Constructions (Circles)

The incenter of a triangle is the intersection point of all the three interior angle bisectors of the triangle.

The circumcenter of a triangle is defined as the point where the perpendicular bisectors of the sides of triangle intersect.

Since,

In the given figure, P is intersection of an angle bisector and perpendicular bisector of a side.

Hence, Option 4 is the correct option.

The incenter of a triangle is the intersection point of all the three interior angle bisectors of the triangle.

Hence, Option 2 is the correct option.

We know that,

A regular hexagon comprises 6 equilateral triangles.

∴ BAC is an equilateral triangle.

∴ AB = BC = CA.

From figure,

⇒ AP < CA

⇒ AP < BC

⇒ Radius < Side of hexagon

Hence, Option 3 is the correct option.

We know that,

A regular hexagon comprises 6 equilateral triangles.

∴ BAC is an equilateral triangle.

∴ AB = BC = CA.

From figure,

AB and AC are radius of the circle.

In a regular hexagon, length of each side is equal.

Thus,

BC = $\frac{18}{6}$ = 3 cm.

∴ AB = AC = 3 cm

Thus, radius of circle = 3 cm.

Hence, Option 3 is the correct option.

The point of intersection of the bisectors of angles of a triangle is called the incenter of the triangle.

Thus, we can say that,

Bisectors of angles of a triangle are concurrent.

So, reason (R) is true.

Since, from figure, P is the incenter, so bisector of angle C will pass through it.

Thus, we can say that,

PC bisects angles C.

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given, in triangle ABC, ∠A = 35° and ∠C = 55°.

The sum of angles in a triangle is always 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 35° + ∠B + 55° = 180°

⇒ 90° + ∠B = 180°

⇒ ∠B = 180° - 90°

⇒ ∠B = 90°.

An angle inscribed in a semicircle is always a right angle (90°).

So, reason (R) is true.

Since ∠ABC is 90°, and AC is the diameter of the circle, the circle must pass through point B.

This is because the angle subtended by the diameter (AC) at any point on the circumference of the circle (in this case, point B) is always a right angle.

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Since PQ is the perpendicular bisector of AB, any point on PQ is equidistant from A and B.

Similarly, since PR is the perpendicular bisector of BC, any point on PR is equidistant from B and C.

Therefore, point P, which lies on both PQ and PR, is equidistant from A, B and C.

This means that P is the center of a circle passing through A, B and C.

The perpendicular bisectors for sides AB, BC, and CA always meet at a single point called the circumcenter of ∆ ABC that is equidistant from all three vertices.

Thus, we can say that,

Perpendicular bisectors of sides of a triangle are concurrent.

So, statement 2 is true.

Consequently, the perpendicular bisector of AC must also pass through P, as P is equidistant from A and C.

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given, pentagon ABCDE is inscribed in a circle with center O.

If a pentagon is regular then the sides subtends equal angles at the center.

If ABCDE is a regular pentagon, then ∠AOB = $\frac{360°}{5}$.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Steps of Construction :

1. Draw a circle with center O and radius 3 cm.

2. From O, take a point P such that OP = 5 cm.

3. Draw a perpendicular bisector of OP which intersects OP at M.

4. With center M and radius OM, draw a circle which intersects the circle with center O at A and B.

5. Join AP and BP. Measure AP and BP.

Hence, AP and BP are required tangents.

On measuring,

AP = BP = 4 cm.

Hence, length of each tangent = 4 cm.

Steps of Construction :

1. Draw a circle with center O, radius = 5 cm and BC as diameter.

2. Draw arcs making an angle of 135° (180° - 45°) at O such that ∠AOB = 135°.

3. At A and B, draw two rays making an angle of 90° at each point which meet other at point P.

Hence, AP and BP are required tangents making an angle of 45° with each other.

Steps of Construction :

1. Draw a line segment BC = 4.5 cm.

2. With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.

3. Join AC and AB.

4. Draw perpendicular bisectors of AC and BC intersecting each other at O.

5. With center O, and radius OA or OB or OC draw a circle which pass through A, B and C. Measure OC.

On measuring, OC = 2.6 cm.

Hence, above is the required circumcircle with radius 2.6 cm.

(i) Steps of construction :

1. Draw a line segment BC = 7 cm.

2. At B, draw a ray BX making an angle of 45° and cut off BE = AB - AC = 1 cm.

3. Join EC and draw the perpendicular bisector of EC intersecting BX at A.

4. Join AC.

Hence, ABC is the required triangle.

(ii) Steps of construction :

1. Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.

