE = AD. (As sides opposite to equal angles are equal)

Hence, proved that △AED is an isosceles triangle.

Since, SP is the bisector of the angle ∠RPT.

So, ∠RPS = ∠SPT

From figure,

∠RPS = ∠RQS (As angle in same segment are equal)

Given, PQRS is a cyclic quadrilateral.

∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

∠QRS = ∠SPT

∴ ∠QRS = ∠RPS

or,

∠QRS = ∠RQS

In △QRS,

∠QRS = ∠RQS

∴ SQ = RS (As sides opposite to equal angles are equal.)

Hence, proved that SQ = RS.

Given, AB = AC

∴ ∠ABC = ∠ACB (As angles opposite to equal sides are equal)

As BCED is a cyclic quadrilateral,

∠ADE = ∠BCE (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)

∴ ∠ADE = ∠ABC

Since these angles are equal and are corresponding this is the property of parallel lines,

∴ DE || BC.

Hence, proved that DE || BC.

Let ABCD be a cyclic parallelogram.

In parallelogram opposite angles are equal.

∴ ∠A = ∠C and ∠B = ∠D.

Sum of opposite angles of a cyclic quadrilateral is 180°

⇒ ∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°.

∴ ∠C = 90°.

Similarly,

⇒ ∠B + ∠D = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = 90°.

∴ ∠D = 90°.

Hence, ∠A = ∠B = ∠C = ∠D = 90°.

In parallelogram opposite sides are equal i.e. AD = BC and AB = CD.

Hence, ABCD is a rectangle as opposite sides are equal and all the angles are equal to 90°.

Let ABCD be a cyclic rhombus.

In rhombus opposite angles are equal.

∴ ∠A = ∠C and ∠B = ∠D.

Sum of opposite angles of a cyclic quadrilateral is 180°

⇒ ∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°.

∴ ∠C = 90°.

Similarly,

⇒ ∠B + ∠D = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = 90°.

∴ ∠D = 90°.

Hence, ∠A = ∠B = ∠C = ∠D = 90°.

In rhombus all sides are equal i.e. AD = BC = AB = CD.

Hence, ABCD is a square as all sides are equal and all the angles are equal to 90°.

In △ODB and △OAC,

∠ODB = ∠C

∠O = ∠O (Common)

∴ △ODB ~ △OAC (AA axiom)

Since, in similar triangles the ratio of the corresponding sides are equal.

$$\begin{aligned}\therefore \frac{OD}{OA} = \frac{OB}{OC} \\[1em] \Rightarrow \frac{6}{OA} = \frac{3}{6 + 2} \\[1em] \Rightarrow OA = \frac{6 \times 8}{3} \\[1em] \Rightarrow OA = 16.\end{aligned}$$

AB = OA - OB = 16 - 3 = 13 cm.

Since, △ODB ~ △OAC

$$\begin{aligned}\therefore \frac{\text{Area of △OAC}}{\text{Area of △ODB}} = \frac{OC^2}{OB^2} \\[1em] \frac{\text{Area of △OAC}}{\text{Area of △ODB}} = \frac{8^2}{3^2} \\[1em] \frac{\text{Area of △OAC}}{\text{Area of △ODB}} = \frac{64}{9} ...(i)\end{aligned}$$

Subtracting 1 from both sides we get,

$$\begin{aligned}\frac{\text{Area of △OAC}}{\text{Area of △ODB}} - 1 = \frac{64}{9} - 1 \\[1em] \frac{\text{Area of △OAC - Area of △ODB}}{\text{Area of △ODB}} = \frac{64 - 9}{9} \\[1em] \frac{\text{Area of quadrilateral CABD}}{\text{Area of △ODB}} = \frac{55}{9} ....(ii)\end{aligned}$$

Dividing (ii) by (i),

$$\begin{aligned}\frac{\frac{\text{Area of quadrilateral CABD}}{\text{Area of △ODB}}}{\frac{\text{Area of △OAC}}{\text{Area of △ODB}}} = \frac{\frac{55}{9}}{\frac{64}{9}} \\[1em] \frac{\text{\text{Area of quadrilateral CABD}}}{\text{Area of △OAC}} = \frac{55}{64}.\end{aligned}$$

Hence, the length of AB = 13 cm and

$\frac{\text{\text{Area of quadrilateral CABD}}}{\text{Area of △OAC}} = \frac{55}{64}.$

Let tangent be drawn from point P.

In a circle with centre O and radius = 3 cm and P is at a distance of 5 cm.

OT = 3 cm and OP = 5 cm

OT ⊥ PT

In right angle △OTP,

By pythagoras theorem,

$$\begin{aligned}OP^2 = OT^2 + PT^2 \\[1em] 5^2 = 3^2 + PT^2 \\[1em] 25 = 9 + PT^2 \\[1em] PT^2 = 25 - 9 \\[1em] PT^2 = 16 \\[1em] PT = \sqrt{16} \\[1em] PT = 4 \text{ cm}.\end{aligned}$$

Hence, the length of tangent = 4 cm.

The below diagram shows the circle and the tangent:

Given point P is 13 cm away from centre C, so CP = 13 cm.

PT = 12 cm

CT = radius of the circle.

Since the tangent at any point of a circle and the radius through the point are perpendicular to each other.

So, CT ⊥ PT

So, in right angled △CPT by pythagoras theorem,

$$\begin{aligned}\Rightarrow CP^2 = CT^2 + PT^2 \\[1em] \Rightarrow 13^2 = CT^2 + 12^2 \\[1em] \Rightarrow 169 = CT^2 + 144 \\[1em] \Rightarrow CT^2 = 169 - 144 \\[1em] \Rightarrow CT^2 = 25 \\[1em] \Rightarrow CT = \sqrt{25} \text{ cm} \\[1em] \Rightarrow CT = 5 \text{ cm}.\end{aligned}$$

Hence, radius of circle = 5 cm.

Since the tangent at any point of a circle and the radius through the point are perpendicular to each other.

So, from figure,

AP ⊥ CP

So, in right angled △CAP by pythagoras theorem,

$$\begin{aligned}\Rightarrow CP^2 = CA^2 + AP^2 \\[1em] \Rightarrow CP^2 = 6^2 + 8^2 \\[1em] \Rightarrow CP^2 = 36 + 64 \\[1em] \Rightarrow CP^2 = 100 \\[1em] \Rightarrow CP = 10\text{ cm}.\end{aligned}$$

From figure, nearest point to P on the circle is D,

PD = CP - CD = 10 - 6 = 4 cm.

Hence, the distance of P from the nearest point of the circle is 4 cm.

Given,

Radius of circle (r) = OA = OB = 9 cm

From figure,

AB = OA + OB = 9 + 9 = 18 cm

PB = PA - AB = 24 - 18 = 6 cm

We know that,

If a secant segment and tangent segment are drawn to a circle from the same external point, the length of the tangent segment is the geometric mean between the length of the secant segment and the length of the external part of the secant segment.

⇒ PQ² = PB × PA

⇒ PQ² = 6 × 24

⇒ PQ² = 144

⇒ PQ = $\sqrt{144}$ = 12 cm.

Hence, length of tangent PQ = 12 cm.

From figure,

AB is the chord of the outer circle which touches the inner circle at P.

OP is the radius of the inner circle and APB is the tangent to the inner circle.

In the right angled triangle OPB, by pythagoras theorem,

$$\begin{aligned}\Rightarrow OB^2 = OP^2 + PB^2 \\[1em] \Rightarrow 13^2 = 5^2 + PB^2 \\[1em] \Rightarrow 169 = 25 + PB^2 \\[1em] \Rightarrow PB^2 = 169 - 25 \\[1em] \Rightarrow PB^2 = 144 \\[1em] \Rightarrow PB = \sqrt{144} \text{ cm} \\[1em] \Rightarrow PB = 12 \text{ cm}.\end{aligned}$$

As perpendicular line from centre bisects the chord of the circle so,

AP = PB = 12 cm.

AB = AP + PB = 12 + 12 = 24 cm.

Hence, the length of chord = 24 cm.

(i) From figure,

OC = OP + PC = 5 + 2.8 = 7.8 cm

Hence, the distance between centres of circle = 7.8 cm when circles touch externally.

(ii) From figure,

OC = OP - CP = 5 - 2.8 = 2.2 cm.

Hence, the distance between centres of circle = 2.2 cm when circles touch internally.

From figure, AP and AQ are the tangents to the circle.

∴ AQ = AP = 4 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure, BP and BR are the tangents to the circle.

∴ BR = BP = 6 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

CQ = CA - AQ = 12 - 4 = 8 cm.

From figure, CQ and CR are the tangents to the circle.

∴ CR = CQ = 8 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

x = BR + CR = 6 + 8 = 14 cm.

Hence, the value of x = 14 cm.

From A, AP and AQ are the tangents to the circle.

∴ AP = AQ = 5 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From C, CR and CS are the tangents to the circle.

∴ CS = CR = 3 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

BS = BC - CS = 7 - 3 = 4 cm.

From B, BS and BP are the tangents to the circle.

∴ BP = BS = 4 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

x = BP + AP = 4 + 5 = 9 cm.

Hence, the value of x = 9 cm.

From A, AP and AS are the tangents to the circle.

∴ AS = AP = 6 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From B, BP and BQ are the tangents to the circle.

∴ BQ = BP = 5 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From C, CQ and CR are the tangents to the circle.

∴ CR = CQ = 3 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From D, DS and DR are the tangents to the circle.

∴ DS = DR = 4 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

Therefore, perimeter of the quadrilateral ABCD

$$\begin{aligned}= AB + BC + CD + DA \\[1em] = AP + BP + BQ + CQ + CR + DR + DS + AS \\[1em] = 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6 \\[1em] = 36 \text{ cm}\end{aligned}$$

Hence, the perimeter of ABCD = 36 cm.

Join OR as shown in the figure below:

From figure,

OS = OR (As both are radius of circle.)

SD = OR (As OSDR form a square.)

∴ SD = OS.

From D, DS and DR are the tangents to the circle.

∴ DR = DS = 10 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From B, BP and BQ are the tangents to the circle.

∴ BQ = BP = 27 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From figure,

CQ = BC - BQ = 38 - 27 = 11 cm.

Now from C, CQ and CR are the tangents to the circle

CR = CQ = 11 cm.

⇒ DC = x = DR + CR = 10 + 11 = 21 cm.

Hence, the length of x = 21 cm.

Since the tangent at any point of a circle and the radius through the point are perpendicular to each other.

So, from figure,

OB ⊥ AB

In right angled △OBA,

$$\begin{aligned}OB^2 + AB^2 = OA^2 \\[1em] r^2 + 15^2 = (r + 7.5)^2 \\[1em] r^2 + 225 = r^2 + 56.25 + 15r \\[1em] r^2 - r^2 + 225 - 56.25 = 15r \\[1em] 15r = 168.75 \\[1em] r = \frac{168.75}{15} \\[1em] r = 11.25 \text{ cm}.\end{aligned}$$

Hence, radius of the circle = 11.25 cm.

From P, PA and PB are the tangents to the circle.

∴ PA = PB = 15 cm. (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From E, EA and ED are the tangents to the circle.

∴ EA = ED (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From C, BC and CD are the tangents to the circle.

∴ BC = CD (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

Now perimeter of triangle PEC,

$$\begin{aligned}= PE + EC + PC \\[1em] = PE + ED + CD + PC \\[1em] = PE + EA + CB + PC \\[1em] (\because \small{ED = EA, CB = CD}) \\[1em] = AP + PB \\[1em] = 15 + 15 \\[1em] = 30 \text{ cm}.\end{aligned}$$

Hence, perimeter of triangle PEC = 30 cm.

Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively,

where BC = a, CA = b and AB = c (as shown in the given figure).

As the lengths of tangents drawn from an external point to a circle are equal

AE = AF, BD = BF and CD = CE

OD ⊥ BC and OE ⊥ CA (∵ tangents is ⊥ to radius)

ODCE is a square of side r

DC = CE = r

AF = AE = AC - EC = b - r and,

BF = BD = BC - DC = a - r

Now,

AB = AF + BF

⇒ c = (b - r) + (a - r)
⇒ c = b + a - 2r
⇒ 2r = a + b - c
⇒ r = $\frac{a + b - c}{2}$.

