Difference between revisions of "1996 AJHSME Problems/Problem 17"
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The area of <math>\triangle PQT</math> is <math>\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ</math>. | The area of <math>\triangle PQT</math> is <math>\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ</math>. | ||
− | If we set the areas equal, the area of <math>\ | + | If we set the areas equal, the area of <math>\triangle PQT</math> is <math>4</math>. Also, note that <math>PQ=2</math>. Plugging those in, we get: |
<math>4 = \frac{1}{2} \cdot PT \cdot 2</math> | <math>4 = \frac{1}{2} \cdot PT \cdot 2</math> | ||
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If <math>PT = 4</math>, and <math>PO = 2</math>, then <math>OT = 2</math>, and <math>T</math> must be <math>2</math> units to the left of the origin. This would be <math>(-2,0)</math>, giving answer <math>\boxed{C}</math>. | If <math>PT = 4</math>, and <math>PO = 2</math>, then <math>OT = 2</math>, and <math>T</math> must be <math>2</math> units to the left of the origin. This would be <math>(-2,0)</math>, giving answer <math>\boxed{C}</math>. | ||
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==See Also== | ==See Also== |
Latest revision as of 11:20, 22 March 2015
Problem
Figure is a square. Point
is the origin, and point
has coordinates (2,2). What are the coordinates for
so that the area of triangle
equals the area of square
?
Solution
The area of is
.
The area of is
.
If we set the areas equal, the area of is
. Also, note that
. Plugging those in, we get:
If , and
, then
, and
must be
units to the left of the origin. This would be
, giving answer
.
See Also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.