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Two circular flower beds have been shown on two sides of a square lawn ABCD of side 56m.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 flower beds.

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4072 m^{2}

4052 m^{2}

4062 m^{2}

4032 m^{2}

answer is D.

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Detailed Solution

Given,
Side of square

ABCD = 56 m

We know the diagonals of a square are equal in lengths. So,
AC = BD
Question Image

Again it is known that
Square Diagonal

(AC)

AC = \sqrt{2} x

× side square

AC = \sqrt{2} \times 56

AC = 56 \sqrt{2} m

Also, we have,

OA = OB = r

(radius of the sector)

= \frac{1}{2} AC = \frac{1}{2} (56 \sqrt{2})

= 28 \sqrt{2} m

Thus, sector

OAB area

= [\frac{90^{\circ}}{360^{\circ}}] π r^{2}

= (\frac{1}{4}) π r^{2}

= \frac{1}{4} \times \frac{22}{7} \times {(\frac{28}{\sqrt{2}})}^{2} m^{2}

= [\frac{1}{4} \times \frac{22}{7} \times 28 \times 28 \times 2] m^{2}

= 1232 m^{2}

Flower bed AB area= area of flower bed CD,

=

Area of sector OAB − area of

△ OAB

\Rightarrow 1232 - \frac{1}{2} \times OB \times OA

\Rightarrow 1232 - \frac{1}{2} \times 28 \sqrt{2} \times 28 \sqrt{2}

\Rightarrow 1232 - 784

= 448 m^{2}

So, overall area = Square ABCD area+ (Flower bed AB area + flower bed CD area)

= (56 \times 56) + 448 + 448

= 4032 m^{2}

So, sum of the areas of the lawn and flower beds 4032

m^{2}

.
Correct option is 4.

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