In a right-angled triangle △ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of △ABP, △ABQ and △ABC are in arithmetic progression. If the area of △ABC is 1.5 times the area of △ABP, the length of PQ, in cm, is
Answer & solution
Answer: 2
Solution
Easy
All three triangles △ABP,△ABQ,△ABC share the same height AB=5 and have bases BP,BQ,BC along BC. So each area is just 21⋅AB⋅(base)=2.5×(base) — meaning the areas are in the same ratio as the bases. Translate both conditions into the bases, find BP and BQ, then PQ=BQ−BP.
Apply the AP condition. Areas in arithmetic progression means the middle one is the average, i.e. 2[△ABQ]=[△ABP]+[△ABC]. Cancelling the common 2.5:
2BQ=BP+BC=8+12=20⇒BQ=10
4
Distance between the two points:
PQ=BQ−BP=10−8=2
PQ=2cm
Equal height collapses everything to the bases: BP,BQ,BC are themselves in AP, with BC=12 and BP=BC/1.5=8. Then BQ is the mean of 8 and 12, namely 10, so PQ=10−8=2 — no area arithmetic needed.
CAT 2023 Slot 1 QA Q2: In a right-angled triangle △ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points o — Solution | TheCATExam