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Difference between revisions of "1966 AHSME Problems/Problem 2"

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<math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math>
 
<math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math>
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== Solution ==
 
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Let the base of the original triangle be <math>b</math> and the height be <math>h</math>. We know the new base is <math>\frac{11b}{10}</math> and the new height is <math>\frac{9h}{10}</math>. Thus we see the area of the original triangle is <math>\frac{bh}{2}</math> and the area of the new triangle is <math>\frac{99bh}{200}</math>. It follows the percentage decrease is <math>1</math>% because <math>\frac{\frac{bh}{2}-\frac{99bh}{200}}{\frac{bh}{2}}=\frac{1}{100}</math>. So our answer is <math>1</math>% <math>\text{decrease}</math> <math>(E)</math>.
 
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== See also ==
 
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{{AHSME box|year=1966|num-b=1|num-a=3}} 
 
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[[Category:Introductory Geometry Problems]]
 
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{{MAA Notice}}
 

Latest revision as of 13:47, 16 July 2024

Problem

When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is

$\text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease}$

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