Difference between revisions of "1989 AHSME Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x= / | + | Setting <math>y=0</math> we have that the <math>x-</math>intercept of the line is <math>x=6/a </math>. Similarly setting <math>x=0</math> we find the <math>y-</math>intercept to be <math>y=6/b</math>. Then <math>(18/ab)=(1/2)(6/a)(6/b)</math> so that <math>18/ab = 6</math>, simplifying we would get <math>ab=3</math>. Hence the answer is <math>\fbox{A}</math>. |
-minor adjustments by Kevinliu08 | -minor adjustments by Kevinliu08 | ||
Revision as of 01:13, 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 
.
-minor adjustments 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. 
