Question

it is proposed to add to a square lawn measuring 50m on a side two circular ends the centre of each circle being the point of intersection of the diagonals of the square find the area of the whole lawn.​

Answer 1

$\blue\bigstar$QUESTION:-

it is proposed to add to a square lawn measuring 50m on a side two circular ends the centre of each circle being the point of intersection of the diagonals of the square find the area of the whole lawn.

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$\pink\bigstar$SOLUTION:-

Let ABCD be a square with side 50m as shown in the above figure.

Let its diagonal intersect at O.

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Then OA=OB=OC=OD [all sides of square are equal]

$\sf And\: \angle AOD=\angle BOC=90^{0}$

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With O as a centre and radius equal to 1/2 AC, arcs AED and BFC have been drawn .

$\sf Diagonal \:AC=(\sqrt{2}\times 50)m$

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$\sf \longrightarrow radius \: of \: each \: arc = \dfrac{1}{2} (diagonal) = (25 \sqrt{2} )m$

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•Now , Area of whole lawn

=(area of the square ABCD)+2(area of the segment with r=$\sf 25\sqrt{2},\theta=90^{0}$)

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$\sf \implies \bigg\{(50 \times 50) + 2 \bigg( \dfrac{ \pi{ r}^{2} \theta}{360} - \dfrac{1}{2} {r}^{2}sin \theta \bigg) \bigg \}$

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$\implies \sf \bigg \{2500 + 2 \times \bigg(3.14 \times ( 25\sqrt{2} )^{2} \times \dfrac{90}{360} - \dfrac{1}{2} \times (25 \sqrt{2} )^{2} sin90^{0} \bigg) \bigg \}$

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$\sf \implies \{2500 + 2 \times (981.25 - 625) \} {m}^{2}$

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$\boxed{ \sf \implies 3212.5 {m}^{2} }$

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Hence ,the required area for whole lawn is 3212.5 m²

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Answer 2

Step-by-step explanation:

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