Difference between revisions of "1989 AHSME Problems/Problem 6"
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== Problem == | == Problem == | ||
− | If <math>a,b>0</math> and the triangle in the first quadrant bounded by the | + | If <math>a,b>0</math> and the triangle in the first quadrant bounded by the coordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math> |
<math> \mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 } </math> | <math> \mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 } </math> | ||
==Solution== | ==Solution== | ||
− | Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6 | + | Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x= \frac{6}{a} </math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y= \frac{6}{b} </math>. Then <math> \frac{18}{ab}=\frac{1}{2}\cdot\frac{6}{a}\cdot\frac{6}{b}</math> so that <math> \frac{18}{ab} = 6</math>, simplifying we would get <math>ab=3</math>. Hence the answer is <math>\fbox{A}</math>. |
− | - | + | -<math>\LaTeX</math> by Kevinliu08 |
== See also == | == See also == |
Latest revision as of 21:12, 8 September 2023
Problem
If and the triangle in the first quadrant bounded by the coordinate axes and the graph of
has area 6, then
Solution
Setting we have that the
intercept of the line is
. Similarly setting
we find the
intercept to be
. Then
so that
, simplifying we would get
. Hence the answer is
.
- by Kevinliu08
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.