Hence, proved that r = $\frac{a + b - c}{2}$

Join OB as shown in figure below:

OM = 5 cm

OM ⊥ AB and M is mid-point of AB,

MB = $\frac{1}{2}AB = \frac{1}{2} \times 24 =$ 12 cm.

In right-angled triangle △OMB,

$$\begin{aligned}OB^2 = OM^2 + MB^2 \\[1em] \Rightarrow OB^2 = 5^2 + 12^2 \\[1em] \Rightarrow OB^2 = 25 + 144 \\[1em] \Rightarrow OB^2 = 169 \\[1em] \Rightarrow OB = \sqrt{169} \text{ cm} \\[1em] OB = 13 \text{ cm.}\end{aligned}$$

As BP is tangent to circle at B, OB ⊥ BP.

In right-angled triangle △OBP,

$$\begin{aligned}OP^2 = OB^2 + BP^2 \\[1em] \Rightarrow OP^2 = 13^2 + 20^2 \\[1em] \Rightarrow OP^2 = 169 + 400 \\[1em] \Rightarrow OP^2 = 569 \\[1em] \Rightarrow OP = \sqrt{569} \text{ cm}\end{aligned}$$

Hence, the length of OP = $\sqrt{569}$ cm.

Three circles with centres A, B and C touch each other externally and the radii of these circles are 2 cm, 3 cm and 4 cm respectively.

From figure,

By joining the centres of circles, triangle ABC is formed in which,

AB = 2 + 3 = 5 cm

BC = 3 + 4 = 7 cm

CA = 4 + 2 = 6 cm.

Therefore, perimeter of the triangle ABC = AB + BC + CA = 5 + 7 + 6 = 18 cm.

Hence, the perimeter of triangle ABC = 18 cm.

Let p, Q, R, S be the points where the circle touches the sides of the quadrilateral as shown in the figure below:

From A, AP and AS are the tangents to the circle.

∴ AP = AS ...(i) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From B, BP and BQ are the tangents to the circle.

∴ PB = BQ ....(ii) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From C, CR and CQ are the tangents to the circle.

∴ CR = CQ ....(iii) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

From D, DR and DS are the tangents to the circle.

∴ DR = DS ....(iv) (∵ if two tangents are drawn to a circle from an external point then the tangents have equal lengths.)

Adding L.H.S. and R.H.S. of equations (i), (ii), (iii) and (iv) we get,

⇒ AP + PB + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = BC + DA

Hence, proved that AB + CD = BC + DA.

Given, AB = 10 cm, BC = 8 cm, AC = 6 cm

Let the radius of the circles with centre A, B and C be x cm, y cm and z cm.

From figure,

AB = 10 cm.

⇒ x + y = 10 cm ....(i)

BC = 8 cm.

⇒ y + z = 8 cm ....(ii)

AC = 6 cm.

⇒ x + z = 6 cm .....(iii)

Adding eqn. (i), (ii) and (iii) we get,

⇒ x + y + y + z + x + z = (10 + 8 + 6) cm
⇒ 2x + 2y + 2z = 24 cm
⇒ 2(x + y + z) = 24 cm
⇒ x + y + z = 12 cm .....(iv)

Now, (iv) - (i) we get,

⇒ x + y + z - (x + y) = (12 - 10) cm
⇒ x + y + z - x - y = 2 cm
⇒ z = 2 cm.

Also, by (iv) - (ii) we get,

⇒ x + y + z - (y + z) = (12 - 8) cm
⇒ x + y + z - y - z = 4 cm
⇒ x = 4 cm

Also, by (iv) - (iii) we get,

⇒ x + y + z - (x + z) = (12 - 6) cm
⇒ x + y + z - x - z = 6 cm
⇒ y = 6 cm.

Hence, the radius of the circle with centre A, B and C are 4 cm, 6 cm and 2 cm.

Let the sides of triangle PQ, QR and PR meet the circle at L, M and N respectively.

In right-angled triangle PQR

$$\begin{aligned}PR^2 = PQ^2 + QR^2 \\[1em] PR^2 = 24^2 + 7^2 \\[1em] PR^2 = 576 + 49 \\[1em] PR^2 = 625 \\[1em] PR= \sqrt{625} \\[1em] PR = 25 \text{ cm}.\end{aligned}$$

From figure,

RM = RN = (∵ tangents drawn from a common external point to a circle are equal.)

RM = RQ - QM = (7 - x) cm.

PL = PN = (∵ tangents drawn from a common external point to a circle are equal.)

PL = PQ - QL = (24 - x) cm.

We can see,

PR = PN + RN = PL + RM.

⇒ 25 = 24 - x + 7 - x
⇒ 25 = 31 - 2x
⇒ 2x = 31 - 25
⇒ 2x = 6
⇒ x = 3.

Hence, the radius of the inscribed circle is 3 cm.

Join OA and OB.

OA ⊥ AP. (∵ tangent at a point and radius through the point are perpendicular to each other.)

So, in right angled triangle OAP,

$$\begin{aligned}OP^2 = OA^2 + AP^2 \\[1em] OP^2 = 5^2 + 12^2 \\[1em] OP^2 = 25 + 144 \\[1em] OP^2 = 169 \\[1em] OP = \sqrt{169} \\[1em] OP = 13 \text{ cm}.\end{aligned}$$

OB ⊥ BP. (∵ tangent at a point and radius through the point are perpendicular to each other.)

So, in right angled triangle OBP,

$$\begin{aligned}OP^2 = BP^2 + OB^2 \\[1em] 13^2 = BP^2 + 3^2 \\[1em] 169 = 9 + BP^2 \\[1em] BP^2 = 160 \\[1em] BP = \sqrt{160} \\[1em] BP = 4\sqrt{10} \text{ cm}.\end{aligned}$$

Hence, the length of BP = $4\sqrt{10}$ cm.

Let x be the radius of the circle with centre C.

Since M is the mid-point of AB hence, AM = MB = 4 cm.

Two semicircles are thus drawn on AB with diameters as AM and MB.

Since radius = $\frac{\text{diameter}}{2}$.

Hence, radius of both the semicircles with diameters AM and MB = 2 cm.

From figure,

CM = MP - PC = (4 - x) cm.

In right angled triangle CMD,

$$\begin{aligned}CD^2 = CM^2 + DM^2 \\[1em] (x + 2)^2 = (4 - x)^2 + 2^2 \\[1em] x^2 + 4 + 4x = 16 + x^2 - 8x + 4 \\[1em] x^2 - x^2 + 4x + 8x + 4 - 16 - 4 = 0 \\[1em] 12x - 16 = 0 \\[1em] 12x = 16 \\[1em] x = \frac{16}{12} \\[1em] x = \frac{4}{3} = 1\frac{1}{3} \text{ cm.}\end{aligned}$$

Hence, the radius of small circle = $1\frac{1}{3}$ cm.

From figure,

OC is radius and AC is tangent, then OC ⊥ AC.

Let radius of each equal circle = r.

(i) From figure,

AO = AO' + O'X + XO and AO' = O'X = XO = r (radius of circle)

AO = r + r + r = 3r.

$\therefore \frac{\text{AO}'}{\text{AO}} = \frac{r}{3r} = \frac{1}{3}.$

Hence, the value of $\frac{AO'}{AO} = \frac{1}{3}$.

(ii) Considering △ADO' and △ACO

∠A = ∠A (Common angles)

∠D = ∠C (Both are equal to 90°)

∴ By AA axiom △ADO' ~ △ACO.

Since triangles are similar hence the ratio of their areas is equal to the ratio of the square of the corresponding sides.

$$\begin{aligned}\frac{\text{Area of △ADO}'}{\text{Area of △ACO}} = \frac{\text{AO}'^2}{\text{AO}^2} \\[1em] = \frac{r^2}{(3r)^2} \\[1em] = \frac{r^2}{9r^2} \\[1em] = \frac{1}{9}.\end{aligned}$$

Hence, the value of $\frac{\text{Area of △ADO'}}{\text{Area of △ACO}} = \frac{1}{9}.$

Let there be two circles with center A and B and radius 12 and 4 cm respectively.

From figure,

TT' is the common tangent.

DT = BT' = 4 cm.

DB = TT' = 15 cm.

In right angled triangle ADB

AD = AT - DT = 12 - 4 = 8 cm

$$\begin{aligned}AB^2 = AD^2 + DB^2 \\[1em] AB^2 = 8^2 + 15^2 \\[1em] AB^2 = 64 + 225 \\[1em] AB^2 = 289 \\[1em] AB = \sqrt{289} \\[1em] AB = 17 \text{ cm}.\end{aligned}$$

Hence, the distance between two centres = 17 cm.

Let there be two circles with centre A and B with radius 8 cm and 3 cm respectively.

Let TT' be the length of common tangent.

From figure,

DT = BT' = 3cm.

AD = AT - DT = 8 - 3 = 5 cm.

In right angled triangle ADB

$$\begin{aligned}AB^2 = AD^2 + DB^2 \\[1em] \Rightarrow 13^2 = 5^2 + DB^2 \\[1em] \Rightarrow DB^2 = 13^2 - 5^2 \\[1em] \Rightarrow DB^2 = 169 - 25 \\[1em] \Rightarrow DB^2 = 144 \\[1em] \Rightarrow DB = \sqrt{144} \\[1em] \Rightarrow DB = 12 \text{ cm}.\end{aligned}$$

Since, TDBT' is a rectangle,

So, TT' = DB = 12 cm.

Hence, the length of direct common tangent is 12 cm.

Join AP and CQ.

AB ⊥ AP (∵ tangent at a point and radius through the point are perpendicular to each other.)

In right angled triangle PAB.

$$\begin{aligned}PB^2 = PA^2 + AB^2 \\[1em] \Rightarrow PB^2 = 6^2 + 8^2 \\[1em] \Rightarrow PB^2 = 36 + 64 \\[1em] \Rightarrow PB^2 = 100 \\[1em] \Rightarrow PB = \sqrt{100} \\[1em] \Rightarrow PB = 10 \text{ cm}.\end{aligned}$$

Considering △PAB and △BCQ,

∠A = ∠C (Each are equal to 90°)

∠ABP = ∠CBQ (Vertically opposite angles are equal)

△PAB ~ △BCQ by AA axiom.

Since triangles are similar hence, the ratio of their corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AP}{CQ} = \frac{PB}{BQ} \\[1em] \Rightarrow \frac{6}{3} = \frac{10}{BQ} \\[1em] \Rightarrow 2 = \frac{10}{BQ} \\[1em] \Rightarrow BQ = \frac{10}{2} \\[1em] \Rightarrow BQ = 5 \text{ cm}.\end{aligned}$$

From figure,

PQ = PB + BQ = 10 + 5 = 15 cm.

Hence, the length of PQ = 15 cm.

The two circles with centres A, B are of radii 6 cm and 3 cm and AB = 15 cm are shown in the figure below:

Given, AB = 15 cm.

Let AP = x, then PB = 15 - x

Considering △ATP and △SBP,

∠T = ∠S (Each are equal to 90°)

∠APT = ∠BPS (Vertically opposite angles are equal)

△ATP ~ △SBP by AA axiom.

Since triangles are similar hence, the ratio of their corresponding sides are equal.

$$\begin{aligned}\therefore \frac{AT}{BS} = \frac{AP}{PB} \\[1em] \Rightarrow \frac{6}{3} = \frac{x}{15 - x} \\[1em] \Rightarrow 3x = 6(15 - x) \\[1em] \Rightarrow 3x = 90 - 6x \\[1em] \Rightarrow 3x + 6x = 90 \\[1em] \Rightarrow 9x = 90 \\[1em] \Rightarrow x = 10.\end{aligned}$$

∴ AP = 10 cm,

From figure,

PB = AB - AP = 15 - 10 = 5 cm.

Now in right-angled triangle ATP,

$$\begin{aligned}AP^2 = AT^2 + TP^2 \\[1em] 10^2 = 6^2 + TP^2 \\[1em] TP^2 = 100 - 36 \\[1em] TP^2 = 64 \\[1em] TP = \sqrt{64} \\[1em] TP = 8 \text{ cm}.\end{aligned}$$

Similarly in right angled triangle PSB,

$$\begin{aligned}PB^2 = BS^2 + PS^2 \\[1em] 5^2 = 3^2 + PS^2 \\[1em] PS^2 = 25 - 9 \\[1em] PS^2 = 16 \\[1em] PS = \sqrt{16} \\[1em] PS = 4 \text{ cm}.\end{aligned}$$

Hence, TS = TP + PS = 8 + 4 = 12 cm.

