mexicanhoney wrote:

The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8

b) 16

c) 4 sq rt 2

d) 8 sq rt 2

e) 16 sq rt 2

Can someone please explain the answer using the 1:1: sq rt 2 ratio? Thanks.

In this case, three sides of the triangle will be in ratio 1:1:sqrt(2)

Now let's multiply these ratio with x and add to get the perimeter of the triangle:

x+x+xsqrt(2) = 16 + 16sqrt(2)

=> 2x + x*sqrt(2) = 16 + 16*sqrt(2)

One way is to solve the equation and get the value of x.

x = 16(1+sqrt(2))/(2+sqrt(2))

= 8sqrt(2)

[Note : We can solve above equation by rationalizing the denominator]

Other way is to make Left Hand Side and Right Hand Side equations in equivalent form as shown below:

2x + x * sqrt(2) =

2x + x * sqrt (2) = 2 (8 sqrt(2)) + (8 sqrt(2)) * sqrt(2)

Clearly x = 8sqrt(2) [By comparing LHS and RHS equations]

Now Hypotenuse = xsqrt(2)

= 8sqrt(2) * sqrt(2) [substituting the value of x]

= 8 * 2

= 16

- Brajesh