CM\] Radius\[=\frac{10}{2}=5\,\,cm\] Now, Area \[=\pi {{r}^{2}}=\pi \times {{\LEFT( 5 \right)}^{2}}=25\pi \] Area of shaded portion\[=\left( 25\pi -8\times 6 \right)\] \[=\left( 25\times \frac{22}{7}-48 \right)=\frac{550-336}{7}=\frac{214}{7}=30.5\,\,c{{m}^{2}}\]

"> CM\] Radius\[=\frac{10}{2}=5\,\,cm\] Now, Area \[=\pi {{r}^{2}}=\pi \times {{\LEFT( 5 \right)}^{2}}=25\pi \] Area of shaded portion\[=\left( 25\pi -8\times 6 \right)\] \[=\left( 25\times \frac{22}{7}-48 \right)=\frac{550-336}{7}=\frac{214}{7}=30.5\,\,c{{m}^{2}}\]

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In the adjoining figure PQRS is a rectangle \[8\,\,cm\times 6\,\,cm,\]inscribed in the circle. The area of the shaded portion will be.

10th Class Mathematics in 10th Class 11 months ago

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\[PR=\sqrt{64+36}=\sqrt{100}=10\,\,CM\] Radius\[=\frac{10}{2}=5\,\,cm\] Now, Area \[=\pi {{r}^{2}}=\pi \times {{\LEFT( 5 \right)}^{2}}=25\pi \] Area of shaded portion\[=\left( 25\pi -8\times 6 \right)\] \[=\left( 25\times \frac{22}{7}-48 \right)=\frac{550-336}{7}=\frac{214}{7}=30.5\,\,c{{m}^{2}}\]

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