A triangle has sides with lengths of 2, 1, and 7. What is the radius of the triangles inscribed circle?

Question

A triangle has sides with lengths of 2, 1, and 7.  What is the radius of the triangles inscribed circle?

Avery Allen · Accepted Answer

It is not possible

A triangle with sides 2,1,7 doesn't exist Using the triangle inequality theorem:Each side of a triangle triangle must be shorter than the sum of the other two sides #(7>2+1)#Wrong! #(2<1+7)#Correct! #(1<2+7)#Correct!

Savannah Aguilar · Answer

undefined To find the radius of the inscribed circle in a triangle with side lengths 2, 1, and 7, you can use the formula for the radius of the inscribed circle:

$$ r = \frac{{\text{Area of the triangle}}}{{\text{Semiperimeter of the triangle}}} $$

First, calculate the semiperimeter $ s $ of the triangle, which is half the sum of the lengths of its sides:

$$ s = \frac{{2 + 1 + 7}}{2} = \frac{10}{2} = 5 $$

Next, use Heron's formula to find the area $ A $ of the triangle. Heron's formula states that for a triangle with sides of lengths $ a $, $ b $, and $ c $, and semiperimeter $ s $, the area $ A $ is given by:

$$ A = \sqrt{s(s - a)(s - b)(s - c)} $$

Substitute the given side lengths into the formula:

$$ A = \sqrt{5(5 - 2)(5 - 1)(5 - 7)} = \sqrt{5(3)(4)(-2)} $$

$$ A = \sqrt{5 \times 3 \times 4 \times 2} = \sqrt{120} = 2\sqrt{30} $$

Now, use the formula for the radius $ r $ of the inscribed circle:

$$ r = \frac{{\text{Area of the triangle}}}{{\text{Semiperimeter of the triangle}}} $$

$$ r = \frac{{2\sqrt{30}}}{{5}} $$

$$ r = \frac{{2\sqrt{30}}}{{5}} \times \frac{{\sqrt{5}}}{{\sqrt{5}}} $$

$$ r = \frac{{2\sqrt{150}}}{{5}} $$

$$ r = \frac{{2\sqrt{25 \times 6}}}{{5}} $$

$$ r = \frac{{2 \times 5 \sqrt{6}}}{{5}} $$

$$ r = \frac{{10 \sqrt{6}}}{{5}} $$

$$ r = 2 \sqrt{6} $$

So, the radius of the triangle's inscribed circle is $ 2 \sqrt{6} $.