From a point A, which is at a distance of 17 cm from the centre C of a circle with radius 8 cm, the pair of tangents AB and AD to the circle are drawn. Find the area of the quadrilateral ABCD.

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Step 1: Given a circle with center C and radius r = 8 cm.
Step 2: Point A is outside the circle such that AC = 17 cm.
Step 3: Tangents AB and AD are drawn from point A to the circle.
Step 4: Since AB and AD are tangents from the same point, AB = AD.
Step 5: Triangle ABC and triangle ADC are right triangles because the radius is perpendicular to the tangent at the point of contact.
Step 6: Let the points of tangency be B and D.
Step 7: In triangle ABC, AC = 17 cm, CB = 8 cm (radius), and AB is tangent.
Step 8: Using Pythagoras theorem in triangle ABC:
AB^2 = AC^2 - CB^2 = 17^2 - 8^2 = 289 - 64 = 225
=> AB = 15 cm
Step 9: The quadrilateral ABCD consists of two congruent triangles ABC and ADC.
Step 10: Area of triangle ABC = (1/2) * base * height = (1/2) * CB * AB = (1/2) * 8 * 15 = 60 cm²
Step 11: Area of quadrilateral ABCD = 2 * area of triangle ABC = 2 * 60 = 120 cm²

Therefore, the area of quadrilateral ABCD is 120 cm².

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