ABCD is a trapezoid with AB<CD and AB parallel to CD. Γ is a circle inscribed in ABCD, such that Γ is tangent to all four sides. If AD=BC=25 and the area of ABCD is 600, what is the radius of Γ?

Answer 1

Extend DC to point E such the CEB is a right triangle. (That is, extend the trapezoid so that it becomes a rectangle.)

Now, in triangle, CEB, draw the altitude to CB, so it meets CB at F.

It is convenient that the hypotenuse CB = 25, since we know that the altitude is the geometric mean of the two segments of the hypotenuse. Thus, the altitude is 15.

Now, each leg of CEB is the geometric mean of the hypotenuse and the portion of CB adjacent to the leg, so

CE=15 and EB=20
Note that CEB is a nice 3-4-5 right triangle.

So, the altitude of ABCD is 20, and on each end there is a difference of 15 between the base lengths.

Area of the two triangle is 2(20*15/2) = 300
That leaves 300 for the area of the inside rectangle, so CD=15.

So, AB=45

aea of trapezoid: 20(45+15)/2=600

So, what is the radius of the circle? 1/2 the altitude of ABCD, or 10.

Answer 2

The answer is incorrect.
AB<CD, so AB is not 45 and CD is not 15