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Difference between revisions of "1995 AJHSME Problems/Problem 6"

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<math>\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(e)}\ 81</math>
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<math>\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(E)}\ 81</math>
  
 
==Solution 1==
 
==Solution 1==

Latest revision as of 23:44, 27 December 2024

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 $\boxed{C}$.

See Also

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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