Difference between revisions of "1987 AHSME Problems/Problem 8"

Latest revision as of 13:02, 1 March 2018

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 $\boxed{\text{C}}$ .

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 \textbf{(D)}\ \text{between 16 and 17}\qquad
 \textbf{(E)}\ 17</math>
+== Solution ==
+Using Pythagoras' Theorem, <math>BD = \sqrt{3^2 + 4^2} = 5</math>, and <math>AD = \sqrt{(13 - 3)^2 + 4^2} = \sqrt{116}</math>. Thus the sum is <math>5 + \sqrt{116}</math>, and as <math>100 < 116 < 121</math>, <math>10 < \sqrt{116} < 11</math>, so that the sum is between <math>5+10</math> and <math>5+11</math>, or <math>15</math> and <math>16</math>, which is answer <math>\boxed{\text{C}}</math>.
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Difference between revisions of "1987 AHSME Problems/Problem 8"

Latest revision as of 13:02, 1 March 2018

Problem

Solution

See also