Question

(a) In the figure (i) given below, two circular flower beds have been shown on the two
sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed
is the point of intersection O of the diagonals of the square lawn, find the sum of
the areas of the lawn and the flower beds.​

Answer 1

Answer:

4032 m²

Step-by-step explanation:

Side of a square ABCD= 56 m (given)

AC = BD (diagonals of a square are equal in length)

Diagonal of a square (AC) =√2×side of a square.

Diagonal of a square (AC) =√2 × 56 = 56√2 m.

OA= OB = 1/2AC = ½(56√2)= 28√2 m.

[Diagonals of a square bisect each other]

Let OA = OB = r m (radius of sector)

Area of sector OAB = (90°/360°) πr²

Area of sector OAB =(1/4)πr²

= (1/4)(22/7)(28√2)² m²

= (1/4)(22/7)(28×28 ×2) m²

= 22 × 4 × 7 ×2= 22× 56= 1232 m²

Area of sector OAB = 1232 m²

Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²

Area of flower bed AB =area of sector OAB - area of ∆OAB

= 1232 - 784 = 448 m²

Area of flower bed AB = 448 m²

Similarly, area of the other flower bed CD = 448 m²

Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD

=(56× 56) + 448 +448

= 3136 + 896= 4032 m²