Here's one approach:AJ2009 wrote:The perimeter of a certain isosceles right triangle is 16+16*sqrroot2. What is the hypotenuse of the triangle?
a. 8
b. 16
c. 4*sqroot2
d. 8*sqroot2
e. 16*sqroot2
help please!
thanks
As soon as you see the sqrt2 and the two 16's, you should be thinking that this triangle might be in the 45-45-90 family or right triangles.
The ratio of the sides of a 45-45-90 triangle is 1:1:sqrt2
The perimeter 16+16sqrt2 has an integer component and a root component.
To get 16sqrt2, we could multiply each side of the base 1:1:sqrt2 triangle by 16, but this would make the sides 16:16:16sqrt2 and the perimeter 32+16sqrt (nope)
At this point, it helps to create a second "base" triangle by multiplying each side of the 1:1:sqrt2 triangle by sqrt 2.
We get sides sqrt2:sqrt2:2 (this is a useful ratio to keep in your pocket)
Now we might see that if we multiply each side by 8 we get the desired dimensions. We get 8sqrt2:8sqrt2:16.
The perimeter is 16+16sqrt2, and so the hypotenuse is 16
