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CAT 2018 Slot 2 — QA Question 20

Similarity of TrianglesEasy

On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is

Answer & solution

Correct answer: 24

Answer: 24

Solution

Easy

Any angle in a semicircle is a right angle, so ∠BPC=∠BQC=90∘\angle BPC=\angle BQC=90^\circ. That makes CPCP and BQBQ the two altitudes of triangle ABCABC from CC and BB. Equate the two area expressions to solve for BQBQ.

1

Right angles at PP and QQ. BCBC is a diameter, so PP and QQ each see it at 90∘90^\circ.

∠BPC=90∘⇒CP⊥AB∠BQC=90∘⇒BQ⊥AC\begin{aligned} &\angle BPC=90^\circ\Rightarrow CP\perp AB\\ &\angle BQC=90^\circ\Rightarrow BQ\perp AC \end{aligned}
2

Two expressions for the area of △ABC\triangle ABC. Using ABAB with altitude CPCP, and ACAC with altitude BQBQ.

Area=12⋅AB⋅CP=12⋅AC⋅BQ\begin{aligned} &\text{Area}=\frac12\cdot AB\cdot CP=\frac12\cdot AC\cdot BQ \end{aligned}
3

Solve for BQBQ. Substitute AB=30AB=30, CP=20CP=20, AC=25AC=25.

BQ=AB⋅CPAC=30⋅2025⇒ BQ=60025=24\begin{aligned} &BQ=\frac{AB\cdot CP}{AC}=\frac{30\cdot 20}{25}\\ &\Rightarrow\ BQ=\frac{600}{25}=24 \end{aligned}
A B C P Q BQ = 24
BQ=24 cmBQ=24\text{ cm}

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CAT 2018 Slot 2 QA Q20: On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. — Solution | TheCATExam