Please help me work through the below problem, thanks in advance!
The perimeter of a certain isosceles right triangle is 16+16√2. What is the hypotenuse of the triangle?
first draw the triangle. If the hypotenuse is "b" and the 2 other equal sides are "a" then the perimeter is:
$\displaystyle b+2a=16+16\sqrt{2}$
You know the angles of an isosceles right triangle, there are 2 45 degree angles $\displaystyle \frac{\pi}{4}$ and one angle $\displaystyle \frac{\pi}{2}$
So the sine of pi over 4 is $\displaystyle sin(\frac{\pi}{4})=\frac{a}{b}$ solve this for "a" and plug that into the original equation for the perimeter and use algebra to solve for "b".
Hello, mtylerrose!
$\displaystyle \text{The perimeter of a certain isosceles right triangle is }16+16\sqrt{2}$
$\displaystyle \text{What is the hypotenuse of the triangle?}$
You should know that the sides of an isosceles right triangle are: .$\displaystyle x,\:x,\:x\sqrt{2}$
The perimeter is: .$\displaystyle x + x + x\sqrt{2} \;=\;16 + 16\sqrt{2}$
. . . . . . . . . . . . . . . $\displaystyle (2 + \sqrt{2})x \:=\:16(1+\sqrt{2}) $
. . . . . . . . . . . . . . . . . . . . . $\displaystyle x \;=\;\dfrac{16(1+\sqrt{2})}{2+\sqrt{2}}$
Rationalize: .$\displaystyle \displaystyle x\;=\;\frac{16(1+\sqrt{2})}{2 + \sqrt{2}} \cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} \;=\;\frac{16(2 - \sqrt{2} + 2\sqrt{2} - 2)}{4 - 2}$
. . . . . . . . . $\displaystyle x \;=\;\dfrac{16\sqrt{2}}{2} \;=\;8\sqrt{2}$
The hypotenuse is: .$\displaystyle x\sqrt{2} \;=\;8\sqrt{2}\cdot\sqrt{2} \;=\;16 $