Posted by **Hanna** on Sunday, December 4, 2011 at 6:53am.

quadrilateral ABCD has right angles at B and D. If ABCD is kite-shaped with AB=AD=20 and BC=CD=15, find the radius of the circle inscribed in ABCD.

- geometry -
**Reiny**, Sunday, December 4, 2011 at 9:52am
Join BD

BD will be a line of symmetry and the centre of the inscribed circle will have to be on AC and have to be the midpoint of BD. Let that point be P

BD^2 = 20^2 + 20^2 - 2(20)(20)cosA

= 800 - 800cosA

also BD^2 = 15^2+15^2-2(15)(15)cosC

= 450-450cosC

so 450-450cosC= 800-800cosA

800cosA -450cosC = 350

but C = 180-A

cosC = cos(180-A) = -cosA

800cosA +450cosA = 350

CosA = .28 (nice)

A = 73.7398°

BD^2 = 800-800cosA = 800-800(.28) = 576

BD = √576 = 24

BP = 12

Let r be the radius of the circle from P to line AB

angle ABD = 53.13..°

and BP, the hypotenuse of that little triangle is

sin 53.13...° = r/12

r = 12sin53.13..° = 12(.8) = 9.6

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