Difference between revisions of "1966 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | <math>\ | + | 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>. |
== See also == | == See also == | ||
Revision as of 23:12, 12 September 2015
Problem
When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is
Solution
Let the base of the original triangle be 
and the height be 
. We know the new base is 
and the new height is 
. Thus we see the area of the original triangle is 
and the area of the new triangle is 
. It follows the percentage decrease is 
% because 
. So our answer is % 
.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
