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Answers
Answer:
4032 m^2
Step-by-step explanation:
Total area = Area of sector OAB + Area of sector OCD + Area of ∆ OAD + Area of ∆ OBC
= 90°/360° × 22/7 ×(28√2)^2 + 90°/360° × 22/7 × (28√2)^2 + 1/4 × 56 × 56 +1/4 × 56 × 56
= 1/4 × 28 × 56 (22/7+22/7 +2+2) m^2
=7×56/7(22+22+14+14) m^2
=4032 m^2
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 beds are 4032 m².