# 1995 AJHSME Problems/Problem 6

## Problem

Figures $I$, $II$, and $III$ are squares. The perimeter of $I$ is $12$ and the perimeter of $II$ is $24$. The perimeter of $III$ is

$[asy] draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle); draw((9,0)--(9,9)); draw((9,6)--(12,6)); label("III",(4.5,4),N); label("II",(12,2.5),N); label("I",(10.5,6.75),N); [/asy]$

$\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81$

## Solution 1

Since the perimeter of $I$ $12$, each side is $\frac{12}{4} = 3$.

Since the perimeter of $II$ is $24$, each side is $\frac{24}{4} = 6$.

The side of $III$ is equal to the sum of the sides of $I$ and $II$. Therefore, the side of $III$ is $3 + 6 = 9$.

Since $III$ is also a square, it has an perimeter of $9\cdot 4 = 36$, and the answer is $\boxed{C}$.

## Solution 2

Let a side of $I$ equal $x$, and let a side of $II$ equal $y$. The perimeter of $I$ is $4x$, and the perimeter of $II$ is $4y$. One side of $III$ has length $x+y$, so the perimeter is $4x+4y$, which just so happens to be the sum of the perimeters of $I$ and $II$, giving us $12+24=36$, or answer $\boxed{C}$.