, two circular flower beds
have been shown on two sides of a square lawn
ABCD of side 56 m. If the centre of each circular
flower bed is the point of intersection 0 of the
| diagonals of the square lawn, find the sum of the
areas of the lawn and the flower beds.
Fig. 12.15
Answers
Answer:
4032 m2
Explanation:
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
Also, Area of the other flower bed CD = 448 m2
∴ Total area = Area of square ABCD + area of flower bed AB + area of flower bed CD
=(56× 56) + 448 +448
= 3136 + 896= 4032 m2
∴ Sum of the Areas of the lawns and the flower beds = 4032 m2.
Answer:Step-by-step explanation:
Side of square = 56cm
ΔAOB
AB²=OA²+OB²
56²=r² + r²
56×56=2r²
56×56÷2
⇒r²=56×28
Area of flower bed
⇒ Area of 2 sectors + area of 2 triangles
2×∅÷360 + 2×1/2×b×h
90÷180πr² + r² ( we get 90÷180 by cancellation)
1÷2 × 22÷7 ×56×28 + 56×28
88×28 + 56×28
⇒(88+56)×28
⇒144×28= 4032cm²