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í¸Example 4: In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularflower bed is the point of intersect ion O of thediagonals of the square lawn, find the sum of theC.areas ofthe lawn and the flower beds.Fig. 12.15 |
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Answer» Given:Side of a square ABCD= 56 mAC = 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. Like my answer if you find it useful! |
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