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From a circular card sheet of radius $$14\ cm,$$ two circles of radius $$3.5\ cm$$ and a rectangle of length $$3\ cm$$ and breadth $$1\ cm$$ are removed (as shown in the adjoining figure). Find the area of the remaining sheet.$$\bigg($$Take $$\pi =\dfrac{22}{7}\bigg)$$

Solution
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Radius of circular sheet(R)$$=14$$cm and Radius of smaller circle(r)$$=3.5$$cm
Length of rectangle(l)$$=3$$cm and breadth of rectangle(b)$$=1$$cm

According to question,
Area of remaining sheet$$=$$Area of circular sheet$$-[$$Area of two smaller circle$$+$$Area of rectangle$$]$$

$$=\pi R^2-[2(\pi r^2)+(l\times b)]$$

$$=\dfrac{22}{7}\times 14\times 14-\left[\left(2\times \dfrac{22}{7}\times 3.5\times 3.5\right)-(3\times 1)\right]$$

$$=22\times 14\times 2-[44\times 0.5\times 3.5+3]$$

$$=616-80$$

$$=536 cm^2$$

Therefore the area of remaining sheet is $$536 cm^2$$.

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