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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
(i) quadrant OACB.
(ii) shaded region.
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 $\frac{49}{8} \mathrm{~cm}^{2}$.