Question

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 $a - b.$

Answer 1

It's 153 pi.

Explanation:

Solution:

Let the triangle be $ABC,$ with $AB = 10,$ $BC = 24,$ and $AC = 26.$

Since $10^2 + 24^2 = 26^2$, we know that $\triangle ABC$ is a right triangle with hypotenuse $\overline{AC}$. The hypotenuse of a right triangle is a diameter of the triangle's circumcircle, so the area of the circumcircle of $ABC$ is $(26/2)^2\pi = 169\pi$.

Turning to the incircle, we start with the fact that $[ABC] = rs$, where $r$ is the inradius and $s$ is the semiperimeter of the triangle. We have $[ABC] = (10)(24)/2 = 120$ and $s = (10+24+26)/2 = 30$, so $r = [ABC]/s = 120/30 = 4$. Therefore, the area of the incircle is $4^2\pi = 16\pi$.

Finally, the area of the circumcircle is $169\pi - 16\pi = \boxed{153\pi}$ greater than the area of the incircle.

Answer 2

a-b = 156π

Explanation:

For the inscribed circle

Formula for calculating the radius of an inscribed circle in a triangle r = Area of the triangle/0.5 of its perimeter

Given a triangle of side 10, 24 and 26.

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2*10*24

Area of the triangle = 120

Perimeter of the triangle = 10+24+26 = 60

radius of the inscribed circle = 120/0.5(60)

r = 120/30

r = 4

Area of the incircle b = πr² = π(4)² = 16π

For the circumcircle

Formula for calculating the radius of a circumcircle outside a triangle

R = abc/4√s(s-a)(s-b)(s-c)

s = a+b+c/2

a, b and c are the length of the 3 sides

s = 10+24+26/2

s = 60/2 = 30

R = 10(24)(26)/4√30(20)(6)(4)

R = 6240/4*120

R = 13

radius of the circumcircle = 13

Area of the circumcircle a = πR² = π(13)² = 169π

a - b = 169π - 16π

a-b = 156π