$B C$ is a diameter of the circle with centre $O$ and radius 5 , as shown. If $A$ lies on the circle and $A O$ is perpendicular to $B C$, the area of triangle $A B C$ is
(A) 6.25
(D) 37.5
(B) 12.5
(E) 50
(C) 25

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-093.jpg?height=309&width=336&top_left_y=1531&top_left_x=1399)
$B C$ is a diameter of the circle with centre $O$ and radius 5 , as shown. If $A$ lies on the circle and $A O$ is perpendicular to $B C$, the area of triangle $A B C$ is
(A) 6.25
(B) 12.5
(C) 25
(D) 37.5
(E) 50

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-404.jpg?height=323&width=368&top_left_y=2294&top_left_x=1388)

## Solution

If $O$ is the centre of the circle with radius 5 this implies that $O B=A C=O C=5$. Thus, we are trying to find the sum of the areas of two identical isosceles right angled triangles with a side length of 5 . The required area is $2\left[\frac{1}{2}(5)(5)\right]$ or 25 .

Answer: (C)