A circle is inscribed in a quadrilateral ABCD in such a manner that the circle is tangent to all the four sides of the quadrilateral. It is given that Angle BAD = 90^{0}= Angle CBA.
Find the radius of the circle given that: BC = 21 units and AD = 28 units.
What would be the radius of the inscribed circle if BC = 36 units and AD = 45 units?
Call the center of the circle O. Erect perpendiculars from there to BC, CD and DA, thus falling on the points of tangency, which we shall call E, F and G respectively.
Draw line OC, which bisects ECF, and line OD, which bisects angle FDG, making triangle OEC congruent to OFC and triangle OFD congruent to OGD.
Angles BCE and CDA are supplementary, so angles OCD and ODC, being half of the former set, are complementary, and angle DOC is therefore a right angle.
Using the pythagorean theorem mOD ^2 = mGD ^2 + mOG ^2 and mOC ^2 = mOE ^2 + mEC ^2. Substituting the known values and using r for the radius of the circle, and applying the pythagorean theorem to triangle DOC:
(28r)^2 + 2r^2 + (21r)^2 = (492r)^2
(The right side came from CD being the sum of parts of two triangles that are each congruent to two other triangles.)
Solving, 98r = 1176, so r = 12
For the second example,
(45r)^2 + 2r^2 + (36r)^2 = (812r)^2
so 162r = 3240 and r = 20

Posted by Charlie
on 20061231 12:52:35 