# How do you find the perimeter of a right triangle with the area 9 inches squared?

Oct 30, 2017

37, 20.2, or 15.7 inches.

#### Explanation:

Area of a triangle (A) = $\frac{1}{2} \cdot \text{base"*"height}$
Perimeter (P) = all sides added up.

Right triangle: right angled triangle?
So from pythagoras, ${a}^{2} + {b}^{2} = {c}^{2}$

We've been given A: $\frac{1}{2} \cdot b \cdot h = 9$
So $b \cdot h = 18$
As we haven't been given any other conditions, except that it's a right angled triangle, there will be multiple correct answers.

As $b \times h = 18$ then b and h = 1 & 18, 2 & 9, or 3 & 6. If they're whole numbers.

Any of these can be right angled triangles, as we haven't been given any parameters for the hypotenuse.

We can still find the perimeters for these different dimensions of the triangle, using pythagoras.

If b=1, h=18, then hypotenuse = $\sqrt{{1}^{2} + {18}^{2}} = 5 \sqrt{13}$, approx 18, and perimeter = 37 inches.

If b=2, h=9, hypotenuse = $\sqrt{{2}^{2} + {9}^{2}} = \sqrt{85}$, approx 9.2, and perimeter = 20.2 inches.

If b=3, h=6, hypotenuse = $\sqrt{{3}^{2} + {6}^{2}} = 3 \sqrt{5}$, approx. 6.7 inches, so perimeter = 15.7 inches

Oct 30, 2017

The minimum possible perimeter occurs when the triangle has sides:

$3 \sqrt{2}$, $3 \sqrt{2}$ and $6$,

giving a perimeter of $6 + 6 \sqrt{2}$ inches.

#### Explanation:

A right triangle of area $9 {\text{in}}^{2}$ is half of a rectangle of area $18 {\text{in}}^{2}$.

If one side of the rectangle is of length $t$ (inches) then the other is of length $\frac{18}{t}$.

By Pythagoras, the length of the diagonal is:

$\sqrt{{t}^{2} + {\left(\frac{18}{t}\right)}^{2}}$

and the perimeter of the right triangle is:

$t + \frac{18}{t} + \sqrt{{t}^{2} + {\left(\frac{18}{t}\right)}^{2}}$

Notice that if $t > 0$ is very small or very large then the perimeter is very large.

The minimum possible value of the perimeter occurs when $t = \frac{18}{t}$

That is, when:

$t = \sqrt{18} = 3 \sqrt{2}$

Then the perimeter is:

$3 \sqrt{2} + 3 \sqrt{2} + \sqrt{18 + 18} = 6 + 6 \sqrt{2}$