It is almost a no-brainer that CAT **Geometry** must be mastered, and there's no escaping this topic in you **online CAT Preparation**. Over the years, the Geometry has been given huge emphasis, which makes it the second most important topic after Arithmetic. Visit **2IIM's CAT Blueprint** to know more about what to expect in the upcoming **CAT exam**. Solve this question from **CAT Previous Year Paper** 2018 and ace your preparation.

Question 4 : 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 [TITA]

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Let ABC be the triangle on which a circle of diameter BC is drawn, intersecting AB and AC at points P and Q respectively.The lengths of AB ,AC and CP are 30cm , 25cm and 20 cm respectively we have to find the length of BQ in cm.

Key thing here is ,this is semicircle so this angle P and Q should be 90° and now we are looking to do Pythagoras theorem

So think about triangle PAC, by Pythagoras theorem

AC^{2} = AP^{2} + PC^{2} we can find that AP = 15

Since AP = 15 we can find BP by

AB = AP + PB

30 = 15 + PB

PB = 15

Now we can look at the triangle BPC, by Pythagoras theorem we can find that BC = 25

Now we have to find BQ .So we can take the area of the triangle formula which is equal to \\frac{1}{2}) × base × height

area of the triangle formula = \\frac{1}{2}) × base × height

\\frac{1}{2}) × AB × PC = \\frac{1}{2}) × AC × BQ

\\frac{1}{2}) × 30 × 20 = \\frac{1}{2}) × 25 × BQ

BQ = 24 cm

The question is **"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 [TITA] "**

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