From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.
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Given:

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut.

To do:

We have to find the area of the remaining portion of the square.

Solution:

Length of the side of the square $ABCD = 4\ cm$

Area of the square $ABCD = 4^2$

$= 16\ cm^2$

Radius of the quadrant at each corner $= 1\ cm$

Area of the quadrant at each corner $=\frac{\pi r^{2} \theta}{360^{\circ}}$

$=\frac{22}{7} \times \frac{1 \times 1 \times 90^{\circ}}{360^{\circ}}$

$=\frac{22}{28} \mathrm{~cm}^{2}$

Area of the 4 sectors at corners $=\frac{4 \times 22}{28}$

$=\frac{22}{7} \mathrm{~cm}^{2}$

Area of the circle at the centre of the square having radius $1 \mathrm{~cm}$

$=\pi r^{2}$

$=\frac{22}{7}(1)^{2}$

$=\frac{22}{7} \mathrm{~cm}^{2}$

Total area to be cut out from the square $=$ Area of the 4 sectors $+$ Area of the circle at the centre

$=\frac{22}{7}+\frac{22}{7}$

$=\frac{44}{7} \mathrm{~cm}^{2}$

Therefore,

Area of the remaining portion $=$ Area of the square $-$ Area to be cut from square

$= 16 - \frac{44}{7}$

$=\frac{112-44}{7}$

$= \frac{68}{7}\ cm^2$

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Updated on: 10-Oct-2022

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