Hence, the length of a transverse common tangent to these circles are 12 cm.

AP and BP are tangents to the circle and OA and OB are radius of the circle.

∴ OA ⊥ AP and OB ⊥ BP.

∴ ∠OAP = ∠OBP = 90°

Sum of angles of a quadrilateral = 360°.

Hence, in quadrilateral OAPB,

⇒ ∠APB + ∠OAP + ∠OBP + ∠AOB = 360°
⇒ 70° + 90° + 90° + ∠AOB = 360°
⇒ 250° + ∠AOB = 360°
⇒ ∠AOB = 360° - 250°
⇒ ∠AOB = 110°.

Hence, the value of ∠AOB = 110°.

(ii) Arc AB subtends ∠AOB at centre and ∠AQB at remaining part of circle.

∴ ∠AOB = 2∠AQB (As angle at centre is double the angle subtended at remaining part of circle.)

⇒ 2∠AQB = 110°
⇒ ∠AQB = $\frac{110°}{2}$ = 55°.

Hence, the value of ∠AQB = 55°.

(iii) Reflex ∠AOB = 360° - ∠AOB = 360° - 110° = 250°.

Arc AB subtends Reflex ∠AOB at centre and ∠ARB at remaining part of circle.

∴ Reflex ∠AOB = 2∠ARB (As angle at centre is double the angle subtended at remaining part of circle.)

⇒ 2∠ARB = 250°
⇒ ∠ARB = $\frac{250°}{2}$ = 125°.

Hence, the value of ∠ARB = 125°.

From figure,

QA and QP are the tangents to the outer circle.

∴ QA = QP .....(i) (∵ the length of the different tangents to a circle from a single point are equal.)

Similarly, from Q, QB and QP are the tangents to the inner circle.

∴ QB = QP .....(ii) (∵ the length of the different tangents to a circle from a single point are equal.)

From (i) and (ii),

QA = QB.

Hence, proved that QA = QB.

In the figure,

AB ⊥ AD. (∵ tangent at a point and radius through the point are perpendicular to each other.)

From figure,

∠ABD = ∠ABC = 50°

In △ABD,

    ∠ABD + ∠BDA + ∠DAB = 180°
⇒ 50° + ∠BDA + 90° = 180°
⇒ ∠BDA + 140° = 180°
⇒ ∠BDA = 180° - 140°
⇒ ∠BDA = 40°.

From figure,

∠ADC = ∠BDA = 40°.

Arc AC subtends ∠AOC at the centre and ∠ADC on point D.

∴ ∠AOC = 2∠ADC (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

∠AOC = 2 × 40° = 80°.

Hence, value of ∠AOC = 80°.

Given,

RS || PQ,

∠TRS = ∠RPQ = 30° (∵ Corresponding angles are equal)

∠RQS = ∠TRS = 30° (∵ Angles in alternate segments are equal.)

Hence, value of ∠RQS = 30°.

(i) ∠QAB = ∠BDA (∵ angles in alternate segment are equal.)

∴ ∠QAB = 30°.

Hence, the value of ∠QAB = 30°.

(ii) In △ADB,

∠DAB = 90° (∵ angle in semi-circle is 90°.)

Since, sum of angles in a triangle is 180°.

⇒ ∠ABD + ∠ADB + ∠DAB = 180°
⇒ ∠ABD + 30° + 90° = 180°
⇒ ∠ABD + 120° = 180°
⇒ ∠ABD = 60°.

From figure,

∠PAD = ∠ABD = 60° (∵ angles in alternate segment are equal)

Hence, the value of ∠PAD = 60°.

(iii) In △BCD,

∠BCD = 90° (As angle in semi-circle is 90°.)

∠CBD = 60°

Since, sum of angles in a triangle is 180°.

⇒ ∠BCD + ∠CBD + ∠CDB = 180°
⇒ 90° + 60° + ∠CDB = 180°
⇒ ∠CDB + 150° = 180°
⇒ ∠CDB = 30°.

Hence, the value of ∠CDB = 30°.

ABCD is a cyclic quadrilateral.

In cyclic quadrilateral, the exterior angle = opposite interior angle.

∴ ∠BCD = ∠EAB = 85°

From figure,

⇒ ∠BCD + ∠BCF = 180° (∵ both are linear pair)
⇒ ∠BCF + 85° = 180°
⇒ ∠BCF = 95°.

Now in △BCF,

Since, sum of angles in a triangle is 180°.

⇒ ∠BCF + ∠BFC + ∠CBF = 180°
⇒ 95° + 50° + ∠CBF = 180°
⇒ ∠CBF + 145° = 180°
⇒ ∠CBF = 35°.

We know, BF is a tangent and BC is a chord.

∴ ∠CAB = ∠CBF = 35° (∵ angles in alternate segment are equal.)

⇒ ∠CAB = 35°.

Hence, the value of ∠CAB = 35°.

From figure,

In △SRT,

SR ⊥ ST (∵ tangent is perpendicular to radius from that point.)

so, ∠TSR = 90°

Since, sum of angles in a triangle = 180°

⇒ ∠TSR + ∠SRT + ∠STR = 180°
⇒ 90° + 65° + x = 180°
⇒ x + 155° = 180°
⇒ x = 25°.

SQ subtends ∠SOQ at the centre and ∠STQ on point D.

∴ ∠SOQ = 2∠STQ (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

y = 2x = 2 × 25° = 50°.

In △OSP,

Since, sum of angles in a triangle = 180

⇒ ∠OSP + ∠SOP + ∠SPO = 180°
⇒ 90° + y + z = 180°
⇒ 90° + 50° + z = 180°
⇒ z + 140° = 180°
⇒ z = 180° - 140° = 40°.

Hence, the value of x = 25°, y = 50° and z = 40°.

(i) We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

In quadrilateral ORPQ,

∠OQP = ∠ORP = 90° [∵ The tangent at any point of a circle and the radius through this point are perpendicular to each other]

∠QPR = 70° [Given]

⇒ ∠OQP + ∠ORP + ∠QPR + ∠QOR = 360° [By angle sum property of quadrilateral]

⇒ 90° + 90° + 70° + ∠QOR = 360°

⇒ 250° + ∠QOR = 360°

⇒ ∠QOR = 110°.

Hence, ∠QOR = 110°.

(ii) Let M be a point on circumference of circle.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended at the circumference.

⇒ ∠QMR = $\frac{1}{2}$ ∠QOR

= $\frac{110°}{2}$

= 55°.

Sum of opposite angles in cyclic quadrilateral is 180°.

⇒ ∠QSR + ∠QMR = 180°.

⇒ ∠QSR = 180° - 55°

⇒ ∠QSR = 125°.

Hence, ∠QSR = 125°.

(i) ∠BCO = ∠ACO = 30° (∵ C is the intersecting point of tangent AC and BC. So, OC divides ∠ACB in two halves.)

Hence, the value of ∠BCO = 30°.

(ii) We know that the tangent at any point of a circle and the radius through the point are perpendicular to each other.

∴ ∠OAC = ∠OBC = 90°.

∴ ∠AOC = ∠BOC (∵ tangents are equally inclined to the line joining the point and the centre of the circle.)

Since sum of angles in a triangle = 180.

In AOC

⇒ ∠AOC + ∠OAC + ∠ACO = 180°
⇒ ∠AOC + 90° + 30° = 180°
⇒ ∠AOC + 120° = 180°
⇒ ∠AOC = 180° - 120° = 60°.

∴ ∠BOC = 60°.

From figure,

∠AOB = ∠AOC + ∠BOC = 60° + 60° = 120°.

Hence, the value of ∠AOB = 120°.

(iii) Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.

∴ ∠AOB = 2∠APB (∵ angle subtended at centre by an arc is double the angle subtended at remaining point of the circle.)

∠APB = $\frac{1}{2}$ x ∠AOB = $\frac{1}{2}$ x 120° = 60°.

Hence, the value of ∠APB = 60°.

(i) From figure,

∠BCD = ∠ABC (∵ alternate angles are equal)

∠BCD = 55°.

Arc BD subtends ∠BOD at the centre and ∠BCD at the remaining part of the circle.

∴ ∠BOD = 2∠BCD (∵ angle subtended at centre by an arc is double the angle subtended at remaining point of circle.)

∠BOD = 2 × 55° = 110°.

Hence, the value of ∠BOD = 110°

(ii) OB and OD are radius and, BP and DP are tangents to the circle.

∴ OB ⊥ BP and OD ⊥ DP.

In quadrilateral OBPD, sum of angles = 360°

∠BOD + ∠ODP + ∠OBP + ∠BPD = 360°
110° + 90° + 90° + ∠BPD = 360°
290° + ∠BPD = 360°
∠BPD = 360° - 290°
∠BPD = 70°.

Hence, the value of ∠BPD = 70°.

From figure,

∠APB = 90° (∵ angles in semicircle is equal to 90)

From figure,

∠APT + ∠APT' = 180° (∵ they form linear pair)

⇒ ∠APT + ∠APB + ∠BPT' = 180°
⇒ ∠APT + 90° + 30° = 180°
⇒ ∠APT + 120° = 180°
⇒ ∠APT = 180° - 120°
⇒ ∠APT = 60°.

Hence, the value of ∠APT = 60°.

(ii) From figure,

BAP = BPT' = 30. (∵ angles in alternate segment are equal.)

Arc BP subtends ∠BOP at the centre and ∠BAP at the remaining part of the circle.

∴ ∠BOP = 2∠BAP (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle.)

∠BOP = 2 × 30° = 60°.

Hence, the value of ∠BOP = 60°.

Given,

AB and AD are bisectors of ∠CAQ and ∠CAP respectively.

Let AB bisects ∠CAQ in two halves of each value x.

∴ ∠CAB = x and ∠BAQ = x.

Let AD bisects ∠CAP in two halves of each value y.

∴ ∠CAD = y and ∠DAP = y.

Since, ∠CAQ and ∠CAP are linear pair so,

⇒ ∠CAQ + ∠CAP = 180°
⇒ ∠CAB + ∠BAQ + ∠CAD + ∠DAP = 180°
⇒ x + x + y + y = 180°
⇒ 2x + 2y = 180°
⇒ x + y = $\frac{180°}{2}$
⇒ x + y = 90°.

∴ ∠CAB + ∠CAD = 90° ⇒ ∠BAD = 90°.

Since angle in semicircle is equal to 90°.

Hence, proved BD is the diameter of the circle.

Given, ∠CAQ : ∠CAP = 1 : 2.

Let ∠CAQ = k so ∠CAP = 2k.

Since ∠CAQ and ∠CAP are linear pair so,

⇒ ∠CAQ + ∠CAP = 180°
⇒ k + 2k = 180°
⇒ 3k = 180°
⇒ k = 60°.

∠CAQ = 60° and ∠CAP = 2 × 60° = 120°.

From figure,

∠ADC = ∠CAQ = 60° (∵ angles in alternate segments are equal)

∠ABC = ∠CAP = 120° (∵ angles in alternate segments are equal)

Since sum of opposite angles in cyclic quadrilateral is 180°.

⇒ ∠D + ∠B = 180°
⇒ 60° + ∠B = 180°
⇒ ∠B = 180° - 60°
⇒ ∠B = 120°.

Similarly,

⇒ ∠A + ∠C = 180°
⇒ 90° + ∠C = 180°
⇒ ∠C = 180° - 90°
⇒ ∠C = 90°.

Hence, the angles of cyclic quadrilateral are ∠A = 90°, ∠B = 120°, ∠C = 90° and ∠D = 60°.

(i) From figure,

OQ ⊥ AC and OR ⊥ AB (∵ OQ and OR are the radii and AC and AB are tangents.)

Now in quadrilateral AROQ,

∠A = 60°, ∠ORA = 90° and ∠OQA = 90°.

∠A + ∠ORA + ∠OQA + ∠QOR = 360°
60° + 90° + 90° + ∠QOR = 360°
240° + ∠QOR = 360°
∠QOR = 360° - 240°
∠QOR = 120°.

Hence, the value of ∠QOR = 120°.

(ii) Arc QR subtends ∠QOR at the centre and ∠QPR at the remaining part of the circle.

