1953 AMC 12 Problem 33

Question

The perimeter of an isosceles right triangle is

2p.

Its area is:

(2+\sqrt2)p

(2-\sqrt2)p

(3-2\sqrt2)p^2

(1-2\sqrt2)p^2

(3+2\sqrt2)p^2

Let each leg have length

s

, so the hypotenuse is

s\sqrt2

Solve

s(2+\sqrt2)=2p

, then use area

\frac{s^2}{2}

If each leg is

s,

then

s(2+\sqrt2)=2p,

so

s=p(2-\sqrt2).

Therefore the area is

\begin{gathered} \frac{s^2}{2} =\frac{p^2(2-\sqrt2)^2}{2}\\ =(3-2\sqrt2)p^2. \end{gathered}

Thus, the correct answer is C.

Problem 33 in other years