# 1987 AHSME Problems/Problem 8

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## Problem

In the figure the sum of the distances $AD$ and $BD$ is $[asy] draw((0,0)--(13,0)--(13,4)--(10,4)); draw((12.5,0)--(12.5,.5)--(13,.5)); draw((13,3.5)--(12.5,3.5)--(12.5,4)); label("A", (0,0), S); label("B", (13,0), SE); label("C", (13,4), NE); label("D", (10,4), N); label("13", (6.5,0), S); label("4", (13,2), E); label("3", (11.5,4), N); [/asy]$ $\textbf{(A)}\ \text{between 10 and 11} \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ \text{between 15 and 16}\qquad\\ \textbf{(D)}\ \text{between 16 and 17}\qquad \textbf{(E)}\ 17$

## Solution

Using Pythagoras' Theorem, $BD = \sqrt{3^2 + 4^2} = 5$, and $AD = \sqrt{(13 - 3)^2 + 4^2} = \sqrt{116}$. Thus the sum is $5 + \sqrt{116}$, and as $100 < 116 < 121$, $10 < \sqrt{116} < 11$, so that the sum is between $5+10$ and $5+11$, or $15$ and $16$, which is answer $\boxed{\text{C}}$.

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