Question

Triangle ABC is inscribed in a circle, such that AC is a diameter of the circle and angle BAC is 45. If the area of triangle ABC is 72 square units, how much larger is the area of the circle than the area of triangle ABC?

Answer 1

If AC is a diameter of the circle, then angle ABC is a right angle. Therefore, triangle ABC is a 45-45-90 triangle, and the base and the height are equal. Assign the variable x to represent both the base and height:
$A=\frac{bh}{2}$
$72=\left(x\right)\left(x\right)/2$
$144=x^2$
$x=12$
Because this is a 45-45-90 triangle, and the two legs are equal to 12, the common ratio tells you that the hypotenuse, which is also the diameter of the circle, is $12\sqrt{2}$. Therefore, the radius is equal to $6\sqrt{2}$ and the area of the circle, $\pi r^2$, equals $72\pi$. The area of the circle is $72\pi -72$ square units larger than the area of triangle ABC.