A triangle has side lengths of 10, 24, and 26. Let a be the area of the circumcircle. Let b be the area of the incircle. Compute a - b.
Thank you so much!!!
Since these sides form a right triangle, the center of the circumcircle will be the midpoint of the hypotenuse.
This makes the radius of the circumcircle one-half the length of the hypotenuse, or 13.
I can't see any easy way to find the radius of the incircle, so I did this.
I placed this on a coordinate axis with vertex C to be at the origin: C = (0, 0).
I placed vertex A at (24, 0) and vertex B at (0,10).
I call the center of the incircle D.
I found the size of anlge A by using the inverse tangent of BC over CA. and then divided the angle by 2 to get
the size of angle(CAD) to be 11.30993247o.
Using this, the equation of AD is: y = -0.2x + 4.8.
Also the equation of CD is: y = x.
Solving these two simultaneously, the point D becomes (4, 4); making the radius of the incircle 4.
From this, I can find the area of the circumcircle and the area of the incircle.