Question

In the figure given, two circular flower beds have been shown on the 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.

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Answer 1

Answer:

solution

$\sf{Using \: Pythagoras \: theorem \: in \: ∆ ABD \: we \: obtain}$

BD² = AB²+AD²

☞BD²= 56²+56²

☞BD²= 2× (56)²

☞BD= 56√2 metre

☞BD= AC = 56√2 m

∴ OA= OB= 1/2 ×AC= 28√2 m

so, the radius of the circle having centre at the point of intersection of diagonals is 28√2 m.

❄️Let A be the centre of one of the circular ends. Then,

❄️A = area of the segment of a angle 90° in a circle of radius 28√2 metre

$\sf{ : \implies \: A = ( \frac{22}{7}\times\frac{9}{360} - sin45 \: cos45)} \times ( {28 \sqrt{2} })^{2} \\ \\ \sf{ : \implies \: A = (\frac{11}{4} - \frac{1}{2} }) \times 28 \times 28 \times 2 \\ \\ \sf{ : \implies \: A = 28 \times 28 \times 2 \times \frac{4}{14} = \color{green} 448 {m}^{2} }$

∴ area of 2 flower bed 2A =2 × 448 = 896 m²

☞Area of square lawn = 56× 56 = 3136 m²

Hence, the total area = (3136 +896) = 4032 m²

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