∴ ∠QOR = 2∠QPR (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

120° = 2∠QPR
∠QPR = $\frac{120°}{2}$ = 60°.

Hence, the value of ∠QPR = 60°.

From figure,

∠ACB = 90 (∵ angles in semicircle is equal to 90.)

Since sum of angles in a triangle = 180.

In △ABC,

⇒ ∠CAB + ∠ACB + ∠CBA = 180°
⇒ 34° + 90° + ∠CBA = 180°
⇒ 124° + ∠CBA = 180°
⇒ ∠CBA = 180° - 124°
⇒ ∠CBA = 56°.

Hence, the value of ∠CBA = 56°.

(ii) From figure,

∠BCQ = ∠CAB = 34°. (∵ angles in alternate segments are equal.)

∠ACQ = ∠ACB + ∠BCQ = 90° + 34° = 124°.

Since sum of angles in a triangle = 180°.

In △ACQ,

⇒ ∠CAQ + ∠ACQ + ∠CQA = 180°
⇒ 34° + 124° + ∠CQA = 180°
⇒ 158° + ∠CQA = 180°
⇒ ∠CQA = 180° - 158°
⇒ ∠CQA = 22°.

Hence, the value of ∠CQA = 22°.

(i) From figure,

OA ⊥ AP and OB ⊥ BP (∵ OA and OB are the radii and AP and BP are tangents.)

Now in quadrilateral AOBP,

∠P = 60°, ∠OAP = 90° and ∠OBP = 90°.

∠P + ∠OAP + ∠OBP + ∠AOB = 360°
60° + 90° + 90° + ∠AOB = 360°
240° + ∠AOB = 360°
∠AOB = 360° - 240°
∠AOB = 120°.

Hence, the value of ∠AOB = 120°.

(ii) Join AB as shown in the figure below:

Considering △OAB,

The triangle is isosceles as OA = OB = radii of the circle so, ∠OAB = ∠OBA = x.

Since sum of angles in a triangle = 180.

In △OAB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°
⇒ 120° + x + x = 180°
⇒ 120° + 2x = 180°
⇒ 2x = 180° - 120°
⇒ 2x = 60°
⇒ x = 30°

Hence, the value of ∠OAB = 30°.

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

120° = 2∠ACB
∠ACB = $\frac{120°}{2}$ = 60°.

Hence, the value of ∠ACB = 60°.

From figure,

XT = YT (∵ tangents from an external point to a circle are of equal length.)

So, △XTY is an isosceles triangle with

∠YXT = ∠XYT = a.

Since sum of angles in a triangle = 180°.

In △XTY,

⇒ ∠YXT + ∠XYT + ∠XTY = 180°
⇒ a + a + 80° = 180°
⇒ 2a + 80° = 180°
⇒ 2a = 180° - 80°
⇒ 2a = 100°
⇒ a = 50°.

From figure,

OX = OZ = radius of the circle

So, △OXZ is an isosceles triangle with

∠OXZ = ∠OZX = b.

Since sum of angles in a triangle = 180°.

In △OXZ,

⇒ ∠OXZ + ∠OZX + ∠XOZ = 180°
⇒ b + b + 140° = 180°
⇒ 2b + 140° = 180°
⇒ 2b = 180° - 140°
⇒ 2b = 40°
⇒ b = 20°.

From figure,

OX ⊥ XT (∵ tangent at a point and radius through the point are perpendicular to each other.)

∴ ∠OXT = 90°

∠OXY + ∠YXT = 90°
∠OXY + 50° = 90°
∠OXY = 90° - 50°
∠OXY = 40°.

From figure,

∠ZXY = ∠OXZ + ∠OXY = 20° + 40° = 60°.

Hence, the value of ∠ZXY = 60°.

From figure,

OP ⊥ PT (∵ tangent at a point and radius through the point are perpendicular to each other.)

∴ ∠OPT = 90°.

Given, ∠QPT = 30°.

From figure,

∠OPT = 90°
∠OPQ + ∠QPT = 90°
∠OPQ + 30° = 90°
∠OPQ = 90° - 30° = 60°.

In △OPQ,

OP = OQ (∵ both are equal to radius of the circle.)

So, the triangle is isosceles. So,

∠OQP = OPQ = 60°.

Since sum of angles in a triangle = 180°.

In △OPQ,

⇒ ∠OQP + ∠OPQ + ∠POQ = 180°
⇒ 60 + 60 + ∠POQ = 180°
⇒ 120 + ∠POQ = 180°
⇒ ∠POQ = 180° - 120°
⇒ ∠POQ = 60°

Reflex ∠POQ = 360° - ∠POQ = 360° - 60° = 300°.

Arc PQ subtends Reflex ∠POQ at the centre and ∠PRQ at the remaining part of the circle.

∴ Reflex ∠POQ = 2∠PRQ (∵ angle subtended at centre by an arc is double the angle subtended at remaining part of circle)

300° = 2° × ∠PRQ
∠PRQ = $\frac{300°}{2}$ = 150°.

(i) Hence, the value of ∠PRQ = 150°.

(ii) Hence, the value of ∠POQ = 60°.

We know that when two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

Given, chords AB and CD of a circle intersect internally at a point P. So,

PA.PB = PC.PD

(i) Given, AP = 6 cm, PB = 4 cm and PD = 3 cm.

We know that,

PA.PB = PC.PD

⇒ 6 × 4 = PC × 3
⇒ 24 = 3PC
⇒ PC = $\frac{24}{3}$ = 8 cm.

Hence, the length of PC = 8 cm.

(ii) Given, AB = 12 cm, AP = 2 cm, PC = 5 cm

PB = AB - AP = 12 - 2 = 10 cm.

We know that,

PA.PB = PC.PD

⇒ 2 × 10 = 5 × PD
⇒ 20 = 5PD
⇒ PD = $\frac{20}{5}$ = 4 cm.

Hence, the length of PD = 4 cm.

(iii) Given, AP = 5 cm, PB = 6 cm and CD = 13 cm

Let PC = x, so PD = 13 - x

We know that,

PA.PB = PC.PD

⇒ 5 × 6 = x(13 - x)
⇒ 30 = 13x - x²
⇒ x² - 13x + 30 = 0
⇒ x² - 10x - 3x + 30 = 0
⇒ x(x - 10) - 3(x - 10) = 0
⇒ (x - 3)(x - 10) = 0
⇒ x - 3 = 0 or x - 10 = 0
⇒ x = 3 or x = 10.

Hence, PC = 3 cm or 10 cm.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ TP² = AT × BT

From figure,

BT = AT - AB = 16 - 12 = 4 cm.

Putting values we get,

⇒ TP² = 16 × 4
⇒ TP² = 64
⇒ TP = $\sqrt{64}$
⇒ TP = 8 cm.

Hence, the length of TP = 8 cm.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ TP² = PC × PD

From figure,

PC = PD + CD
= 5 + 7.8
= 12.8 cm.

⇒ TP² = 12.8 × 5
⇒ TP² = 64
⇒ TP = $\sqrt{64}$
⇒ TP = 8 cm.

Similarly,

⇒ TP² = AP × BP
⇒ 8² = AP × 4
⇒ 64 = 4AP
⇒ AP = $\frac{64}{4}$
⇒ AP = 16 cm.

(i) From figure,

AB = AP - BP = 16 - 4 = 12 cm.

Hence, the length of AB = 12 cm.

(ii) The length of tangent PT = 8 cm.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PT² = PA × PB

(i) Putting values in above equation:

⇒ 8² = 5 × PB
⇒ 8² = 5PB
⇒ PB = $\frac{64}{5}$
⇒ PB = 12.8 cm.

AB = PB - PA = 12.8 - 5 = 7.8 cm.

Hence, the length of AB = 7.8 cm.

(ii) We know,

PB = AB + PA = 13.5 + 4.5 = 18 cm.

PT² = PA × PB
⇒ PT² = 4.5 × 18
⇒ PT² = 81
⇒ PT = $\sqrt{81}$
⇒ PT = 9 cm.

Hence, the length of PT = 9 cm.

(i) In △ACD and △DCB

∠C = ∠C (Common angles)

∠CAD = ∠CDB (Angles in alternate segments are equal)

∴ △ACD ~ △DCB (By AA axiom.)

Hence, proved that △ACD ~ △DCB.

(ii) Since triangles are similar hence, the ratio of their corresponding sides are equal.

$$\begin{aligned}\frac{AC}{DC} = \frac{DC}{BC} \\[1em] DC^2 = AC \times BC \\[1em] DC^2 = (AB + BC) \times BC \\[1em] DC^2 = (7 + 9) \times 9 \\[1em] DC^2 = 16 \times 9 \\[1em] DC^2 = 144 \\[1em] DC = \sqrt{144} \text{ cm} \\[1em] DC = 12 \text{ cm}.\end{aligned}$$

Hence, the length of DC = 12 cm.

Given, PA : AB = 1 : 3.

Let PA = k, so AB = 3k.

PB = PA + AB = k + 3k = 4k.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PT² = PA × PB

⇒ 6² = k × 4k

⇒ 36 = 4k²

⇒ k² = $\frac{36}{4}$

⇒ k = $\sqrt{9}$ cm

⇒ k = 3 cm.

PB = 4k = 4(3) = 12 cm.

Hence, the length of PB = 12 cm.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AQ² = AP × AB .....(Eq. 1)

Given Q is mid-point of AC, so AQ = $\frac{AC}{2}$.

$\Rightarrow \Big(\frac{AC}{2}\Big)^2 = AP \times AB$

Since, AC = AB.

$$\begin{aligned}\Rightarrow \Big(\frac{AB}{2}\Big)^2 = AP \times AB \\[1em] \Rightarrow \frac{AB^2}{4} = AP \times AB \\[1em] \Rightarrow AB^2 = 4 \times AP \times AB\end{aligned}$$

Dividing both sides by AB,

AB = 4 x AP.

Hence, proved that AB = 4AP.

We know that when two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

Given, chords AB and CD of a circle intersect externally at a point P. So,

PA.PB = PC.PD .....(Eq. 1)

Let PA = a, so PC = a. (∵ PA = PB)

Putting these value in Eq. 1 we get,

a.PB = a.PD

Dividing both sides by a we get,

PB = PD.

Let PB = PD = b

From figure,

AB = PA - PB = a - b.

CD = PC - PD = a - b.

Hence, proved that AB = CD.

From figure,

∠ACB = ∠BAT = 45° (∵ angles in alternate segments are equal)

Since sum of angles in a triangle = 180°.

In △ABC,

⇒ ∠ACB + ∠CAB + ∠ABC = 180°
⇒ 45° + 65° + ∠ABC = 180°
⇒ 110° + ∠ABC = 180°
⇒ ∠ABC = 180° - 110°
⇒ ∠ABC = 70°.

Hence, the value of ∠ABC = 70°.

Join OA, OB and CB. In △ATC,

Ext. ∠CAB = ∠ATC + TCA (∵ external angle in a triangle is equal to the sum of opposite interior angles.)

Ext. ∠CAB = 36° + 48° = 84°.

From figure,

∠ABC = ∠TCA = 48° (∵ angles in alternate segment are equal.)

Since sum of angles in a triangle = 180°.

In △ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180°
⇒ 48° + 84° + ∠ACB = 180°
⇒ 132° + ∠ACB = 180°
⇒ ∠ACB = 180° - 132°
⇒ ∠ACB = 48°.

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)

∠AOB = 2 × 48° = 96°.

Hence, the angle subtended by AB at the center of the circle is 96°.

In △ABC,

AB = AC (Given)

∴ ∠C = ∠B (∵ angles opposite to equal sides are equal.)

From figure,

∠TAC = ∠B (∵ angles in alternate segment are equal.)

But ∠B = ∠C

∴ ∠TAC = ∠C

But angles ∠TAC and ∠C are alternate angles. Since, they are equal

Hence, proved that AT || BC.

The figure below shows a rectangle ABCD with its sides touching the circle at points P, Q, R and S.

We know that,

Length of tangents from an external point to the circle are equal.

Hence,

AP = AS .....(Eq. 1)

BP = BQ .....(Eq. 2)

CR = CQ .....(Eq. 3)

DR = DS .....(Eq. 4)

Adding the above 4 equations,

⇒ AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC

But AB = CD and AD = BC (As opposite sides of a rectangle are equal.)

⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC.

∴ AB = BC = CD = DA.

