we know if we draw a Ʇ which AD is bisect BC in two equal part BD = DC = 12 Now apply PGT in ∆ADC AD = 24 If 0 in centre of circle. then OD = (24 – r) Now apply PGT in ∆OBD we get r2 = (12)² + (24 – r)2 r2 – (24 – r )2 = 12² (r + 24 – r )(r – 24 + r) = 144 2r = 24 + 6 = 2r = 30 = r = 15
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