Difference between revisions of "1989 AHSME Problems/Problem 6"
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<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> | ||
| − | 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>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>A</math>. | + | ==Solution== |
| + | 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>6=(1/2)(6/a)(6/b)</math> so that <math>ab=3</math>. Hence the answer is <math>\mathrm{(A)}</math>. | ||
Revision as of 23:11, 29 May 2013
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 
. Hence the answer is 
.
