Question

Write an expression in simplest form for the perimeter of a right triangle with leg lengths of 12a5 and 9a5.

Answer 1

Given:

The lengths of legs of a right triangle are $12a^5$ and $9a^5$.

To find:

The perimeter of a right triangle.

Solution:

In a right angle triangle,

$Hypotenuse=\sqrt{Leg_1^2+Leg_2^2}$

$Hypotenuse=\sqrt{(12a^5)^2+(9a^5)^2}$

$Hypotenuse=\sqrt{144a^{10}+81a^{10}}$

$Hypotenuse=\sqrt{225a^{10}}$

On further simplification, we get

$Hypotenuse=\sqrt{(15a^{5})^2}$

$Hypotenuse=15a^5$

Now, the perimeter of the triangle is the sum of all of its sides.

$Perimeter=Leg_1+Leg_2+Hypotenuse$

$Perimeter=12a^5+9a^5+15a^5$

$Perimeter=36a^5$

Therefore, the perimeter of the right triangle is $36a^5$.