In right-angled triangle ABC, AB=6 cm and BC=8 cm, with AB perpendicular to BC. Points P and Q lie on BC, and the areas of triangles ABP, ABQ and ABC are in arithmetic progression. If the area of triangle ABC is 1.5 times the area of triangle ABP, find PQ in cm.
Answer and solution
Correct answer: B
Solution
Answer: B — 4/3 cm
1. Area ABC=½×6×8=24 cm². Area ABP=24/1.5=16 cm².
2. All three triangles use bases on BC and height AB=6. Thus area=3×base and BP=16/3.
3. Their areas in the stated order are in AP, so their bases BP, BQ, BC are in AP: 2BQ=BP+BC.
4. BQ=(16/3+8)/2=20/3. Therefore PQ=BQ−BP=4/3 cm, option B.
Why this method works:
Every triangle has vertex A and a base on BC, so all share the perpendicular height AB. Their areas therefore scale directly with BP, BQ, BC. An arithmetic progression of areas gives an arithmetic progression of those bases in the same order.
Independent check and common trap:
The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.
Apply: Same height means BP: BQ: BC is an AP. BP=(2/3)BC=16/3; Q is halfway from P to C, so PQ=(8−16/3)/2=4/3.
Check: The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.
1. Area ABC=½×6×8=24 cm². Area ABP=24/1.5=16 cm².
2. All three triangles use bases on BC and height AB=6. Thus area=3×base and BP=16/3.
3. Their areas in the stated order are in AP, so their bases BP, BQ, BC are in AP: 2BQ=BP+BC.
4. BQ=(16/3+8)/2=20/3. Therefore PQ=BQ−BP=4/3 cm, option B.
Why this method works:
Every triangle has vertex A and a base on BC, so all share the perpendicular height AB. Their areas therefore scale directly with BP, BQ, BC. An arithmetic progression of areas gives an arithmetic progression of those bases in the same order.
Independent check and common trap:
The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.
G Strategy shortcut
Spot: Geometry and Arithmetic Progression.Apply: Same height means BP: BQ: BC is an AP. BP=(2/3)BC=16/3; Q is halfway from P to C, so PQ=(8−16/3)/2=4/3.
Check: The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.
G Strategy: Spot: Geometry and Arithmetic Progression. Apply: Same height means BP: BQ: BC is an AP. BP=(2/3)BC=16/3; Q is halfway from P to C, so PQ=(8−16/3)/2=4/3. Check: The three areas are 16,20,24 square centimetres; the bases are 16/3,20/3,24/3 centimetres. Their consecutive differences equal 4/3. In particular P and Q lie inside BC and PQ=4/3, consistent with the stated geometry.