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# In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If $OD = 2\ cm$, find the area of the (i) quadrant OACB.(ii) shaded region."

Given:

OACB is a quadrant of a circle with centre O and radius 3.5 cm.

$OD = 2\ cm$

To do:

We have to find the area of the

Solution:

(i) Radius of the quadrant $OACB = 3.5\ cm$

Area of the quadrant  $\mathrm{OACB}=\frac{\pi r^{2} \theta}{360^{\circ}}$

$=\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90^{\circ}}{360^{\circ}}$

$=\frac{22 \times 35 \times 35 \times 90^{\circ}}{7 \times 360^{\circ} \times 100}$

$=\frac{77}{8} \mathrm{~cm}^{2}$

The area of the quadrant OACB is $\frac{77}{8} \mathrm{~cm}^{2}$.

(ii) $OD = 2\ cm$

$OB = 3.5\ cm$

Therefore,

Area of the triangle $\mathrm{OBD}=\frac{1}{2} \times \mathrm{OB} \times \mathrm{OD}$

$=\frac{1}{2} \times 3.5 \times 2$

$=3.5 \mathrm{~cm}^{2}$

Area of the shaded region $=$ Area of the quadrant $-$ Area of the triangle $\mathrm{OBD}$

$=\frac{77}{8}-\frac{35}{10}$

$=\frac{77}{8}-\frac{7}{2}$

$=\frac{77-28}{8}$

$=\frac{49}{8} \mathrm{~cm}^{2}$

The area of the shaded region is $\frac{49}{8} \mathrm{~cm}^{2}$.

Updated on: 10-Oct-2022

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