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gatecse
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in Quantitative Aptitude
Feb 22, 2021
recategorized
Apr 11, 2021
by Lakshman Patel RJIT

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4 votes

In the above figure, $\textsf{O}$ is the center of the circle and, $\textsf{M}$ and $\textsf{N}$ lie on the circle. The area of the right triangle $\textsf{MON}$ is $50\;\text{cm}^{2}$. What is the area of the circle in $\text{cm}^{2}?$

- $2\pi$
- $50\pi$
- $75\pi$
- $100\pi$

Migrated from GO Mechanical 1 year ago by gatecse

3 votes

Best answer

Let the radius of the circle be $r\;\text{cm}.$

The area of triangle $ = \dfrac{1}{2} \times \text{base} \times \text{height}$

The area of the right triangle $\text{MON} = \dfrac{1}{2} \times r \times r = 50$

$\implies r^{2} = 100$

$\implies r = 10\;\text{cm}$

Now, the area of circle $ = \pi r^{2} = \pi (10)^{2} = 100\pi\;\text{cm}^{2}.$

So, the correct answer is $(D).$

The area of triangle $ = \dfrac{1}{2} \times \text{base} \times \text{height}$

The area of the right triangle $\text{MON} = \dfrac{1}{2} \times r \times r = 50$

$\implies r^{2} = 100$

$\implies r = 10\;\text{cm}$

Now, the area of circle $ = \pi r^{2} = \pi (10)^{2} = 100\pi\;\text{cm}^{2}.$

So, the correct answer is $(D).$