Question

plz help with the question below. i will mark you as a brainlist plzzz

Answer 1

Answer:

Given:

Side of a square ABCD= 56 m

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 (ràdius 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

Hence, the sum of the areas of the lawns and the flower be 4032 m²