Asked by Matty

A circle with a 4-inch radius is centered at A, and a circle with a 9-inch radius is centered at B, where A and B are 13 inches apart. There is a segment that is tangent to the small circle at P and to the large circle at Q. It is a common external tangent of the two circles. What kind of quadrilateral is PABQ? What are the lengths of its sides?

Answers

Answered by Steve
Draw the figure. Since AP and BQ are parallel, PABQ is a trapezoid.

Draw PR to intersect BQ at R.

AP=BR=4 so RQ=5

So, the sides are
PA=4
AB=PR=13
BQ=9
QP = √(5^2+13^2) = √194
Answered by Joyce
I agree with your answers for the length of PA, AB, PR, BQ except for QP. to solve for QP you'll do the √(13^2-5^2)=12. Since PABQ is a trapezoid, AB will be one of its opposite sides which will also act as the hypotenuse of the right-angled triangle within
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