Hence, proved that ABCD is a square.

From figure,

PT is a tangent and PA is chord.

∠APT = ∠ABP (∵ angles in alternate segments are equal.) ...(i)

BDCA is a cyclic quadrilateral as all the vertices lie on the circumference of the circle.

In cyclic quadrilateral the exterior angle is equal to the opposite interior angle.

∴ ∠ABP = ∠ACD ....(ii)

From (i) and (ii),

∠APT = ∠ACD

The angles ∠APT and ∠ACD are alternate angles, but since they are equal,

Hence, proved that CD || PT.

In △ACB,

AC = BC (Radius of the same circle)

∴ ∠BAC = ∠ABC ....(i)

PAQ is tangent and AC is the chord of the circle.

∠PAC = ∠ABC (∵ angles in alternate segment are equal) ....(i)

From (i) and (ii)

∠BAC = ∠PAC

Hence, proved that AC bisects ∠PAB.

In △OAB,

OA = OB (∵ both are radius of the common circle.)

So, △OAB is a isosceles triangle with,

∠OBA = ∠OAB = 32°.

Since sum of angles in a triangle = 180°.

In △OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°
⇒ 32° + 32° + ∠AOB = 180°
⇒ 64° + ∠AOB = 180°
⇒ ∠AOB = 180° - 64°
⇒ ∠AOB = 116°.

Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)

⇒ 116° = 2y
⇒ y = $\frac{116°}{2}$
⇒ y = 58°.

From figure,

∠ABT = ∠ACB = 58° (∵ angles in alternate segments are equal.)

∴ x = 58°.

Hence, the value of x = 58° and y = 58°.

(i) From figure,

From M, MA and MP are the tangents.

∴ MA = MP.....(i) (∵ length of the different tangents to a circle from a single point are equal.)

Similarly,

From M, MB and MP are the tangents.

∴ MB = MP.....(ii) (∵ length of the different tangents to a circle from a single point are equal.)

From (i) and (ii),

MA = MB.

Hence, proved that M bisects AB.

(ii) Since MA = MP

Hence in triangle APM,

∠MAP = ∠MPA ....(i) (∵ angles opposite to equal sides are equal.)

Since MB = MP

Hence in triangle BPM,

∠MPB = ∠MBP ....(ii) (∵ angles opposite to equal sides are equal.)

Adding equations (i) and (ii)

⇒ ∠MAP + ∠MPB = ∠MPA + ∠MBP
⇒ ∠MAP + ∠MBP = ∠APB

Since sum of angles in a triangle = 180°

In triangle APB

⇒ ∠APB + ∠MAP + ∠MBP = 180°

Putting value of ∠MAP + ∠MBP = ∠APB in above equation

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = $\frac{180°}{2}$ = 90°.

Hence, proved that ∠APB = 90°.

Join PC and QD.

We know that,

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ PC ⊥ CD and QD ⊥ CD

⇒ ∠PCD = 90° and ∠QDC = 90°

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

⇒ ∠CPR = 2 x ∠CAR = 2 x 38° = 76°

⇒ ∠CPR = ∠CPQ = 76°

CDQP is quadrilateral.

∴ ∠PCD + ∠QDC + ∠DQP + ∠CPQ = 360°

⇒ 90° + 90° + ∠DQP + 76° = 360°

⇒ 256° + ∠DQP = 360°

⇒ ∠DQP = 360° - 256°

⇒ ∠DQP = 104°

From figure,

⇒ ∠DQR = ∠DQP

⇒ ∠DQR = 104°

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

⇒ ∠DQR = 2 x ∠DBR

⇒ 104° = 2 x ∠DBR

⇒ ∠DBR = $\frac{104°}{2}$

⇒ ∠DBR = 52°.

Hence, ∠DBR = 52°.

From figure,

OA = OB (Radius of the circle.)

So, △OAB is an isosceles triangle with ∠OBA = ∠OAB (As angles opposite to equal sides are equal.)

∠OBA = 40°.

Since, sum of angles in a triangle = 180°.

⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 40° + 40° + ∠AOB = 180°
⇒ 80° + ∠AOB = 180°
⇒ ∠AOB = 180° - 80°
⇒ ∠AOB = 100°.

Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.

∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)

100° = 2∠ACB
∠ACB = $\frac{100°}{2}$ = 50°.

Hence, Option 1 is the correct option.

Cyclic quadrilateral ABCD is shown in the figure below:

The sum of opposite angles in a quadrilateral = 180°.

∴ ∠ADC + ∠ABC = 180°

140° + ∠ABC = 180°
∠ABC = 180° - 140° = 40°.

In △ABC,

∠ACB = 90° (∵ angles in semicircle = 90°.)

Since, sum of angles in a triangle = 180°.

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 40° + 90° + ∠BAC = 180°
⇒ 130° + ∠BAC = 180°
⇒ ∠BAC = 180° - 130°
⇒ ∠BAC = 50°.

Hence, Option 2 is the correct option.

In the figure,

OA = OB (Radius of the circle.)

So, △OAB is an isosceles triangle with ∠OBA = ∠BAO (∵ angles opposite to equal sides are equal.)

∠OBA = 60°.

In a triangle the exterior angle is equal to the sum of opposite interior angle.

∴ ∠AOC = ∠BAO + ∠OBA = 60° + 60° = 120°.

Arc AC subtends ∠AOC at centre and ∠ADC at remaining part of circle.

∠AOC = 2∠ADC (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)

120° = 2∠ADC

∠ADC = $\frac{120°}{2}$ = 60°.

Hence, Option 3 is the correct option.

From figure,

In △OPQ,

OP = OQ = PQ = (Radius of the circle.)

Hence, △OPQ is an equilateral triangle.

∴ ∠POQ = 60° (∵ all angles of an equilateral triangle = 60°.)

Arc PQ subtends ∠POQ at centre and ∠PRQ at remaining part of circle.

∠POQ = 2∠PRQ (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)

60° = 2∠PRQ

∠PRQ = $\frac{60°}{2}$ = 30°.

Hence, Option 3 is the correct option.

From figure,

∠ADB = ∠ACB = 2x (∵ ∵ angles in same segment are equal.)

Join OA as shown in the figure below:

Arc AB subtends ∠AOB at centre and ∠ADB at remaining part of circle.

∠AOB = 2∠ADB = 2(2x) = 4x (∵ angle subtended at centre is double the angle subtended at remaining part of circle.)

In △OAB,

OA = OB (Radius of the circle.)

So, △OAB is an isosceles triangle with ∠OBA = ∠OAB (∵ angles opposite to equal sides are equal.)

∠OAB = 3x.

Since, sum of angles in a triangle = 180°.

⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 3x + 3x + 4x = 180°
⇒ 10x = 180°
⇒ x = 18°

Hence, Option 1 is the correct option.

Given, the point P is 13 cm from O, the centre of the circle as shown in the figure below:

Radius of the circle (OQ) = 5 cm

PQ and PR are tangents from P to the circle.

PQ ⊥ OQ (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ OQP = 90°.

So, in △OQP,

$$\begin{aligned}OP^2 = OQ^2 + PQ^2 \\[1em] 13^2 = 5^2 + PQ^2 \\[1em] PQ^2 = 13^2 - 5^2 \\[1em] PQ^2 = 169 - 25 \\[1em] PQ^2 = 144 \\[1em] PQ = 12 \text{ cm}\end{aligned}$$

Area of △OPQ = $\frac{1}{2} \times PQ \times OQ$ = $\frac{1}{2} \times 12 \times 5 = 30$ cm².

Similarly,

PR ⊥ OR (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠ORP = 90°.

So, in △ORP,

$$\begin{aligned}OP^2 = OR^2 + PR^2 \\[1em] 13^2 = 5^2 + PR^2 \\[1em] PR^2 = 13^2 - 5^2 \\[1em] PR^2 = 169 - 25 \\[1em] PR^2 = 144 \\[1em] PR = 12 \text{ cm}\end{aligned}$$

Area of △POR = $\frac{1}{2} \times PR \times OR$ = $\frac{1}{2} \times 12 \times 5 = 30$ cm².

Area of quadrilateral PQOR = Area of △POR + Area of △OPQ = 30 + 30 = 60 cm².

Hence, Option 1 is the correct option.

Since, ABCD is a cyclic quadrilateral.

We know that,

The sum of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠BAD + ∠BCD = 180°

⇒ (2x + 5)° + (x + 10)° = 180°

⇒ 3x° + 15° = 180°

⇒ 3x° = 180° - 15°

⇒ 3x° = 165°

⇒ x° = $\frac{165°}{3}$

⇒ x° = 55°

⇒ x = 55.

Hence, Option 3 is the correct option.

OR ⊥ PR (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠ORP = 90°.

Since, sum of angles in a triangle = 180°.

⇒ ∠ORP + ∠POR + ∠OPR = 180°
⇒ 90° + 55° + ∠OPR = 180°
⇒ 145° + ∠OPR = 180°
⇒ ∠OPR = 180° - 145°
⇒ ∠OPR = 35°.

∠QPO = ∠OPR = 35° (∵ the tangents are equally inclined to the line joining the point and the centre of the circle.)

From figure,

∠QPR = ∠OPR + ∠QPO = 35° + 35° = 70°.

Hence, Option 3 is the correct option.

Join OA as shown in the figure below:

OA ⊥ PA (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠OAP = 90°.

Given, PA ⊥ PB

∴ ∠APB = 90°.

∵ the tangents are equally inclined to the line joining the point and the centre of the circle.

∠APO = $\frac{1}{2}$ x ∠APB = 45°.

Since, sum of angles in a triangle = 180°.

In △OAP,

⇒ ∠APO + ∠OAP + ∠AOP = 180°
⇒ 45° + 90° + ∠AOP = 180°
⇒ 135° + ∠AOP = 180°
⇒ ∠AOP = 180° - 135°
⇒ ∠AOP = 45°.

Since, ∠AOP = ∠APO hence, △OAP is an isosceles triangle with OA = AP = 5 cm.

In right angled triangle △OAP,

$$\begin{aligned}OP^2 = OA^2 + AP^2 \\[1em] OP^2 = 5^2 + 5^2 \\[1em] OP^2 = 25 + 25 \\[1em] OP^2 = 50 \\[1em] OP = \sqrt{50} \\[1em] OP = 5\sqrt{2} \text{ cm}\end{aligned}$$

Hence, Option 4 is the correct option.

The figure is shown below:

Since tangent and radius at point of contact of a circle are perpendicular to each other. Hence,

XAY ⊥ AO

Given XAY || to CD hence,

CD ⊥ AB.

In right angled triangle OEC,

OE = AE - AO = 8 - 5 = 3 cm.

⇒ OC² = OE² + CE² (By pythagoras theorem)
⇒ 5² = 3² + CE²
⇒ CE² = 25 - 9
⇒ CE² = 16
⇒ CE = 4 cm.

Similarly in right angled triangle OED,

⇒ OD² = OE² + ED² (By pythagoras theorem)
⇒ 5² = 3² + ED²
⇒ ED² = 25 - 9
⇒ ED² = 16
⇒ ED = 4 cm.

⇒ CD = CE + ED = 4 + 4 = 8 cm.

Hence, Option 4 is the correct option.

From figure,

AB is chord to the bigger circle which is tangent to the smaller circle at C.

OC ⊥ AC (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠ACO = 90°.

In △ACO,

$$\begin{aligned}\Rightarrow OA^2 = OC^2 + AC^2 \\[1em] \Rightarrow 5^2 = 4^2 + AC^2 \\[1em] \Rightarrow 25 = 16 + AC^2 \\[1em] \Rightarrow 25 - 16 = AC^2 \\[1em] \Rightarrow AC^2 = 9 \\[1em] \Rightarrow AC = 3 \text{ cm}.\end{aligned}$$

Length of chord AB = 2 × AC = 2 × 3 = 6 cm.

Hence, Option 2 is the correct option.

We know that,

Angle between tangent and the chord at the point of contact is equal to angle of the alternate segment.

∴ ∠PQS = ∠PST = 30°

In △ PQS,

By angle sum property of triangle,

⇒ ∠PQS + ∠QPS + ∠PSQ = 180°

⇒ 30° + 60° + ∠PSQ = 180°

⇒ ∠PSQ + 90° = 180°

⇒ ∠PSQ = 180° - 90° = 90°.

