Another Problem using Heron's

Geometry Level 3

Let $\triangle ABC$ have a circle inscribed inside of it such that the circle is tangent at points $X$ , $Y$ , and $Z$ along sides $AB$ , $AC$ , and $BC$ , respectively. Let $AX=3$ , $BZ=4$ , and $CY=5$ . Find the area of $\triangle ABC$ .

If your answer can be expressed as $a\sqrt{b}$ where $a$ and $b$ are positive integers with $b$ squarefree, find $a+b$ .

The answer is 17.

1 solution

Jake Lai
May 19, 2015

Let's draw the triangle.

Since convergent tangents of a circle are equal, it must be the case that the triangle has such lengths:

Hence, we proceed with Heron's.

$s = \frac{7+8+9}{2} = 12$

$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \boxed{12\sqrt{5}}$

Great! Exactly how I wanted it to be solved. Although here is my question to you: is it a coincidence that the semiperimeter is the sum $3+4+5$ , and the semiperimeter minus each of the side lengths work out to be 3, 4, and 5? Obviously there is the simple answer, but is there anything more to Heron's formula?

If so, try and find a derivation of Heron's using the exradii.

Finn Hulse - 6 years ago

Same , used instead:- $\dfrac 1 4 *\sqrt{ (7+8+9)*(-7+8+9)*(7-8+9)*(7+8-9)}$

Niranjan Khanderia - 6 years ago

Another Problem using Heron's

The answer is 17.

1 solution

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