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CAT 2023 Slot 1 — QA Question 2

Basics of TrianglesEasy

In a right-angled triangle △ˆ†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 △ˆ†ABP,  △ˆ†ABQ and △ˆ†ABC are in arithmetic progression. If the area of △ˆ†ABC is 1.5  times the area of △ˆ†ABP, the length of PQ, in cm, is    

Answer & solution

Correct answer: 2

Answer: 2

Solution

Easy

All three triangles △ABP, △ABQ, △ABC\triangle ABP,\ \triangle ABQ,\ \triangle ABC share the same height AB=5AB=5 and have bases BP, BQ, BCBP,\ BQ,\ BC along BCBC. So each area is just 12⋅AB⋅(base)=2.5×(base)\tfrac12\cdot AB\cdot(\text{base})=2.5\times(\text{base}) — meaning the areas are in the same ratio as the bases. Translate both conditions into the bases, find BPBP and BQBQ, then PQ=BQ−BPPQ=BQ-BP.

A B C P Q 5 12
1

Express each area using common height AB=5AB=5:

[△ABP]=12⋅AB⋅BP=2.5 BP[△ABQ]=12⋅AB⋅BQ=2.5 BQ[△ABC]=12⋅AB⋅BC=2.5⋅12=30\begin{aligned} &[\triangle ABP]=\tfrac12\cdot AB\cdot BP=2.5\,BP\\ &[\triangle ABQ]=\tfrac12\cdot AB\cdot BQ=2.5\,BQ\\ &[\triangle ABC]=\tfrac12\cdot AB\cdot BC=2.5\cdot 12=30 \end{aligned}
2

Use [△ABC]=1.5 [△ABP][\triangle ABC]=1.5\,[\triangle ABP] to find BPBP:

2.5⋅12=1.5×2.5 BP ⇒ 12=1.5 BP ⇒ BP=82.5\cdot 12=1.5\times 2.5\,BP\ \Rightarrow\ 12=1.5\,BP\ \Rightarrow\ BP=8
3

Apply the AP condition. Areas in arithmetic progression means the middle one is the average, i.e. 2 [△ABQ]=[△ABP]+[△ABC]2\,[\triangle ABQ]=[\triangle ABP]+[\triangle ABC]. Cancelling the common 2.52.5:

2 BQ=BP+BC=8+12=20 ⇒ BQ=102\,BQ=BP+BC=8+12=20\ \Rightarrow\ BQ=10
4

Distance between the two points:

PQ=BQ−BP=10−8=2PQ=BQ-BP=10-8=2
PQ=2 cmPQ=\mathbf{2}\ \text{cm}

Equal height collapses everything to the bases: BP, BQ, BCBP,\,BQ,\,BC are themselves in AP, with BC=12BC=12 and BP=BC/1.5=8BP=BC/1.5=8. Then BQBQ is the mean of 88 and 1212, namely 1010, so PQ=10−8=2PQ=10-8=2 — no area arithmetic needed.

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CAT 2023 Slot 1 QA Q2 — Basics of Triangles solution | TheCATExam