Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Q 3. The lengths of two parallel chords of a circle are 6 cm and 8 cm . If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Answers (1)

best_answer

Given : $A B=8 \mathrm{~cm}, C D=6 \mathrm{~cm}, O M=4 \mathrm{~cm}$ and $A B \| C D$.
To find: Length of ON
Construction: Draw $O M \perp C D$ and $O N \perp A B$
Proof:

Proof: CD is a chord of a circle and $O M \perp C D$
Thus, $C M=M D=3 \mathrm{~cm} \quad$ (perpendicular from centre bisects chord) and $\mathrm{AN}=\mathrm{NB}=4 \mathrm{~cm}$
Let $M N$ be x .
So, ON = $4-\mathrm{x}$
$(\mathrm{MN}=4 \mathrm{~cm})$
In $\triangle$ OCM, using Pythagoras,
$O C^2=C M^2+O M^2$--------1

and

In $\triangle$ OAN, using Pythagoras,
$O A^2=A N^2+O N^2$
From 1 and 2,
$C M^2+O M^2=A N^2+O N^2$
$\Rightarrow 3^2+4^2=4^2+(4-x)^2$
$\Rightarrow 9+16=16+16+x^2-8 x$
$\Rightarrow 9=16+x^2-8 x$
$\Rightarrow x^2-8 x+7=0$

$\Rightarrow x^2-7 x-x+7=0$
$\Rightarrow x(x-7)-1(x-7)=0$
$\Rightarrow(x-1)(x-7)=0$
$\Rightarrow x=1,7$
So, $\mathrm{x}=1$ (since $x \neq 7>O M$ )
$\mathrm{ON}=4-\mathrm{x}=4-1=3 \mathrm{~cm}$

Hence, the second chord is 3 cm away from the centre.

Posted by

seema garhwal

View full answer

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date 2025 Live: Bihar Class 10, 12 model papers out, time table soon; updates on CBSE, UP

Latest Question