2. From O, draw perpendicular OL to BC.

3. With O as center and OL as radius draw a circle, touching sides of △ABC. Measure OL.

On measuring,

OL = 1.8 cm.

Hence, above is the required incircle of △ABC with radius = 1.8 cm.

Steps of construction :

1. Draw a line segment AB = 6 cm.

2. At A, draw a ray (AX) making an angle of 60° with AB.

3. With B as center and radius = 7.5 cm draw an arc which intersects AX ray at C.

4. Join BC. ABC is the required triangle.

5. Draw the perpendicular bisectors of AB and AC intersecting each other at O.

6. With center O and radius OA, OB or OC, draw a circle passing through A, B and C.

7. From O, draw OD ⊥ AB.

Proof :

In right △OAD and △OBD,

⇒ OA = OB (Radius of same circle)

⇒ OD = OD (Common)

⇒ ∠ODA = ∠ODB (Both = 90°)

∴ △OAD ≅ △OBD (By R.H.S. axiom)

By C.P.C.T.

⇒ AD = BD.

Hence, proved that AD = BD.

Steps of construction :

1. Draw a line segment BC = 4 cm.

2. At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.

3. From E, draw EY, perpendicular to CX.

4. From C, draw a ray making an angle 45° with CB, which intersects EY at A.

5. Join AB. Hence, ABC is the required triangle.

6. Draw perpendicular bisectors of sides AB and BC intersecting each other at O.

7. With center O, and radius OA or OB or OC draw a circle passing through A, B and C. Measure OB.

On measuring,

OB = 2 cm.

Hence, above is the required circumcircle of △ABC with radius = 2 cm.

Steps of construction :

1. Draw a line segment BC = 6 cm.

2. With B as center and radius 8 cm draw an arc.

3. With C as center and radius 5 cm draw another arc which intersects first arc at A.

4. Join AB and AC. ABC is the required triangle.

5. Draw angle bisectors of B and A intersecting each other at I. I is the incenter.

6. Through I, draw ID ⊥ AB.

7. From D as center cut off DP = DQ = $\frac{2}{2}$ = 1 cm.

8. With I as center and radius IP or IQ, draw a circle cutting off 2 cm chords from each side of triangle.

Each interior angle of a regular hexagon = $\Big(\frac{2n - 5}{n}\Big) \times 90°$

$$\begin{aligned}= \frac{2 \times 6 - 4}{6} \times 90° \\[1em] = 8 \times 15° \\[1em] = 120°.\end{aligned}$$

Steps of construction :

1. Draw a regular hexagon ABCDEF with each side equal to 5 cm and each interior angle = 120°.

2. Draw the perpendicular bisectors of AB and AF which intersect each other at O.

3. With O as center and OA as radius draw a circle which will pass through all the vertices of the regular hexagon.

Hence, above is the required circumcircle of regular hexagon.

Each interior angle of a regular hexagon = $\Big(\frac{2n - 5}{n}\Big) \times 90°$

$$\begin{aligned}= \frac{2 \times 6 - 4}{6} \times 90° \\[1em] = 8 \times 15° \\[1em] = 120°.\end{aligned}$$

Steps of construction :

1. Draw a regular hexagon ABCDEF with each side equal to 5.8 cm and each interior angle = 120°.

2. Draw the bisectors of interior angles at A and B which intersect each other at point I.

3. From point I , draw IP perpendicular to AB.

4. With I as center and IP as radius, draw a circle which will touch all the sides of the regular hexagon drawn.

Hence, above is the required incircle of regular hexagon.

(i) Steps of construction :

1. Draw a line segment BC = 5.5 cm

2. From B draw a ray BX such that ∠XBC = 120°.

3. From B draw an arc of 6 cm cutting BX at A.

4. Join AC. ABC is the required triangle.

5. Construct perpendicular bisectors of AB and BC, such that they intersect at O.

6. With O as center and OA as radius draw a circle passing through A, B and C.

Hence, above is the required circumcircle of triangle ABC.

(ii) Steps of construction :

1. Draw perpendicular bisector of BC, such that it intersects the circle at D.

2. Join CD and AD. ABCD is the cyclic quadrilateral.

Hence, above is the required cyclic quadrilateral.

Steps of construction :

1. Draw a line BC = 6 cm.

2. At B, draw a ray BX making an angle of 120° with BC. With B as center and radius 3.5 cm, cut off AB = 3.5 cm.

3. Join AC. ABC is the required triangle.

4. Draw perpendicular bisector of BC which cuts BC at point O. With O as center and radius = OB, draw a circle.

5. Draw angle bisector of ∠ABC which meets the circle at point P. Thus, point P is equidistant from AB and BC.

Measure ∠BCP.

