Use ruler and compasses for the following question taking a scale of 10 m = 1 cm.
A park in the city is bounded by straight fences AB, BC, CD and DA.
Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD.
Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.
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We know that,
The locus of a point equidistant from two intersecting lines is pair of bisectors of the angles between the two lines.
The locus of a point which is equidistant from two given points is actually the perpendicular bisector of the segment that joins the two points.
Given,
Scale : 10 m = 1 cm
BC = 63 m = $\frac{63}{10}$ = 6.3 cm.
AB = 50 m = $\frac{50}{10}$ = 5 cm.
Steps of construction :
Draw a line BC = 6.3 cm.
Draw ∠ABC = 75° such that AB = 5 cm.
Draw BE, angle bisector of ∠ABC.
Construct ∠BAF = 90°, intersecting BE at D.
Join ABCD.
Construct XY, the perpendicular bisector of AB, intersecting BE at P.