Hence, Option 4 is the correct option.

In the given figure,

PA and PB are tangents to the circle with centre O.

∠APB = 50°

Since sum of opposite angles of quadrilateral = 180°.

∴ ∠AOB + ∠APB = 180°
⇒ ∠AOB + 50° = 180°
⇒ ∠AOB = 180° - 50° = 130°.

In △OAB,

OA = OB (Radius of the same circle)

Hence, △OAB is an isosceles triangle with ∠OAB = ∠OBA.

Since, sum of angles of a triangle = 180°.

In △OAB,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ ∠OAB + ∠OAB + 130° = 180°
⇒ 2∠OAB = 180° - 130°
⇒ 2∠OAB = 50°
⇒ ∠OAB = 25°.

Hence, Option 1 is the correct option.

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

∴ BF = BD = 4 cm, CE = CD = 3 cm and AF = AE.

Given, CA = 8 cm.

From figure,

⇒ CA = AE + CE
⇒ 8 = AE + 3
⇒ AE = 8 - 3 = 5 cm.

We know AF = AE = 5cm.

From figure,

AB = AF + BF = 5 + 4 = 9 cm.

Hence, Option 4 is the correct option.

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

∴ BP = BR = 6 cm, CQ = CP = 5 cm and AQ = AR = 4 cm.

Perimeter of △ABC ⇒ AB + BC + CA
⇒ AR + BR + BP + CP + AQ + CQ
⇒ 4 + 6 + 6 + 5 + 4 + 5
⇒ 30 cm.

Hence, Option 3 is the correct option.

The circle with centre O and PQ as tangent is shown in the figure below:

We know that,

OP ⊥ PQ (∵ tangent through a point and radius from that point are perpendicular to each other)

Given, △OPQ is an isosceles triangle.

Since, ∠OPQ = 90° hence, the other two angles will be equal to each other.

∴ ∠QOP = ∠OQP

We know that sum of angles in a triangle = 180°.

In △OPQ,

⇒ ∠OPQ + ∠QOP + ∠OQP = 180°
⇒ ∠OPQ + ∠QOP + ∠QOP = 180°
⇒ 90° + 2∠QOP = 180°
⇒ 2∠QOP = 180° - 90
⇒ ∠QOP = $\frac{90°}{2}$
⇒ ∠QOP = 45°.

Hence, Option 3 is the correct option.

Join OP.

We know that the tangents are equally inclined to the line joining the point and the centre of the circle.

∴ ∠APO = $\frac{∠\text{APB}}{2} = \frac{40°}{2}$ = 20°.

AP ⊥ OA.

∴ ∠OAP = 90°.

In right angle triangle OAP,

⇒ ∠APO + ∠AOP + ∠OAP = 180°
⇒ 20° + ∠AOP + 90° = 180°
⇒ ∠AOP + 110° = 180°
⇒ ∠AOP = 180° - 110° = 70°.

As the tangents subtends equal angles at centre.

∴ ∠BOP = ∠AOP = 70°.

∠AOB = ∠AOP + ∠BOP = 70° + 70° = 140°.

Arc AB subtends ∠AOB at center and ∠ACB on the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\frac{1}{2} \times$ ∠AOB

⇒ ∠ACB = $\frac{1}{2} \times 140°$

⇒ ∠ACB = 70°.

Hence, Option 2 is the correct option.

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

From figure,

PA and PB are the tangents to the first circle.

∴ PA = PB = 3.8 cm

PA and PC are tangents to the second circle.

∴ PC = PA = 3.8 cm

From figure,

BC = PB + PC = 3.8 + 3.8 = 7.6 cm.

Hence, Option 1 is the correct option.

We know that if two tangents are drawn from an external point to a circle then, the lengths of the tangents are equal.

From figure,

PA and PD are the tangents to the circle from P.

∴ PA = PD

QB and QA are the tangents to the circle from Q.

∴ QB = QA

Hence,

⇒ PD + BQ
⇒ PA + QA
⇒ PQ.

Hence, Option 1 is the correct option.

From figure,

BQ is chord and PQR is a tangent.

∠BQR = ∠A (∵ angles in alternate segment are equal.)

As AB || PQR

∠BQR = ∠B (∵ alternate angles are equal)

∴ ∠A = ∠B = 70°

We know that sum of angles in a triangle = 180°.

In △AQB,

⇒ ∠A + ∠B + ∠AQB = 180°
⇒ 70° + 70° + ∠AQB = 180°
⇒ 140° + ∠AQB = 180°
⇒ ∠AQB = 180° - 140°
⇒ ∠AQB = 40°.

Hence, Option 2 is the correct option.

We know that if two chords of a circle intersect internally or externally, then the products of the lengths of segments are equal.

Hence, from figure,

PA.PB = PC.PD .....(i)

Given,

PC = 15 cm and CD = 7 cm

From figure,

PD = PC - CD = 15 - 7 = 8 cm.

Let BP = x cm then AB = (12 - x) cm

Putting values in equation (i),

⇒ PA.PB = PC.PD
⇒ 12.x = 15.8
⇒ 12x = 120
⇒ x = $\frac{120}{12}$ = 10 cm.

AB = 12 - x = 12 - 10 = 2 cm.

Hence, Option 1 is the correct option.

Let l and m be tangents to circle with center O, touching the circle at point A and B.

From figure,

Shortest distance between tangents = OA + OB = R + R = 2R.

Hence, Option 2 is the correct option.

In △ PST,

⇒ PS = PT (Tangents from an external point to a circle are equal in length)

⇒ ∠PST = ∠PTS = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠PST + ∠PTS + ∠SPT = 180°

⇒ a + a + 80° = 180°

⇒ 2a = 180° - 80°

⇒ 2a = 100°

⇒ a = $\frac{100°}{2}$ = 50°.

From figure,

⇒ ∠QST = ∠STP = 50° (Alternate angles are equal)

Hence, Option 4 is the correct option.

The circumcenter of a triangle is defined as the point equidistant from the three vertices of the triangle.

Hence, Option 2 is the correct option.

Let ABC be the triangle and O is the circumcenter of triangle.

Circumcenter is the center of circle which passes through all vertices of triangle.

∴ OA = OB = OC (Radius of circle)

Hence, Option 3 is the correct option.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠ABC = 90°.

In △ APQ and △ ABC,

⇒ ∠APQ = ∠ABC (Both equal to 90°)

⇒ ∠PAQ = ∠BAC (Common angles)

∴ △ APQ ~ △ ABC (By A.A. axiom)

We know that,

The ratio of area of similar triangles is equal to the ratio of the square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △ APQ}}{\text{Area of △ ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \frac{18}{\text{Area of △ ABC}} = \frac{AP^2}{AB^2} \\[1em] \Rightarrow \text{Area of △ ABC} = \frac{AB^2}{AP^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \frac{(3 + 4)^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \frac{7^2}{3^2} \times 18 \\[1em] \Rightarrow \text{Area of △ ABC} = \frac{49}{9} \times 18 = 98 \text{ cm}^2.\end{aligned}$$

From figure,

Area of QPBC = Area of △ ABC - Area of △ APQ = 98 - 18 = 80 cm².

Hence, Option 3 is the correct option.

From figure,

⇒ ∠QOT + ∠TOP = 180° (Linear Pair)

⇒ 110° + ∠TOP = 180°

⇒ ∠TOP = 180° - 110° = 70°.

⇒ ∠OPT = 90° (Tangent is perpendicular to radius at point of intersection)

In △ TOP,

⇒ ∠OPT + ∠TOP + ∠PTO = 180°

⇒ 90° + 70° + x°= 180°

⇒ x° + 160° = 180°

⇒ x° = 180° - 160° = 20°.

Hence, Option 1 is the correct option.

A cyclic quadrilateral is defined as a quadrilateral whose all four vertices lie on the circumference of a circle. In other words, it is a quadrilateral that is inscribed in a circle.

So, reason (R) is true.

In the given figure, O is center of the circle. Hence, it does not lie on the circumference of the circle.

Therefore, OABC is not a cyclic quadrilateral.

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given,

AB is the diameter of the circle.

We know that,

Angle in a semi-circle is a right angle.

So, reason (R) is true.

∠APB = 90°

We know that,

Side opposite to 90° is hypotenuse.

Thus, in triangle APB, AB is the hypotenuse.

Using pythagoras theorem in ΔAPB,

⇒ Hypotenuse² = Base² + Perpendicular²

⇒ AB² = BP² + AP²

So, assertion (A) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

From figure,

∠CBE and ∠CBA forms a linear pair.

⇒ ∠CBE + ∠CBA = 180°

⇒ 108° + ∠CBA = 180°

⇒ ∠CBA = 180° - 108°

⇒ ∠CBA = 72°

ABCD is a cyclic quadrilateral, opposite angles are supplementary.

So, reason (R) is true.

⇒ ∠CBA + ∠ADC = 180°

⇒ 72° + ∠ADC = 180°

⇒ ∠ADC = 180° - 72°

⇒ ∠ADC = 108°.

So, assertion (A) is true.

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

We know that,

In a cyclic quadrilateral, an exterior angle is equal to the opposite interior angle.

So, reason (R) is true.

In case of assertion (A) : An exterior angle of a cyclic quadrilateral is equal to an interior angle.

This statement is incorrectly stated as the exterior angle is equal to the opposite interior angle.

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

According to alternate Segment theorem,

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by the chord in the alternate segment of the circle.

So, reason (R) is true.

∴ ∠QRP = ∠QPT (By alternate segment theorem)

⇒ ∠QRP = 50°

In ΔPQR, according to angle sum property,

∴ ∠PQR + ∠PRQ + ∠QPR = 180°

⇒ 45° + 50° + ∠QPR = 180°

⇒ 95° + ∠QPR = 180°

⇒ ∠QPR = 180° - 95°

⇒ ∠QPR = 85°

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given,

QR = OP ⇒ QR = OQ

⇒ ∠QOR = ∠ORQ = 20° (∵ angles opposite equal sides of a triangle are equal)

Exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ ∠OQP = ∠QOR + ∠ORQ = 20° + 20° = 40°.

As OP = OQ, ∠OPQ = ∠OQP

⇒ ∠OPQ = 40°
⇒ ∠OPR = 40°.

Exterior angle in a triangle is equal to the sum of two opposite interior angles.

∴ x = ∠TOP = ∠OPR + ∠ORP = 40° + 20° = 60°.

Hence, the value of x = 60°.

Since ABC is an equilateral triangle so,

∠A = ∠B = ∠C = 60°.

From figure,

∠BDC = ∠BAC (∵ angles in alternate segments are equal.)

∴ ∠BDC = 60°.

BDCE is a cyclic quadrilateral. Hence, sum of the opposite angles = 180°.

⇒ ∠BDC + ∠BEC = 180°
⇒ 60° + ∠BEC = 180°
⇒ ∠BEC = 180° - 60° = 120°.

Hence, the value of ∠BDC = 60° and ∠BEC = 120°.

In △AOD, ∠AOD = 90°.

OA = OD (Radii of the semi-circle)

∠OAD = ∠ODA (∵ angles opposite equal side are equal.)

We know that sum of angles in a triangle = 180°.

In △OAD,

⇒ ∠AOD + ∠OAD + ∠ODA = 180°
⇒ 90° + ∠OAD + ∠OAD = 180°
⇒ 90° + 2∠OAD = 180°
⇒ 2∠OAD = 180° - 90°
⇒ ∠OAD = $\frac{90°}{2}$ = 45°.

From figure,

∠BAD = ∠OAD = 45°.

Arc AD subtends ∠AOD at the centre and ∠ACD at the remaining part of the circle.

∠AOD = 2∠ACD (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

⇒ 90° = 2∠ACD
⇒ ∠ACD = $\frac{90°}{2}$ = 45°.

Hence, the value of ∠BAD = 45° and ∠ACD = 45°.

We know that angle between the radius and tangent at the point of contact is right angle.

∴ ∠A = 90°.

Also in △OBE, OB = OE = radius of the circle.

∴ ∠B = ∠OEB .....(i)

In △ABD,

⇒ ∠A + ∠B + ∠ADB = 180°
⇒ 90° + ∠B + 55° = 180°
⇒ ∠B + 145° = 180°
⇒ ∠B = 180° - 145° = 35°.

∴ ∠OEB = 35°.

