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

Question

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

Prepp Team · Accepted Answer

To find the area of the circle, we need to first understand the relationship between the right triangle $ \triangle MON $ and the circle.

Given:

$ O $ is the center of the circle.
$ M $ and $ N $ lie on the circle.
The area of $ \triangle MON $ is $ 50 \, \text{cm}^2 $.

Since $ \triangle MON $ is a right triangle with hypotenuse $ MN $ lying on the circle, $ O $ must be the midpoint of the hypotenuse $ MN $. According to the properties of the circle, this implies that the hypotenuse $ MN $ is the diameter of the circle.

The formula for the area of a right triangle is:

Area = \frac{1}{2} \times \text{base} \times \text{height}

In $ \triangle MON $, the base $ MO $ and height $ ON $ are both radii of the circle. Let's denote the radius as $ r $.

Thus, we have:

Area = \frac{1}{2} \times r \times r = 50

\frac{1}{2} r^2 = 50

r^2 = 100

Now, the area of the circle is given by the formula:

\text{Area of Circle} = \pi r^2

Substituting $ r^2 = 100 $:

\text{Area of Circle} = \pi \times 100 = 100\pi \, \text{cm}^2

Therefore, the area of the circle is $ 100\pi \, \text{cm}^2 $, which matches the option $ 100\pi $.