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CAT 2019 Slot 1 — QA Question 27

Basics of CirclesEasy

AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

Answer & solution

  • A

    7.8

  • B

    8.5

  • C

    9.3

  • 9.1

Solution

Easy

An angle in a semicircle is a right angle, so both △APB\triangle APB and △AQB\triangle AQB are right-angled at PP and QQ. Apply Pythagoras in △APB\triangle APB to find APAP (hence AQAQ), then again in △AQB\triangle AQB to get QBQB.

A B P Q

Radius =5⇒=5\Rightarrow diameter AB=10AB=10. Given PB=6PB=6 and AP=2 AQAP=2\,AQ. Let AQ=xAQ=x, so AP=2xAP=2x.

1

Right angle at PP. ABAB is a diameter, so ∠APB=90∘\angle APB=90^\circ. Pythagoras in △APB\triangle APB:

AB2=AP2+PB2⇒ 102=(2x)2+62(AP=2x, PB=6)⇒ 100=4x2+36 ⇒ x2=16 ⇒ x=4\begin{aligned} &AB^2=AP^2+PB^2\\ &\Rightarrow\ 10^2=(2x)^2+6^2\quad\text{(}AP=2x,\ PB=6)\\ &\Rightarrow\ 100=4x^2+36\ \Rightarrow\ x^2=16\ \Rightarrow\ x=4 \end{aligned}
2

Right angle at QQ. Likewise ∠AQB=90∘\angle AQB=90^\circ. Pythagoras in △AQB\triangle AQB with AQ=x=4AQ=x=4:

AB2=AQ2+QB2⇒ 102=42+QB2(from step 1, x=4)⇒ QB2=84 ⇒ QB=84≈9.17\begin{aligned} &AB^2=AQ^2+QB^2\\ &\Rightarrow\ 10^2=4^2+QB^2\quad\text{(from step 1, }x=4)\\ &\Rightarrow\ QB^2=84\ \Rightarrow\ QB=\sqrt{84}\approx 9.17 \end{aligned}
QB=84≈9.1 cmQB=\sqrt{84}\approx 9.1\ \text{cm}
CAT 2019 Slot 1 QA Q27: AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB — Solution | TheCATExam