Hence, ∠BCP = 30°.

(i) Steps of construction :

1. Draw a line segment AB = 5 cm.

2. With A as center cut an arc of 3 cm on AB to obtain C.

3. With A as center and radius = 2.5 cm draw an arc.

4. With C as center and radius = 2.5 cm draw an arc cutting the previous arc and mark the point O.

5. With O as center and radius = 2.5 cm draw a circle.

Hence, above is the required circle.

(ii) Steps of construction :

1. Join OB.

2. Draw the perpendicular bisector of OB, let it meet OB at point M.

3. With M as center and radius equal to OM, draw a circle to cut the previous circle at points P and Q.

4. Join PB and QB. Hence, PB and QB are required tangents. Measure PB and QB.

On measuring PB = QB = 3.2 cm.

Hence, length of each tangent = 3.2 cm.

Steps of construction :

1. Draw a line segment AB = 7 cm.

2. Construct AX such that ∠XAB = 60°.

3. With A as center and radius = 5 cm cut arc on AX and mark it as point C.

4. Join BC. ABC is the required triangle.

5. Draw AY and BZ, angle bisector of A and B.

6. Let AY and BZ meet at point O.

7. Draw OD ⊥ AB.

8. With O as center and OD as radius draw a circle.

(i) Hence, AY is the locus of points equidistant from AB and AC.

(ii) Hence, BZ is the locus of points equidistant from BA and BC.

Steps of construction :

1. Draw a line segment BC = 6.8 cm.

2. Mark D as the mid-point of BC.

3. With B as center and radius = 5 cm draw an arc.

4. With D as center and radius = 4.4 cm draw an arc cutting another arc at A.

5. Join AC. Hence, ABC is the required triangle.

6. Draw angle bisectors of B and C. Let it meet at point I.

7. Draw IM ⊥ BC.

8. With I as center and IM as radius draw a circle.

Hence, above is the required incircle of the triangle.

Steps of construction :

1. Draw two concentric circles of radius 4 cm and 6 cm with O as center.

2. Take a point P on the outer circle.

3. Join OP.

4. Draw perpendicular bisectors of OP, let it meet OP at point M.

5. With M as centre and OM as radius cut arcs at inner circle at point A and B.

6. Join PA and PB. Measure PA and PB.

On measuring,

PA = PB = 4.5 cm.

Hence, PA and PB are tangents and length of each tangent = 4.5 cm.

Steps of construction :

1. Draw BC = 7.2 cm.

2. Draw a ray BX, such that ∠XBC = 90°.

3. With B as center and radius = 6 cm cut an arc on BX and mark point A.

4. Join AC.

5. Take a point D on AC such that BD ⊥ AC.

6. Draw perpendicular bisectors of BD and BC which intersect at I.

7. With I as center and IC as radius draw a circle passing through B, D and C. Measure IC.

Hence, above is the required circumcircle with radius = 3.6 cm.

Steps of Construction

1. Draw a line segment AB = 6 cm.

2. At point B, construct an angle of 60° with BA.

3. On this ray, mark point C such that BC = 5 cm.

4. Join AC. Thus, △ABC is constructed.

5. Produce AB beyond B. On the produced line, mark point D such that AD = 9 cm.

6. Produce AC beyond C. On the produced line, mark point E such that AE = 8 cm.

7. Since the required circle touches AB produced, BC, and AC produced, it is the excircle opposite vertex A.

8. Draw the external angle bisectors of ∠B and ∠C. Let them intersect at point O.

9. From O, draw a perpendicular to BC, meeting BC at P.

10. With centre O and radius OP, draw a circle. The circle so obtained touches AB produced, BC, and AC produced.

Steps of construction :

1. Draw a line segment BC = 6 cm.

2. At point B, construct an angle of 120° with BC.

3. On the ray forming the angle, mark point A such that AB = 3.5 cm.

4. Join AC. Thus, △ABC is constructed.

5. To draw the circle with BC as diameter, construct the perpendicular bisector of BC.

6. Let the perpendicular bisector meet BC at O. Then O is the midpoint of BC.

7. With centre O and radius OB or OC, draw a circle. This is the required circle with BC as diameter.

8. Now, to find a point P on the circle which is equidistant from A and C, construct XY, the perpendicular bisector of AC.

9. Let XY, the perpendicular bisector of AC meet the circle at point P.

10. Then PA = PC, because every point on the perpendicular bisector of AC is equidistant from A and C.