Difference between revisions of "1989 AHSME Problems/Problem 6"
Kevinliu08 (talk | contribs) (→Solution) |
Kevinliu08 (talk | contribs) (→Solution) |
||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
− | 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}\frac{6}{a}\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>. | + | 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 | -<math>LATEX</math> by Kevinliu08 |
Revision as of 02:18, 8 November 2022
Problem
If and the triangle in the first quadrant bounded by the co-ordinate 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.