An inscribed circle problem (Posted on 2006-12-31)

	Submitted by K Sengupta
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Let r = radius of the circle inscribed in the trapezoid ABCD where: Angle BAD = 90 degrees = Angle CBA Therefore, AB = 2r Let BC = b, AD = a Now, we know that when a quadrilateral is circumscribed about a circle, so that the circle is tangent to all the four sides of the quadrilateral, then the sum of the lengths of opposite sides of the quadrilateral are equal. (Reference: http://www.artofproblemsolving.com/Books/IntroGeom/exc2.pdf; # 12.4.5) Consequently, CD = a + b – 2r Let us draw CK perpendicular to AD. Then, CK = AB = 2r; KD = a-b. Hence, from Triangle CKD: CD^2 = CK^2 + KD^2 Or, (a+b-2r)^2 = 4*r^2 + (a-b)^2 Or, r = ab/(a+b) For Part A: Substituting a = 28; b = 21, we obtain: r = 12. For Part B: Substituting a = 45; b = 36, we obtain: r = 30. ------------------------------------------------------------------------------- Also refer to the respective methodologies adopted by Charlie and Bractals, leading to the same solution, in the Comments section.

	Subject	Author	Date
	solution	Charlie	2006-12-31 12:52:35
	Solution	Bractals	2006-12-31 12:23:31