2
Step 1: Understand the problem Given: Right triangle ABC with right angle at A Altitude AB = 5 cm, base BC = 12 cm Points P and Q on BC such that areas of triangles ABP, ABQ, and ABC are in arithmetic progression (AP) Area of ABC = 1.5 × area of ABP Find: Length of PQ. Step 2: Find the area of triangle ABC Area of ABC = (1/2) × AB × BC = (1/2) × 5 × 12 = 30 cm² Step 3: Let areas form an AP Let: Area of ABP = x Area of ABQ = y Area of ABC = z = 30 Given: Areas in AP: x, y, z And: z = 1.5 × x → 30 = 1.5 × x → x = 20 AP relation: y - x = z - y → 2y = x + z → 2y = 20 + 30 → y = 25 So: Area ABP = 20 cm² Area ABQ = 25 cm² Area ABC = 30 cm² Step 4: Relate area to base lengths Area of triangle ABR = (1/2) × AB × BR → BR = (2 × Area) / AB BP = (2 × 20) / 5 = 40 / 5 = 8 cm BQ = (2 × 25) / 5 = 50 / 5 = 10 cm Step 5: Find PQ PQ = |BQ - BP| = 10 - 8 = 2 cm Answer: Length of PQ = 2 cm