SSC CGL 18 Aug 2021 Shift 3 Paper

Question : 68 of 100

Marks: +1, -0

In a circle with centre

O, A B

and

C D

are parallel chords on the opposite sides of a diameter. If

A B=12 {cm}

C D=18 {cm}

and the distance between the chords

A B

and

C D

15 {cm}

, then find the radius of the circle (in

{cm})

Solution:

Given, Chord

A B=12 {cm}

C D=18 {cm}

,
So

A M=M B=6 {cm}

and

C N=N D=9 {cm}

∴ \ln △ O B M

r^{2} \;\; =x^{2}+(6)^{2}

⇒ \;\; r^{2} \;\; =x^{2}+36

. . . (i)
In

△ O D N

r^{2} \;\; =(15-x)^{2}+(9)^{2}

⇒ \;\; r^{2} \;\; =(15-x)^{2}+81

. . . (ii)
Equating Eqs. (i) and (ii),

x^{2}+36=(15-x)^{2}+81

⇒ x^{2}+36=225+x^{2}-30 x+81

⇒ \;\; 30 x=225-36+81=270

⇒ \;\; x=\;{270}/{30} ⇒ x=9

Value of

x

put in Eq. (i)

r^{2}=(9)^{2}+36

⇒ \;\; r^{2}=81+36=117

⇒ \;\; r=√{117}=√{13 × 9}

∴

Radius of circle

=3 √{13} {cm}

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