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
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?
2 answers
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
2 answers