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Q.

In the given figure, ABCD is a square of 7 cm and A, B, C and D are the centres of equal circles which touch externally in pairs. The area of the coloured region is

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a

115.5cm2

b

164.5cm2

c

66.5cm2

d

266.5cm2

answer is B.

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Detailed Solution

ABCD is a square with a side length of 7cm.
A, B, C, and D are the centres of four equal circles, each of which touches two other circles on the outside.
To determine: the total area of the figure.
The distances between the respective centres are denoted by AB,BC,CD, and AD.
All angles are right angles because ABCD is a square, and AB=BC=CD=AD=7cm.
The distance between the centres of two circles that are touching on the outside is now the sum of their radii.
⇒Radius of circle =

\frac{7}{2}

cm
⇒Area of a circle = πr2,
⇒Area of a circle =

π \times {(\frac{7}{2})}^{2}

⇒Area of a circle =

(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) c m^{2}

⇒Area of a circle =

(11 \times \frac{7}{2}) c m^{2}

⇒Area of a circle =

\frac{77}{2} c m^{2}

Since, there are 3 circles whose area we require to find.
⇒Area of all the circles we require =

(3 \times \frac{77}{2}) c m^{2}

⇒Area of all the circles we require =

\frac{231}{2} c m^{2} = 115.5 cm 2 \dots \dots \dots \dots \dots . (1) \Rightarrow Area = {(Side)}^{2}

\Rightarrow Area = {(7)}^{2} = 49 c m^{2}

………………..(2)
Adding the equations (1) and (2), we will get:-
⇒Area of the shaded region = (115.5 + 49) square cm = 164.5

c m^{2}

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