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

Question

Eva Adamson · Accepted Answer

The minimum possible perimeter occurs when the triangle has sides:

#3sqrt(2)#, #3sqrt(2)# and #6#,
giving a perimeter of #6+6sqrt(2)# inches.
A right triangle of area #9 "in"^2# is half of a rectangle of area #18 "in"^2#. If one side of the rectangle is of length #t# (inches) then the other is of length #18/t#. By Pythagoras, the length of the diagonal is: #sqrt(t^2+(18/t)^2)# and the perimeter of the right triangle is: #t + 18/t + sqrt(t^2+(18/t)^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 = 18/t# That is, when: #t = sqrt(18) = 3sqrt(2)# Then the perimeter is: #3sqrt(2) + 3sqrt(2) + sqrt(18+18) = 6+6sqrt(2)#

Hazel Anglin · Answer

37, 20.2, or 15.7 inches.

Area of a triangle (A) = #1/2*"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: #1/2*b*h=9# So #b*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)=5sqrt13#, 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)= 3sqrt(5)#, approx. 6.7 inches, so perimeter = 15.7 inches

Genesis Allen · Answer

undefined To find the perimeter of a right triangle with an area of 9 square inches, you can use the formula for the area of a triangle: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $. Since the triangle is a right triangle, you can also use the Pythagorean theorem: $ a^2 + b^2 = c^2 $, where $ a $ and $ b $ are the lengths of the legs of the triangle and $ c $ is the length of the hypotenuse.

First, solve for the base and height using the area formula:
$ 9 = \frac{1}{2} \times \text{base} \times \text{height} $
$ 18 = \text{base} \times \text{height} $

Next, use the Pythagorean theorem to find the lengths of the sides:
Let $ a $ and $ b $ be the legs of the triangle, and $ c $ be the hypotenuse. Since the area is 9 square inches, the product of the base and height is 18.

For a right triangle, $ a^2 + b^2 = c^2 $.
Choose two numbers whose product is 18 and whose sum of squares is the smallest. These numbers are 3 and 6.

So, $ a = 3 $ inches and $ b = 6 $ inches.

Finally, find the perimeter by adding the lengths of all three sides:
$ \text{Perimeter} = a + b + c $
$ \text{Perimeter} = 3 + 6 + c $
$ \text{Perimeter} = 9 + c $

To find $ c $, use the Pythagorean theorem again:
$ 3^2 + 6^2 = c^2 $
$ 9 + 36 = c^2 $
$ 45 = c^2 $
$ c = \sqrt{45} $
$ c = 3\sqrt{5} $ inches

Therefore, the perimeter of the right triangle with an area of 9 square inches is $ 9 + 3\sqrt{5} $ inches.