geometry
posted by Hanna .
quadrilateral ABCD has right angles at B and D. If ABCD is kiteshaped with AB=AD=20 and BC=CD=15, find the radius of the circle inscribed in ABCD.

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+15^22(15)(15)cosC
= 450450cosC
so 450450cosC= 800800cosA
800cosA 450cosC = 350
but C = 180A
cosC = cos(180A) = cosA
800cosA +450cosA = 350
CosA = .28 (nice)
A = 73.7398°
BD^2 = 800800cosA = 800800(.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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