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OP = √(12² + 5²) = √169 = 13 cm.
r² = 20² − 16² = 144. Therefore, r = 12 cm.
∠OPQ = 90° − (180° − 50°)/2 = 25°.
∠OPQ = 90° − (180° − 80°)/2 = 50°.
From the tangent geometry, ∠PTQ = 2∠OPQ = 2 × 35° = 70°.
∠OPQ = ½∠PTQ = 50°.
For a tangential quadrilateral, AB + CD = AD + BC. Thus 8 + 7 = AD + 11, giving AD = 4 cm.
AB + CD = AD + BC. Therefore 13 + 11 = AD + 9, so AD = 15 cm.
7 + 9 = 12 + AD. Hence AD = 4 cm.
In a parallelogram circumscribing a circle, the tangential quadrilateral property makes all four sides equal. Hence it is a rhombus.
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