# Deriving hypotenuse from perimeter of a triangle

• Dec 12th 2010, 09:40 AM
mtylerrose
Deriving hypotenuse from perimeter of a triangle

The perimeter of a certain isosceles right triangle is 16+16√2. What is the hypotenuse of the triangle?
• Dec 12th 2010, 09:48 AM
snowtea
Right triangle with side lengths a, b, c (where c is hypotnuse)
Isosceles means a=b

perimeter = a + b + c = a + a + c
Pythagorean theorem says: a^2 + b^2 = a^2 + a^2 = c^2

You have 2 equations and 2 unknowns. Solve for c.
• Dec 12th 2010, 09:53 AM
first draw the triangle. If the hypotenuse is "b" and the 2 other equal sides are "a" then the perimeter is:

$b+2a=16+16\sqrt{2}$

You know the angles of an isosceles right triangle, there are 2 45 degree angles $\frac{\pi}{4}$ and one angle $\frac{\pi}{2}$

So the sine of pi over 4 is $sin(\frac{\pi}{4})=\frac{a}{b}$ solve this for "a" and plug that into the original equation for the perimeter and use algebra to solve for "b".
• Dec 12th 2010, 09:55 AM
Snowtea's solution is easier, I didn't catch that...
• Dec 12th 2010, 10:02 AM
Soroban
Hello, mtylerrose!

Quote:

$\text{The perimeter of a certain isosceles right triangle is }16+16\sqrt{2}$
$\text{What is the hypotenuse of the triangle?}$

You should know that the sides of an isosceles right triangle are: . $x,\:x,\:x\sqrt{2}$

The perimeter is: . $x + x + x\sqrt{2} \;=\;16 + 16\sqrt{2}$

. . . . . . . . . . . . . . . $(2 + \sqrt{2})x \:=\:16(1+\sqrt{2})$

. . . . . . . . . . . . . . . . . . . . . $x \;=\;\dfrac{16(1+\sqrt{2})}{2+\sqrt{2}}$

Rationalize: . $\displaystyle x\;=\;\frac{16(1+\sqrt{2})}{2 + \sqrt{2}} \cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} \;=\;\frac{16(2 - \sqrt{2} + 2\sqrt{2} - 2)}{4 - 2}$

. . . . . . . . . $x \;=\;\dfrac{16\sqrt{2}}{2} \;=\;8\sqrt{2}$

The hypotenuse is: . $x\sqrt{2} \;=\;8\sqrt{2}\cdot\sqrt{2} \;=\;16$

• Dec 12th 2010, 10:17 AM
mtylerrose
Thanks everyone! I had no idea where to start, but now I am starting to understand.

Soroban,

How did you know to rationalize the way you did?
• Dec 12th 2010, 12:46 PM