Question 389086
If you recall the 6-8-10 triangle, you can see that the length is 8 and the width is 6, and the perimeter is 28.


If not, we can observe that if x is the length and y is the width, then


{{{sqrt(x^2 + y^2) = 10}}}


{{{x^2 + y^2 = 100}}}


We want to find 2(x + y). If we take x + y and square it, we obtain


{{{(x+y)^2 = x^2 + 2xy + y^2}}}. Note that {{{x^2 + y^2 = 100}}} and {{{xy = 48}}}. Substituting, this becomes:


{{{(x+y)^2 = 100 + 2(48) = 196}}}


{{{x+y = 14}}}


The perimeter is 2(x+y) = 28. Isn't it nice that we didn't even have to solve for the length and the width to find the perimeter?