Question

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 following figure). Find the area of the remaining sheet. (Take $\pi =\frac{22}{7}$)

Answer 1

Since, we know that the area of circle $=\pi \times r^2$
Area of bigger circle $=\frac{22}{7}\times 14\times 14$
$=44\times 14$
$=616c{m}^2$
Thus, the area of bigger circle is $616$ $c{m}^2$

Similar way, the area of $2$ smaller circles $=2\pi r^2$
$=2\times \frac{22}{7}\times 3.5\times 3.5$
$=44\times 0.5\times 3.5$
$=77$ $c{m}^2$
Thus, the area of $2$ smaller circle is $77c{m}^2$

Area of rectangle $=Length\times Breadth$
$=3\times 1=3$ $c{m}^2$

Thus, the area of rectangle is $3c{m}^2$

Now, remaining area of sheet $=$ Area of bigger circle $-$ Area of both smaller circle $-$ Area of rectangle

$=616-77-3=536c{m}^2$

Hence, area of remaining sheet is $536c{m}^2$

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