Test Index

Circles Part 1

Section: Mathematics

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Question : 27 of 100

Marks: +1, -0

In the circle given below, let

O A=1

O B=13

P Q \perp O B

. Then, the area of the triangle

P Q B

(in square units) is

[26 Feb 2021 Shift 1]

$24 \sqrt{2}$
$24 \sqrt{3}$
$26 \sqrt{23}$
$26 \sqrt{2}$

Solution:

Given,

O A=1

unit,

O B=13

unit Since,

O B

is diameter of circle. Then,

radius(r)=\frac{13}{2}=6.5

units

Draw a line joining points

P

and

C

, where

C

is the centre of the given circle.

Then,

P C=

radius of circle

=6.5

units

O C=

radius of circle

=6.5

units

Now,

A C=O C-O A=6.5-1=5.5

unit

Then, using Pythagoras theorem,

\; (P A)^{2}\;\; =(P C)^{2}-(A C)^{2}

\;\; =(6.5)^{2}-(5.5)^{2}

\;\; =(6.5-5.5)(6.5+5.5)

\;\; =(1)(12)=12

\;\;\therefore\;\; P A\;\; =\sqrt{12}

\;\text{ Then, }\; P Q\;\; =2 P A=2 \sqrt{12}

\;\text{ Hence, area of }\; \triangle P Q B=\; \frac{1}{2} \times \; \text{ Base }\; \times (\; \text{ Height }\;)

\;\; =\; \frac{1}{2} \times (P Q) \times (A B)

\;\; =\; \frac{1}{2} \times (P Q) \times (O B-O A)

\;\; =\; \frac{1}{2} \times (2 \sqrt{12}) \times (13-1)

\;\; =12 \sqrt{12}=24 \sqrt{3} \;\text{ sq units }\;

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