A circle with centre O and radius 8 cm is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, BC = 30 cm and BS = 24 cm, then find the length DC.
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
Given: Circle with centre O, radius = 8 cm, inscribed in quadrilateral ABCD. AD ⊥ DC, BC = 30 cm, BS = 24 cm.
Step 1: Using equal tangents from external point B:
BS = BR = 24 cm (tangents from B)
So, CR = BC − BR = 30 − 24 = 6 cm
Step 2: Equal tangents from C:
CR = CQ = 6 cm
Step 3: Since AD ⊥ DC, angle D = 90°. The circle touches DC at Q and AD at P, with OQ ⊥ DC and OP ⊥ AD (radius ⊥ tangent). So OPDQ is a square with side = radius = 8 cm.
Therefore, DQ = 8 cm
Step 4:
$$DC = DQ + QC = 8 + 6 = \boxed{14 \text{ cm}}$$
Source: Chapter 10, Sections 10.3 (Theorem 10.2 — equal tangents from external point)
---
Explanation
- Key property used: Tangents from an external point to a circle are equal in length (Theorem 10.2).
- From B: BS = BR; from C: CR = CQ — both follow from this theorem.
- Since ∠D = 90° and both tangent sides at D meet at right angles with the radius, OPDQ forms a square of side equal to the radius (8 cm), giving DQ = 8 cm.
- Examiners award marks for: identifying equal tangent pairs, finding CR = CQ = 6 cm, justifying DQ = 8 cm (square argument), and the final addition.