Question

PQRS is a square lawn with side PQ=42 meters. Two circular flower beds are there on the sides PS and QR with Centre at O the intersection of its diagonal. Find the total area of the flower beds.​

Answer 1

In square, each angle is right angle.

In triangle SPQ

QS2=PS2+PQ2

=42×42+42×42

=2(42×42)

QS=42root2

Diagonals bisect each other perpendicularly

$radius = \frac{42 \sqrt{2} }{2}$

$= 21 \sqrt{2}$

Area of shaded region=2(area of segment)

$=2( \frac{e}{360} \times \pi {r}^{2} - \frac{1}{2} {r}^{2}sine)$

$= 2( \frac{90}{360} \times \frac{22}{7} \times 21 \sqrt{2} \times 21 \sqrt{2} - \frac{1}{2} \times 21 \sqrt{2} \times 21 \sqrt{2} \times 1$)

=2(693-441)

=2×252

=504sq.meters

$e \: stands \: for \: theta$

All the best