Two circular flower beds 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 O of the
diagonals of the square lawn, find the sum of the
areas of the lawn and the flower beds.
Answers
Answer:
Diagonal of squar (AC)=√2xside of a square .
Diagonal of a square (AC)=√2x56=56 √2m.
OA=OB=1/2AC=1/2 (56√2)=28√2 m.Hence the sum of areas of the lawn and the flower bed are 4032 m2
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