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.

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 = $\dfrac{63}{10}$ = 6.3 cm.

AB = 50 m = $\dfrac{50}{10}$ = 5 cm.

Steps of construction :

1. Draw a line BC = 6.3 cm.

2. Draw ∠ABC = 75° such that AB = 5 cm.

3. Draw BE, angle bisector of ∠ABC.

4. Construct ∠BAF = 90°, intersecting BE at D.

5. Join ABCD.

6. Construct XY, the perpendicular bisector of AB, intersecting BE at P.