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Areas Related to Circles

Short Answer Type

141. Find the area of the shaded part in the given figure. Take π = 3.14.

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142. Find the area of the shaded region in the given fig.

Fig. 12.77.

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143. A sheet of paper is in the form of a rectangle ABCD in which AB = 35 cm and AD = 28 cm. A semicircular portion with BC as diameter is cut off. Find the area of the remaining portion.

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144. Find the area and the length of the boundary of the shaded region in the given figure, it being given that the diameter of larger semicircle is 56 cm.

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145. Find the areas of the shaded region in the given figure.

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146. In the given fig., ABCD is a square inscribed in a circle. Find the area of the shaded region. (π = 3.14)

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147. Semicircular lawns are attached to both edges along the breadth of a rectangular lawn measuring 50 m × 35 m. Find the total area enclosed.

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148. An athlete track 14 m wide consists of two straight sections 120 m long joining semicircular ends whose inner radius is 35 m. Calculate the area of the shaded region.

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149. Find the area of the shaded design, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (π = 3.14)

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150. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field to be ploughed at the rate of Rs. 0.50 per m². Find the cost of ploughing the field.

$open parentheses bold Take bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses$

Length of circular field = $fraction numerator Total space cost over denominator Rate end fraction$

$equals space 5280 over 24 equals 220 space straight m$
Let r be the radius of the field, then Circumference = 220 m

$rightwards double arrow space space space space space space 2 πr space equals space 220 rightwards double arrow space space space space space space space space straight r space equals space fraction numerator 220 over denominator 2 straight pi end fraction rightwards double arrow space space space space space space space space straight r space equals space fraction numerator 220 space straight x space 7 over denominator 2 space straight x space 22 end fraction rightwards double arrow space space space space space space space space straight r space equals space 35 space$
$rightwards double arrow$ Radius of the field is 35m.
Now area of circular field = $bold πr to the power of bold 2$

$equals open parentheses 22 over 7 straight x space 35 space straight x space 35 close parentheses straight m squared equals space left parenthesis 22 space straight x space 5 space straight x space 35 right parenthesis space straight m squared equals space 3850 space straight m squared$
Hence, the cost of plaughing the field
= 3850 x 0.50
= 1925