Question

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.
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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