Question

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 O 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

Solution:-

Area of Lawn + Area of flower bed = Area of lawn (square of 56m side) + Area of flower bed AB and CD

Area of lawn = Area of square of side 56cm.

$= \dashrightarrow \sf{ {(side)}^{2} = {(56)}^{2} = 3136 {m}^{2} }$

∴Area of lawn is 3316m²

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Now, Finding the area of flower bed AB & AC.

Area of flower bed AB = Area of segment AB = Area of sector OAB - Area of ΔOAB.

Let's find the length of OA & OB.

We know that the diagonals of a square bisect each other at right angles,

∴∠AOB = 90° & OA = OB

Let OA = OB = x

Now, let's consider ΔOAB

➻By Pythagoras theorem,

$\dashrightarrow \sf{ {(hypotenuse)}^{2} = {(height)}^{2} + {(base)}^{2} }$

$\dashrightarrow \sf{AB^2 = OA^2 + OB^2}$

➻We know that AB = 56m

$\dashrightarrow \sf{ {(56)}^{2} = {(x)}^{2} + {(x)}^{2} }$

$\dashrightarrow \sf{ {(2x)}^{2} = {(56)}^{2} }$

$\dashrightarrow \sf{ {(x)}^{2} = 56 \times 28 \: \: \: \: \qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \red{ 1}}$

➻We know the area of sector,

$\dashrightarrow \sf{ \frac{ \theta}{360} \times \pi {x}^{2} }$

Given:-

• θ = 90°
• r = x

➻Area of sector OAB,

$\dashrightarrow \sf{ \frac{ 90}{360} \times \pi{x}^{2} }$

$\dashrightarrow \sf{ \frac{90}{360}\pi = 56 \times 28 }$

$\dashrightarrow \sf{ \frac{1}{4} \times \frac{22}{7} \times 56 \times 28}$

$\dashrightarrow \sf{22 \times 56} = {1232m}^{2}$

∴Area of sector OAB = 1232m²

Take:-

• x² = 56 × 28 (from 1)
• π = 22/7

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Now let's find the area of ΔOAB.

➻The diagonals of square divide the square into two congruent triangles

∴ΔAOB ≌ ΔBOC ≌ ΔAOD ≌ ΔAOD

➻Thus, all the triangles are similar and We know that congruent triangles have the same area.

∴ar(ΔAOB) = ar(ΔBOC) = ar(ΔAOD)

Which is 1/4th of the area of square ABCD.

➺ 4ar(AOB) = ar(ABCD)

➺ ar(AOB) = 1/4ar (ABCD)

➺ ar(AOB) = 1/4 × 3136 = 784m²

∴Area of ΔAOB = 784m²

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Area of flower bed = Area of sector OAB - Area of ΔAOB

➙ 1232 - 784

➙ 448m²

As flower bed CD is same as AB, We can say that the area of flower bed CD is similar to area of AB = 448m²

∴Area of lawn = Area of flower bed

➼ 3316 + (448 + 448)

➼ 3316 + 896

➼ 4032m²

∴The area of lawn is 4032m²

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