Difference between revisions of "1989 AHSME Problems/Problem 6"
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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>. | 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> | + | -<math>\LaTeX</math> by Kevinliu08 |
== See also == | == See also == | ||
Latest revision as of 20:12, 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. 
