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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>\traingle PQT</math> is <math>4</math>.  Also, note that <math>PQ=2</math>.  Plugging those in, we get:
+
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>
Line 38: Line 38:
  
 
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 10:20, 22 March 2015

Problem

Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates (2,2). What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$?

[asy] pair O,P,Q,R,T; O = (0,0); P = (2,0); Q = (2,2); R = (0,2); T = (-4,0); draw((-5,0)--(3,0)); draw((0,-1)--(0,3)); draw(P--Q--R); draw((-0.2,-0.8)--(0,-1)--(0.2,-0.8)); draw((-0.2,2.8)--(0,3)--(0.2,2.8)); draw((-4.8,-0.2)--(-5,0)--(-4.8,0.2)); draw((2.8,-0.2)--(3,0)--(2.8,0.2)); draw(Q--T); label("$O$",O,SW); label("$P$",P,S); label("$Q$",Q,NE); label("$R$",R,W); label("$T$",T,S); [/asy]

NOT TO SCALE

$\text{(A)}\ (-6,0) \qquad \text{(B)}\ (-4,0) \qquad \text{(C)}\ (-2,0) \qquad \text{(D)}\ (2,0) \qquad \text{(E)}\ (4,0)$

Solution

The area of $\square OPQR$ is $2^2 = 4$.

The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$.

If we set the areas equal, the area of $\triangle PQT$ is $4$. Also, note that $PQ=2$. Plugging those in, we get:

$4 = \frac{1}{2} \cdot PT \cdot 2$

$PT = 4$

If $PT = 4$, and $PO = 2$, then $OT = 2$, and $T$ must be $2$ units to the left of the origin. This would be $(-2,0)$, giving answer $\boxed{C}$.

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

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