question_answer

In the given figure, ABC is a right angled triangle, right angled at B. BC = 21cm and AB = 28 cm. Width AC as diameter of a semi-circle and width BC as radius a quarter circle are drawn. What is the area of the shaded portion?

A)  \[425\,\,\text{c}{{\text{m}}^{2}}\]

B)  \[425.75\,\,\text{c}{{\text{m}}^{2}}\]

C)  \[428\,\,\text{c}{{\text{m}}^{2}}\]

D)  \[\text{428}\text{.75}\,\,\text{c}{{\text{m}}^{\text{2}}}\]

Correct Answer: D

Solution :

In \[\Delta ABC,\]

\[A{{C}^{2}}=\sqrt{{{28}^{2}}+{{21}^{2}}}=\sqrt{784+441}=\sqrt{1225}\]

\[\Rightarrow \]   \[AC=35\,\,\text{cm}\]

Area of shaded portion = Area of semi-circle ACE

+ Area of A ABC -Area of quadrant circle BCD

\[=\frac{\pi {{r}^{2}}}{2}+\frac{1}{2}\times BC\times BA-\frac{\pi }{4}\times r_{1}^{2}\]

\[=\frac{22}{7}\times \frac{1}{2}\times \frac{35}{2}\times \frac{35}{2}+\frac{1}{2}\times 21\times 28-\frac{22}{7\times 4}\times 21\times 21\]\[=\frac{5\times 11\times 35}{4}+\frac{1}{2}[21\times 28-33\times 21]\]\[=\frac{1925}{4}+\frac{1}{2}(-105)=48.125-5250=428.75\,\,\text{c}{{\text{m}}^{2}}\]