# Finding the Area of a Circle with Inscribed Triangle

This video shows how to find the area of a semi-circle with an inscribed triangle. The area of a semi-circle is 1/2 πr^{2}, and the area of a triangle is 1/2 bh. To find the area of a semi-circle with an inscribed triangle, you need to subtract the area of the triangle from the area of the semi-circle.

## Area of a Circle with Inscribed Triangle

A right triangle with leg lengths 10 and 24 is inscribed in a semicircle, with its hypotenuse as the circle’s diameter. What is the area of the semicircle excluding the triangle? This word excluding means that we’re going to be subtracting.

We’re finding the area of the semicircle minus the area of the triangle. The area of a circle is Pi times radius squared, but a semicircle is only half of a circle, so the area of the semicircle is half Pi times radius squared. The area of a triangle is half base times height, and to find the area of the semicircle excluding the triangle, we’re going to take the area of the semicircle, and subtract the area of the triangle.

To find the area of the semicircle all we need to know is the radius, which we don’t have, but we can find it. Since, in the problem, it says that the hypotenuse of the triangle is the circle’s diameter. If we find the hypotenuse of the right triangle, then we’ll have the diameter of the circle, and the radius is just half of that.

To find the missing side of a right triangle we use **Pythagorean theorem**. A squared plus B squared equals C squared. The A and the B are the legs of the right triangle, which are given, 10 and 24. 10 squared plus 24 squared is C squared.

10 squared, 10 times 10 is 100, plus 24 squared, 24 times 24. 4 times 4 is 16, carry the 1, 4 times 2 is 8, plus 1 is 9. (0 placeholder) 2 times 4 is 8, 2 times 2 is 4, and then add. 6, 9 plus 8 is 17, carry the 1, 576 is equal to C squared.

676 is equal to C squared, square root both sides, and C is equal to 26. That means our diameter is 26. Now you may have already known that without doing Pythagorean theorem, because this is a Pythagorean triple, 5, 12, 13, and it’s just doubled.

5 times 2 is 10, 12 times 2 is 24, then 13 times 2 is 26, so you may not have needed to do Pythagorean theorem. Now that we have the diameter, we can find the area of the semicircle, 1/2 Pi times radius squared. The area of the semicircle is 1/2 times Pi, times the radius, which is half the diameter, so 13 squared.

The area of the semicircle is half of 13 squared, times Pi. 13 squared is 169, so it’s half of 169, times Pi. Half of 169 is 84 and 5 tenths Pi. Now we need to find the area of the triangle. The area of the triangle is half times the base, times the height.

The base of a triangle can be any of its sides, but the height must be perpendicular to the base, so our base and height are 10 and 24, since they’re perpendicular to each other. Half of 10, times 24. Half of 24 is 12, and 12 times 10 is 120, so the area of the triangle is 120 units squared.

Since we’re finding the area of the semicircle excluding the triangle, we’re going to take the area of our semicircle and subtract the area of the triangle, so the area is 84 and 5 tenths Pi, minus 120.

Since these are not like terms, you could leave this as your answer, or we could find what 84 and 5 tenths Pi is approximately, and then subtract 120 from that. 84 and 5 tenths Pi is approximately 265 and 5 tenths, minus 120, gives us 145 and 5 tenths units squared. The area of the semicircle excluding the triangle is 145 and 5 tenths units squared.