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$",(10.5,6.75),N); [/asy]$

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

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 $\boxed{C}$ .

1995 AJHSME (Problems • Answer Key • Resources)
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1995 AJHSME Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See Also