Difference between revisions of "1950 AHSME Problems/Problem 46"
m |
|||
| Line 10: | Line 10: | ||
==Solution== | ==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)}\ \text{The area of the triangle is 0}}</math>. | 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)}\ \text{The area of the triangle is 0}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=97s | ||
| + | |||
| + | ~MathProblemSolvingSkills.com | ||
==See Also== | ==See Also== | ||
Latest revision as of 15:49, 15 July 2023
Contents
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 
.
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=97s
~MathProblemSolvingSkills.com
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 45 |
Followed by Problem 47 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
