Question

PQRS is a square lawn with side PQ = 42m. Two circular flower beds are there on the sides PQ andQRwith cente at O , the intersection of its diagonals. Find the total area of the flower beds (shaded parts) not able to upload the pic.. PLSANYONE ANSWER IT...

Answer 1

I found the figure. 

Given:
2 shapes : square and circle
The circle is cut in half to form a semi-circle. The square is placed between the two semi-circle. Two semi-circles are the shaded part. 

I will find the area of the shaded part which forms 1 circle.

Area of a circle = π r²

Side of the square lawn serves as the diameter of the circle. Radius is half of the diameter. 

r = d/2 ⇒ 42meters ÷ 2 = 21 meters

Area of circle = 3.14 * (21m)²
A = 3.14 * 441m²
A = 1,384.74 m²

Area of the shaded parts is 1,384.74 square meters.

Answer 2

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