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=√(Leg_1^2+Leg_2^2)`

`Hypotenuse=√((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`.