For the triangle below, let x be the area of the circumcircle, and let y be the area of the incircle. Compute x-y

The image is a right triangle with sides B, C, A, with C being the right angle. Line BA is 26, Line BC is 10, and Line AC is 24.

Hazelnut Jul 14, 2023

#1**0 **

The area of the circumcircle of a triangle is given by:

K = r * s

where r is the circumradius of the triangle and s is the semiperimeter of the triangle.

The area of the incircle of a triangle is given by:

k = r * s * _S_

where r is the inradius of the triangle, s is the semiperimeter of the triangle, and S is the semi-sum of the triangle's side lengths.

In this case, the semiperimeter of the triangle is s = (26 + 10 + 24) / 2 = 30.

The semi-sum of the triangle's side lengths is S = (26 + 10 + 24) / 2 = 30.

The circumradius of the triangle is r = s / 2 * S = 30 * 30 / (2 * 30) = 13.

The inradius of the triangle is r = s / (2 * S) = 30 / (2 * 30) = 4.

Therefore, the area of the circumcircle is K = 13 * 30 = 390pi, and the area of the incircle is k = 4 * 30 * 30 / (2 * 30) = 120pi.

Therefore, x - y = 390pi - 120pi = 270pi.

Guest Jul 14, 2023

#2**+1 **

Right scalene triangle.

Sides: a = 10 b = 24 c = 26

Area: T = 120

Perimeter: p = 60

Semi-perimeter: s = 30

**Inradius: r = 4 Circumradius: R = 13**

**Area of a circle = Pi x radius^2**

**x==Area of the circumcircle ==13^2 Pi==169Pi**

**y==Area of the incircle ==4^2Pi ==16Pi**

**x - y ==169Pi - 16Pi ==153Pi**

Guest Jul 15, 2023