Difference between revisions of "1950 AHSME Problems/Problem 46"
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| − | If you double sides <math>AB</math> and <math> | + | ==Problem== |
| + | In triangle <math>ABC</math>, <math>AB=12</math>, <math>AC=7</math>, and <math>BC=10</math>. If sides <math>AB</math> and <math>AC</math> are doubled while <math>BC</math> remains the same, then: | ||
| + | |||
| + | <math>\textbf{(A)}\ \text{The area is doubled} \qquad\\ | ||
| + | \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ | ||
| + | \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ | ||
| + | \textbf{(D)}\ \text{The median is unchanged} \qquad\\ | ||
| + | \textbf{(E)}\ \text{The area of the triangle is 0}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | If you double sides <math>AB</math> and <math>AC</math>, they become <math>24</math> and <math>14</math> respectively. If <math>BC</math> remains <math>10</math>, then this triangle has area <math>0</math> because <math>{14} + {10} = {24}</math>, so two sides overlap the third side. Therefore the answer is <math>\boxed{\textbf{(E)}\ The area of the triangle is 0}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1950|num-b=45|num-a=47}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 07:41, 29 April 2012
Problem
In triangle 
, 
, 
, and 
. If sides 
and 
are doubled while 
remains the same, then:
Solution
If you double sides and , they become 
and 
respectively. If remains 
, then this triangle has area 
because 
, so two sides overlap the third side. Therefore the answer is 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 45 |
Followed by Problem 47 | |
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| All AHSME Problems and Solutions | ||
