Question

Let $I$ be the incenter of triangle $ABC$. If $AB = AC$, $BC = 30,$ and $IC = 10 \sqrt{3}$, then find the inradius of the triangle.

Answer 1

The inradius of triangle ABC is 5√3.

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Explanation:

The incenter of a triangle is the point where the angle bisectors intersect.

The inradius of a triangle is the radius of the incircle.

The incircle is tangent to all three sides of the triangle and is centered at the triangle's incenter.

Given information:

• AB = AC
• BC = 30
• IC = 10√3

As AB = BC, triangle ABC is an isosceles triangle with a base of 30 units. This means that IB = IC = 10√3.

This creates the isosceles triangle BIC, where IB = IC = 10√3, and BC = 30.

The height of triangle BIC is the inradius (r) of triangle ABC. Therefore, to find the inradius, we need to find the height of triangle BIC.

As triangle BIC is an isosceles triangle, we can use Pythagoras Theorem to calculate its height, since its height is the perpendicular bisector of base BC.

`\boxed{\begin{minipage}{9 cm}\underline{Pythagoras Theorem} \\\\$a^2+b^2=c^2$\\\\where:\\ \phantom{ww}$\bullet$ $a$ and $b$ are the legs of the right triangle. \\ \phantom{ww}$\bullet$ $c$ is the hypotenuse (longest side) of the right triangle.\\\end{minipage}}`

For triangle BIC, one leg is the height (r), the other leg is half the base (BC/2) and the hypotenuse is IC. Therefore:

`\begin{aligned}r^2+\left((BC)/(2)\right)^2&=IC^2\\\\r^2+\left((30)/(2)\right)^2&=(10√(3))^2\\\\r^2+\left(15\right)^2&=(10√(3))^2\\\\r^2+225&=300\\\\r^2&=75\\\\r&=√(75)\\\\r&=√(5^2\cdot3)\\\\r&=√(5^2)√(3)\\\\r&=5√(3)\end{aligned}`

Therefore, the inradius of triangle ABC is 5√3.