Question

In figure, PQRS is square lawn with side PQ = 42 metre. Two circular flower beds are
there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the
total area of the two flower beds (shaded parts).
​

Answer 1

Answer:

Area of square lawn = (side)²=(42x42)m²

Let OP = OS = xm

So, x² + x² = (42)²

⇒ 2x² = 42 × 42

⇒ x² = 21 × 42

Now,

area of sector POS =

\frac{90}{360} \times \pi { \times x}^{2} = \frac{1}{4} \times (\pi \times {x}^{2} ) .. equation..(1)

360

90

×π×x

2

=

4

1

×(π×x

2

)..equation..(1) \frac{1}{4} \times \frac{22}{7} \times 21 \times (42)cm {}^{2} .... (ii)

4

1

×

7

22

×21×(42)cm

2

....(ii)

Also,

Area of Δ POS =

\frac{1}{4}

4

1

x area of square of lawn PQRS

\frac{1}{4} \times (42 \times 42) {}^{2} m {}^{2}

4

1

×(42×42)

2

m

2

... (∠POQ = 90°) (iii)

So,

Area of flower bed PSP = Area of sector POS - Area of Δ POS

( \frac{1}{4} \times \frac{22}{7} \times 21 \times 42) - ( \frac{1}{4} \times 42 \times 42)(

4

1

×

7

22

×21×42)−(

4

1

×42×42)

[from equation ii and iii]

\frac{1}{4} \times 21 \times 42 \times ( \frac{22}{7} - 2)

4

1

×21×42×(

7

22

−2)

\frac{1}{4} \times 21 \times 42 \times ( \frac{8}{7} )

4

1

×21×42×(

7

8

)

Therefore, area of the two flower beds =

2 \times \frac{1}{4} \times 21 \times 42 \times ( \frac{8}{7} )2×

4

1

×21×42×(

7

8

)

= 504 m2

Hence , the total area of the flower beds 504 m².

Step-by-step explanation:

hope it helps you!