Question 1198770: In the figure, P and Q are the centers of two tangent circles and the line PQ intersects the circles at Point A and B. The larger circle touches the side CD of the rectangle ABCD at point T. If the area of ABCD is 15, what is the area of the triangle PQT?
Found 3 solutions by lotusjayden, ikleyn, math_tutor2020:
Answer by lotusjayden(18)   (Show Source):
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
In the figure, P and Q are the centers of two tangent circles and the line PQ intersects the circles
at Point A and B. The larger circle touches the side CD of the rectangle ABCD at point T.
If the area of ABCD is 15, what is the area of the triangle PQT?
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It is clearly seen from the plot (even by an unarmed eye) that PQ is half of AB,
or, in other words, AB is twice as long as PQ. (*)
In addition, the height QT of triangle PQT has the same length as the side BC
of the rectangle ABCD.
From it, it is clear that the area of the triangle PQT is 1/4 of the area
of rectangle ABCD.
ANSWER. The area of triangle PQT is  square units.
Solved.
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*) It is written in support to my statement (*) above.
PQ = r + R; AB = r + r + R + R = 2*(r+R), or AB = 2*PQ.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
I'll reference the diagram @lotusjayden has posted.
AB*BC = 15 since the rectangle ABCD has area 15 square units.
AB = AP+PR+RQ+QB
where R is the point of tangency between circles P and Q.
AP = PR as they are radii of circle P
RQ = QB as they are radii of circle Q
So,
AB = AP+PR+RQ+QB
AB = PR+PR+RQ+RQ
AB = 2(PR+RQ)
AB = 2*PQ
PQ = 0.5*AB
Then,
area of triangle PQT = 0.5*base*height
area of triangle PQT = 0.5*PQ*QT
area of triangle PQT = 0.5*(0.5*AB)*BC
area of triangle PQT = 0.25*AB*BC
area of triangle PQT = 0.25*(area of rectangle ABCD)
area of triangle PQT = 0.25*(15)
area of triangle PQT = 3.75 square units
This converts to the improper fraction 15/4.
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