Thread: Deriving hypotenuse from perimeter of a triangle

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?

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.

first draw the triangle. If the hypotenuse is "b" and the 2 other equal sides are "a" then the perimeter is:



You know the angles of an isosceles right triangle, there are 2 45 degree angles and one angle

So the sine of pi over 4 is solve this for "a" and plug that into the original equation for the perimeter and use algebra to solve for "b".

Snowtea's solution is easier, I didn't catch that...

Hello, mtylerrose!

You should know that the sides of an isosceles right triangle are: .


The perimeter is: .

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . .


Rationalize: .

. . . . . . . . .


The hypotenuse is: .

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?

Originally Posted by 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?

That's a general technique he used that's taught in algebra courses. You always multiply by the conjugate of the denominator to rationalize an expression like that.