From figure,

∠DEC = ∠OEB = 35° (∵ vertically opposite angles are equal.)

∠EDC + ∠ADE = 180° (∵ both form a linear pair)

∠EDC + 55° = 180°
∠EDC = 180° - 55°
∠EDC = 125°.

In △EDC,

⇒ ∠DEC + ∠EDC + ∠DCE = 180°
⇒ 35° + 125° + x° = 180°
⇒ x° + 160° = 180°
⇒ x° = 180° - 160° = 20°.

In △AOC,

⇒ ∠AOC + ∠OAC + ∠ACO = 180°
⇒ y° + 90° + x° = 180°
⇒ y° + 90° + 20° = 180°
⇒ y° + 110° = 180°
⇒ y° = 180° - 110° = 70°.

Hence, the value of x = 20 and y = 70.

We know that in a cyclic quadrilateral the sum of opposite interior angles = 180°.

In cyclic quadrilateral BCDE,

⇒ ∠BCD + ∠BED = 180°
⇒ 140° + ∠BED = 180°
⇒ ∠BED = 180° - 140° = 40°.

∠AEB = 90°. (∵ angle in semicircle = 90°.)

From figure,

∠AED = ∠AEB + ∠BED = 90° + 40° = 130°.

In cyclic quadrilateral AEDB,

⇒ ∠AED + ∠DBA = 180°
⇒ 130° + ∠DBA = 180°
⇒ ∠DBA = 180° - 130° = 50°.

Given chord AE = ED

∴ ∠DBE = ∠EBA

From figure,

⇒ ∠DBA = ∠DBE + ∠EBA
⇒ 50 = ∠DBE + ∠DBE
⇒ 2∠DBE = 50°
⇒ ∠DBE = $\frac{50°}{2}$ = 25°.

or, ∠EBD = 25°.

In △OEB, OE = OB (Radii of the same circle.)

∠OEB = ∠EBO = ∠DBE

But these are alternate angles.

∴ OE || BD.

Hence, the value of ∠AED = 130° and ∠EBD = 25°.

We know that sum of angles in a triangle = 180°.

In △ABC,

∠ACB + ∠BAC + ∠ABC = 180°
∠ABC = 180° - (∠ACB + ∠BAC) .....(i)

In the circle arc AC subtends Reflex ∠AOC at centre and ∠ABC at remaining part of the circle.

∴ Reflex ∠AOC = 2∠ABC (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

From (i)

Reflex ∠AOC = 2(180° - (∠ACB + ∠BAC))

We know

Reflex ∠AOC = 360° - ∠AOC.

or,

360° - ∠AOC = 2(180° - (∠ACB + ∠BAC))
360° -∠AOC = 360° - 2(∠ACB + ∠BAC)
∠AOC = 360° - (360° - 2(∠ACB + ∠BAC))
∠AOC = 2(∠ACB + ∠BAC).

Hence, proved that ∠AOC = 2(∠ACB + ∠BAC).

From figure,

∠BEC = ∠BDC (∵ angles in same segment are equal)

Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∴ ∠BOC = 2∠BEC (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

⇒ ∠BOC = ∠BEC + ∠BEC = ∠BEC + ∠BDC

⇒ ∠BOC = ∠BEC + ∠BDC .....(i)

In △OBE,

⇒ ∠EOD = ∠EBO + ∠BEO ....(ii) (∵ exterior angle is equal to the sum of two opposite interior angles.)

From figure,

∠EBO = ∠EBD and ∠BEO = ∠BEC and Exterior angle ∠EOD = y.

Putting these values in eqn (ii) we get,

⇒ y = ∠EBD + ∠BEC
⇒ ∠BEC = y - ∠EBD .....(iii)

In △ABD,

∠BDC = ∠BAD + ∠ABD ....(iv) (∵ exterior angle is equal to the sum of two opposite interior angles.)

From figure,

∠ABD = ∠EBD and ∠BAD = x.

Putting these values in eqn (iv) we get,

∠BDC = x + ∠EBD ......(v)

Putting value of ∠BEC and ∠BDC from eqn (iii) and (v) respectively in (i) we get,

⇒ ∠BOC = ∠BEC + ∠BDC
⇒ ∠BOC = y - ∠EBD + x + ∠EBD
⇒ ∠BOC = x + y

From figure,

∠BOC = z.

∴ z = x + y.

Hence, proved that x + y = z.

(i) Join AD.

From figure,

∠BDC = ∠BAC = 25°. (∵ angles in same segment are equal.)

∠ADB = 90° (∵ angle in semicircle = 90°.)

∠ADC = ∠ADB + ∠BDC = 90° + 25° = 115°.

Hence, the value of ∠ADC = 115°.

(ii) ∠ACD = ∠CAB = 25° (∵ alternate angles are equal)

Since sum of angles in a triangle = 180°.

In △ADC,

⇒ ∠ADC + ∠DAC + ∠ACD = 180°
⇒ 115° + ∠DAC + 25° = 180°
⇒ ∠DAC + 140° = 180°
⇒ ∠DAC = 180° - 140° = 40°.

Hence, the value of ∠DAC = 40°.

Since sum of angles in a triangle = 180°.

In △ADP,

⇒ ∠ADP + ∠DAP + ∠DPA = 180°
⇒ 75° + ∠DAP + 50° = 180°
⇒ ∠DAP + 125° = 180°
⇒ ∠DAP = 180° - 125° = 55°.

From figure,

∠BAD = ∠DAP = 55°.

In cyclic quadrilateral sum of opposite angles = 180°

In ABCD,

∴ ∠ADC + ∠CBA = 180°
⇒ 75° + ∠CBA = 180°
⇒ ∠CBA = 180° - 75° = 105°.

In △ABQ,

⇒ ∠ABQ + ∠BAQ + ∠AQB = 180°

From figure, ∠BAQ = ∠BAD and ∠ABQ = ∠CBA or,

⇒ 105° + 55° + ∠AQB = 180°
⇒ ∠AQB + 160° = 180°
⇒ ∠AQB = 180° - 160° = 20°.

From figure,

∠CQD = ∠AQB = 20°.

Hence, the value of ∠BAD = 55° and ∠CQD = 20°.

Join AC and BD.

In △ABD and △CBD,

AB = CD (Given)

BD = BD (Common side)

∠BAD = ∠BCD (∵ angles in same segment are equal.)

∴ △ABD ≅ △CBD. (By SSA axiom of congruency.)

∴ BC = AD (As corresponding parts of congruent triangles are congruent.)

Hence, proved that BC = AD.

Since, AB = AC.

Hence, in △ABC,

∠ACB = ∠ABC = 50°.

Since sum of angles in a triangle = 180°.

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 50° + 50° + ∠BAC = 180°
⇒ ∠BAC + 100° = 180°
⇒ ∠BAC = 180° - 100° = 80°.

From figure,

∠BDC = ∠BAC = 80° (∵ angles in same segment are equal.)

In cyclic quadrilateral sum of opposite angles = 180°,

Hence in BDCE,

⇒ ∠BDC + ∠BEC = 180°
⇒ 80° + ∠BEC = 180°
⇒ ∠BEC = 180° - 80° = 100°.

Hence, the value of ∠BDC = 80° and ∠BEC = 100°.

Let T be the point of contact of the tangent from point P to the circle with centre O.

From figure,

OT ⊥ PT (As tangent and radius from point of contact are perpendicular to each other.)

In right-angled triangle OPT

⇒ OP² = OT² + PT²
⇒ 13² = OT² + 12²
⇒ 169 - 144 = OT²
⇒ OT² = 25
⇒ OT = 5 cm.

From figure,

⇒ OA = OT = 5 cm (Radius of the circle.)

⇒ PA = OP - OA = 13 - 5 = 8 cm.

Hence, the distance of P from the nearest point of circle = 8 cm.

Let two circles touch each other at point P and T is a point on common tangent as shown in the figure below:

As tangents drawn from an external point to a circle are equal in length.

From T, TA and TP are tangents to the circle with centre O'.

TA = TP .....(i)

From T, TB and TP are tangents to the circle with centre O.

TB = TP .....(ii)

From (i) and (ii),

TA = TB.

Hence, proved that tangents drawn to two circles from any point on common tangent are equal in length.

The figure is shown below:

(i) In △AOP and △BOP,

OP = OP (Common sides)

OA = OB (Radius of the circle)

∠OAP = ∠OBP (Both are equal to 90° as tangents and radius on point of contact are perpendicular to each other.)

∴ △OAP ≅ △OBP (S.A.S. axiom of congruency)

(As Corresponding parts of congruent triangles are congruent)

∴ ∠AOP = ∠BOP and ∠APO = ∠BPO.

Hence, proved that ∠AOP = ∠BOP.

(ii) In △APM and △BPM,

PM = PM (Common side)

∠APM = ∠BPM (Proved above)

AP = BP (∵ tangents from an exterior point to a circle are equal in length)

∴ △APM ≅ △BPM (S.A.S. axiom of congruency)

(Congruent parts of congruent triangles are congruent.)

∴ AM = BM and ∠AMP = ∠BMP

But ∠AMP + ∠BMP = 180°

∴ ∠AMP = ∠BMP = 90°.

Hence, proved that OP is perpendicular bisector of AB at M.

In △APC and △BQC

∠PCA = ∠QCB (∵ vertically opposite angles are equal)

∠APC = ∠BQC (∵ both are equal to 90 as radius and tangent to a circle at the point of contact are perpendicular to each other.)

∴ △APC ~ △BQC (By AA axiom of similarity)

Since triangles are similar hence the ratio of their corresponding sides are equal.

$\therefore \frac{AP}{BQ} = \frac{PC}{CQ}$

Hence, proved that AP : BQ = PC : CQ

Join OQ as shown in the figure below:

In △OAQ,

OA = OQ (Radius of the same circle.)

∠OAQ = ∠OQA.

Given QA || PO

∴ ∠OAQ = ∠POB (∵ corresponding angles are equal.)

and ∠OQA = ∠QOP (∵ alternate angles are equal.)

But ∠OAQ = ∠OQA,

∴ ∠POB = ∠QOP

Now in △OPQ and △OBP

OP = OP (Common sides)

OQ = OB (Radius of the same circle.)

∠QOP = ∠POB

∴ △OPQ ≅ △OBP (S.A.S. axiom of congruency)

As corresponding parts of congruent triangles are congruent,

∴ ∠OQP = ∠OBP

But ∠OQP = 90°

∴ ∠OBP = 90°

∴ PB is the tangent of the circle.

Hence, proved that PB is the tangent of the circle.

Since radius and tangent at the point of contact of a circle are perpendicular to each other.

∴ ∠AMP = ∠BNQ = 90°.

In right angled triangle △AMP

    AP² = AM² + PM² (By pythagoras theorem)
⇒ AM² = AP² - PM²
⇒ AM² = 17² - 15²
⇒ AM² = 289 - 225
⇒ AM² = 64
⇒ AM = $\sqrt{64}$
⇒ AM = 8 cm.

Similarly in right angled triangle △BNQ

    BQ² = BN² + NQ² (By pythagoras theorem)
⇒ BN² = BQ² - NQ²
⇒ BN² = 13² - 12²
⇒ BN² = 169 - 144
⇒ BN² = 25
⇒ BN = $\sqrt{25}$
⇒ BN = 5 cm.

From figure the distance between A and B is equal to the sum of their radius = 8 + 5 = 13 cm.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

From figure,

PA.PB = PC.PD .....(i)

PA = PB + AB = 7 + 9 = 16 cm.

Putting values in equation (i) we get,

⇒ 16 × 7 = PC × 6
⇒ 112 = PC × 6
⇒ PC = $\frac{112}{6}$

⇒ PC = $\frac{56}{3}$.

From figure,

CD = PC - PD = $\frac{56}{3} - 6 = \frac{56 - 18}{3} = \frac{38}{3} = 12\frac{2}{3}.$

Hence, the value of CD = $12\frac{2}{3}$ cm.

We know that if a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PA.PB = PT² .....(i)

From figure,

PB = PA - AB = 16 - 12 = 4 cm.

Putting values in equation (i),

16 x 4 = PT²
PT² = 64
PT = $\sqrt{64}$ = 8 cm.

Join OT as shown in the figure below:

In △OTP,

OT ⊥ TP (∵ tangents and radius at the point of contact are perpendicular to each other.)

