Question

A chord of a circle is $12cm$ in length and its distance from the center is $8cm$. Find the length of the chord of the same circle which is at a distance of $6cm$ from the center.
A. $30cm$
B. $24cm$
C. $16cm$
D. $18cm$

Answer 1

Hint: First we will make the diagram of the circle according to the question. Here is a diagram of the circle; we have to use Pythagoras theorem i.e. in a given triangle which has three sides$a,b,c$ where $b$base is, $a$ is perpendicular and $c$ is the hypotenuse. So according to the Pythagoras theorem ${c^2} = {a^2} + {b^2}$ to find the radius of the circle. Thus we get the length of the second chord and easily find the length of the diameter.

Complete step by step answer:

As per the question, draw the diagram:

A chord of a circle is $12cm$in length and its distance from the center is $8cm$.
Assume that the center of the circle is $O, AB$ is the one chord of the circle, whose midpoint is$M$
AMO and $BMO$ are both a right-angled triangle.
Now
$AM$ $ = \dfrac{{12}}{2} = 6cm$ \[\]
$AM$ $ = 8cm$
So r is the radius of the circle.
Now
\[
  AO = \sqrt {A{M^2} + O{M^2}} \\
   \Rightarrow \sqrt {{6^2} + {8^2}} \\
   \Rightarrow \sqrt {36 + 64} \\
   \Rightarrow \sqrt {100} \\
   \Rightarrow 10cm \\
 \]
Now the chord CD is 6 cm from the center of circle O.
Let F be the midpoint of CD. Then angle CFO and DFO are both right-angled triangles.
$OF$ =$6cm$
$CO = DO$ $10cm$
So,
$CF = DF$
So
$CF = $
$
  \sqrt {C{O^2} - O{F^2}} \\
   \Rightarrow \sqrt {{{10}^2} - {6^2}} \\
   \Rightarrow \sqrt {100 - 36} \\
   \Rightarrow \sqrt {64} \\
   \Rightarrow 8cm \\
 $
Now $D$ =$2 \times 8 = 16cm$
Hence the correct answer in option $C$.

Note: First of all we have to remember the definition of a circle, all the parameters used in the circle. We have to remember about Pythagoras theorem. In the given question we have to find the diameter of the circle, for this first we have to calculate the radius then multiply it by $2$ as given in the solution hint. Thus we get the correct answer.