| In the given figure, OACBO represents a quadrant of a circle of radius 4.5 cm with centre O. Then, the area of shaded portion is [take \[\pi =22/7\]] |
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A) \[15.91c{{m}^{2}}\]
B) \[9.16\text{ }c{{m}^{2}}\]
C) \[6.75\text{ }c{{m}^{2}}\]
D) \[22.66\text{ }c{{m}^{2}}\]
Correct Answer: B
Solution :
Area of quadrant \[OACBO=\frac{1}{4}\pi {{r}^{2}}\] \[=\frac{1}{4}\times \frac{22}{7}\times {{\left( 4.5 \right)}^{2}}=15.91\,c{{m}^{2}}\] and area of \[\Delta BOD=\frac{1}{2}\times BO\times OD\] \[=\frac{1}{2}\times 4.5\times 3=6.75c{{m}^{2}}\] Hence, area of shaded portion = Area of quadrant OACBO - Area of \[\Delta BOD\] \[=15.91-6.75\text{ }=9.16\text{ }c{{m}^{2}}\]
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