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| − | == Problem ==
| + | 太阳 月亮 星星 很多 许多 需要 必须 东西 现在 过去 将来 一定 以后 如果 要是 就是 觉得· 时间 世界 地球 一会儿 一些· |
| − | When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is
| + | 好像 告诉 |
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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>
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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}}
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Latest revision as of 20:38, 29 September 2023
太阳 月亮 星星 很多 许多 需要 必须 东西 现在 过去 将来 一定 以后 如果 要是 就是 觉得· 时间 世界 地球 一会儿 一些·
好像 告诉