In right angled triangle OTP,

OP² = OT² + PT² (By pythagoras theorem)
(OD + DP)² = OT² + PT²

Since, OD = OT = radius of circle = r.

(r + 2)² = r² + 8²
r² + 4 + 4r = r² + 64
r² - r² + 4r = 64 - 4
4r = 60
r = 15 cm.

Hence, the length of PT = 8 cm and radius of circle = 15 cm.

We know that if a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PA.PB = PT² .....(i)

From figure,

PA = AB + PB = 8 + 6 = 14 cm.

Putting values in equation (i),

14 x 6 = PT²
PT² = 84
PT = $\sqrt{84} = 2\sqrt{21}$ cm.

Joining OT as shown in the figure below:

In △OTP,

OT ⊥ TP (∵ tangents and radius at a point of contact are perpendicular to each other.)

In right angled triangle OTP,

$$\begin{aligned}OP^2 = OT^2 + PT^2 (\text{By pythagoras theorem}) \\[1em] (OD + DP)^2 = OT^2 + PT^2 \\[1em]\end{aligned}$$

Since, OD = OT = radius of circle = r.

$$\begin{aligned}\Rightarrow (r + 4)^2 = r^2 + (\sqrt{84})^2 \\[1em] \Rightarrow r^2 + 16 + 8r = r^2 + 84 \\[1em] \Rightarrow r^2 - r^2 + 8r = 84 - 16 \\[1em] \Rightarrow 8r = 68 \\[1em] \Rightarrow r = \frac{68}{8} \\[1em] \Rightarrow r = 8.5 \text{ cm.}\end{aligned}$$

Hence, the radius of the circle = 8.5 cm and length of tangent = $2\sqrt{21}$ cm.

From figure,

OP = OQ = radius of circle = 6 cm.

XP = XO + OP = 10 + 6 = 16 cm.

XQ = XO - OQ = 10 - 6 = 4 cm.

We know that,

If two chords of a circle intersect externally, then the products of the length of segments are equal.

From figure,

XA $\times$ XB = XP $\times$ XQ

5 $\times$ XA = 16 $\times$ 4

XA = $\frac{64}{5}$ = 12.8 cm.

AB = XA - XB = 12.8 - 5 = 7.8 cm

Let XT be the tangent to the circle as shown in the figure below:

We know that if a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ XP.XQ = XT² .....(i)

XT² = 16 x 4 = 64

XT = $\sqrt{64}$ = 8 cm.

Hence, the length of AB = 7.8 cm and length of tangent = 8 cm.

From figure,

∠ADC = ∠CBP = 40°. (∵ angles in alternate segments are equal.)

Since sum of angles in a triangle = 180°.

In △ADP,

∠DAP + ∠APD + ∠ADP = 180°

From figure, ∠ADP = ∠ADC = 40°.

⇒ p° + q° + 40° = 180°
⇒ p° + q° = 180° - 40°
⇒ p° + q° = 140° .....(i)

Join AC and BD as shown in the figure below:

∠CQB = ∠AQD = 2q° (∵ vertically opposite angles are equal.)

Given C, P, B, Q are concyclic.

∴ ∠CPB + ∠CQB = 180°
⇒ q° + 2q° = 180°
⇒ 3q° = 180°
⇒ q° = 60°.

Putting value of q in equation (i) we get,

⇒ p° + 60° = 140°
⇒ p° = 140° - 60° = 80°.

Hence, the value of p = 80 and q = 60 and the relation between p and q is p + q = 140.

∠AOB = 130°.

∠AOB + ∠BOC = 180° (∵ these angles form linear pair.)

130° + ∠BOC = 180°
∠BOC = 180° - 130° = 50°.

(i) Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∠BOC = 2∠BEC (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ 50° = 2∠BEC
⇒ ∠BEC = $\frac{50°}{2} = 25°.$

Hence, the value of ∠BEC = 25°.

(ii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

⇒ ∠AOB = 2∠ACB (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ 130° = 2∠ACB

⇒ ∠ACB = $\frac{130°}{2} = 65°.$

Hence, the value of ∠ACB = 65°.

(iii) Given, CD || EB,

∠ECD = ∠CEB = 25°. (∵ alternate angles are equal.)

From figure,

∠BCD = ∠ACB + ∠ACE + ∠ECD = 65° + 20° + 25° = 110°.

Hence, the value of ∠BCD = 110°.

(iv) EBCD is a cyclic quadrilateral hence the sum of opposite interior angles = 180°.

∴ ∠BED + ∠BCD = 180°
⇒ ∠BEC + ∠CED + ∠BCD = 180°
⇒ 25° + ∠CED + 110° = 180°
⇒ ∠CED + 135° = 180°
⇒ ∠CED = 180° - 135° = 45°.

Hence, the value of ∠CED = 45°.

(i) In △BCD, BC = CD

∴ ∠CBD = ∠BDC (∵ angles opposite to equal sides are equal.)

Since sum of angles in a triangle = 180°.

In △BCD,

∠BCD + ∠CBD + ∠CDB = 180°
⇒ 120° + ∠CBD + ∠CBD = 180°
⇒ 120° + 2∠CBD = 180°
⇒ 2∠CBD = 180° - 120°
⇒ ∠CBD = $\frac{60°}{2}$ = 30°.

As ∠CBD = ∠BDC,

∴ ∠BDC = 30°.

Hence, the value of ∠BDC = 30°.

(ii) From figure,

∠BDC = ∠BEC (∵ angles in same segment are equal.)

∴ ∠BEC = 30°

Hence, the value of ∠BEC = 30°.

(iii) Given AB = CB

∴ ∠BEC = ∠AEB (∵ equal chords subtend equal angles.)

∴ ∠AEB = 30°

Hence, the value of ∠AEB = 30°.

(iv) Arc AB subtends ∠AOB at the centre and ∠AEB at the remaining part of the circle.

⇒ ∠AOB = 2∠AEB (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ ∠AOB = 2 × 30°
⇒ ∠AOB = 60°.

Hence, the value of ∠AOB = 60°.

(i) Arc AX subtends ∠AOX at the centre and ∠APX at the remaining part of the circle.

⇒ ∠AOX = 2∠APX (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ ∠AOX = 2 × 30°
⇒ ∠AOX = 60°.

Hence, the value of ∠AOX = 60°.

(ii) From figure,

∠XPY = 90° (∵ angle in semicircle = 90°.)

∴ ∠APY = ∠XPY - ∠APX = 90° - 30° = 60°.

Hence, the value of ∠APY = 60°.

(iii) From figure,

∠APB = 90° (∵ angle in semicircle = 90°.)

∴ ∠BPY = ∠APB - ∠APY = 90° - 60° = 30°.

Hence, the value of ∠BPY = 30°.

(iv) In △AOX,

OA = OX (Radius of the same circle.)

∴ ∠OAX = ∠OXA

Since sum of angles of a triangle = 180°

∴ ∠AOX + ∠OAX + ∠OXA = 180°
⇒ 60° + ∠OAX + ∠OAX = 180°
⇒ 2∠OAX = 180° - 60°
⇒ 2∠OAX = 120°
⇒ ∠OAX = 60°.

Hence, the value of ∠OAX = 60°.

(i) ∠CDB = ∠CBP (∵ angles in alternate segments are equal.)

∴ ∠CDB = 25° ....(i)

Similarly, ∠CDA = ∠CAP = 40° (∵ angles in alternate segments are equal.)

∴ ∠ADB = ∠CDA + ∠CDB = 40° + 25° = 65°.

Hence, the value of ∠ADB = 65°.

(ii) Arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.

⇒ ∠AOB = 2∠ADB (As angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ ∠AOB = 2 × 65°
⇒ ∠AOB = 130°.

Hence, the value of ∠AOB = 130°.

(iii) ACBD is a cyclic quadrilateral.

∴ ∠ACB + ∠ADB = 180° (∵ sum of opposite angles = 180°.)

⇒ ∠ACB + 65° = 180°
⇒ ∠ACB = 180° - 65° = 115°.

Hence, the value of ∠ACB = 115°.

(iv) From figure,

⇒ ∠AOB + ∠APB = 180°
⇒ 130° + ∠APB = 180°
⇒ ∠APB = 180° - 130° = 50°.

Hence, the value of ∠APB = 50°.

(a) Join AE.

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

∴ ∠AOB = 2∠AEB

⇒ 80° = 2∠AEB

⇒ ∠AEB = $\frac{80°}{2}$ = 40°.

We know that,

Angle in a semi-circle is a right angle.

⇒ ∠AEC = 90°

From figure,

⇒ ∠BEC = ∠AEC - ∠AEB = 90° - 40° = 50°.

Hence, ∠BEC = 50°.

(b) From figure,

⇒ ∠ECD = ∠CEB = 50° (Alternate angles are equal)

We know that,

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

⇒ ∠AOB = 2∠BCA

⇒ 80° = 2∠BCA

⇒ ∠BCA = $\frac{80}{2}$ = 40°.

From figure,

⇒ ∠BCD = ∠BCA + ∠ACE + ∠ECD = 40° + 20° + 50° = 110°.

Hence, ∠BCD = 110°.

(c) As sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° - ∠BCD = 180° - 110° = 70°.

From figure,

⇒ ∠BED = ∠BEC + ∠CED

⇒ 70° = 50° + ∠CED

⇒ ∠CED = 70° - 50° = 20°.

Hence, ∠CED = 20°.

(a) In △ APB and △ CPD,

⇒ ∠APB = ∠CPD (Vertically opposite angles are equal)

⇒ ∠BAP = ∠DCP (Angles in same segment are equal)

∴ △ APB ~ △ CPD (By A.A. axiom)

Hence, proved that △ APB ~ △ CPD.

(b) We know that,

Corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{CD}{AB} = \frac{PD}{PB} \\[1em] \Rightarrow \frac{CD}{9} = \frac{2}{3} \\[1em] \Rightarrow CD = 9 \times \frac{2}{3} \\[1em] \Rightarrow CD = 6\text{ cm}.\end{aligned}$$

Hence, CD = 6 cm.

(c) We know that,

Ratio of area of similar triangles is equal to the ratio of square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APB}}{\text{Area of △CPD}} = \frac{PB^2}{PD^2} \\[1em] = \frac{3^2}{2^2} \\[1em] = \frac{9}{4} \\[1em] = 9 : 4.\end{aligned}$$

Hence, area △ APB : area △ CPD = 9 : 4.

(a) In △ PTB and △ PAT,

⇒ ∠PTA = ∠PBT (Alternate segment theorem)

⇒ ∠TPA = ∠BPT (Common angle)

∴ △ PTB ~ △ PAT (By A.A. axiom)

Hence, proved that △ PTB ~ △ PAT.

(b) We know that,

If a chord and a tangent intersect externally, then the product of the lengths of segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

⇒ PA × PB = PT²

⇒ PA × (PA + AB) = PT²

⇒ 16 × (16 + AB) = 20²

⇒ 16 × (16 + AB) = 400

⇒ 16 + AB = 25

⇒ AB = 25 - 16 = 9 cm.

Hence, AB = 9 cm.

Let the regular pentagon be ABCDE.

Since it's regular, and all interior angles are equal to 108°.

∴ ∠ABC = ∠BCD = ∠CDE = ∠DEA = ∠EAB

Interior angle of a regular polygon is given by :

$\frac{n - 2}{n} \times 180°$

So, each interior angle of a regular pentagon = $\frac{5 - 2}{5} \times 180° = \frac{3}{5} \times 180°$ - 108°.

In triangle AED,

AE = ED [Sides of a regular pentagon are equal]

∴ ∠EAD = ∠EDA [Angles opposite to equal sides are equal] ......(1)

⇒ ∠AED + ∠EAD + ∠EDA = 180°

⇒ 108° + ∠EAD + ∠EAD = 180°

⇒ 2∠EAD = 180° - 108°

⇒ 2∠EAD = 72°

⇒ ∠EAD = $\frac{72°}{2}$

⇒ ∠EAD = 36°

∴ ∠EDA = 36°.

From figure,

⇒ ∠BAD = ∠BAE - ∠EAD = 108° - 36° = 72°.

In quadrilateral ABCD,

∠BAD + ∠BCD = 72° + 108° = 180°.

Since, sum of opposite angles = 180°,

which is possible when the quadrilateral is cyclic quadrilateral.

Hence, proved that any four vertices of a regular pentagon are concylic.