Question

In a circle with radius 13cm, the length of chord is 24cm. then, the distance of chord from the centre of the circle is:

Answer 1

Hint: The rough figure that represents the given information is shown below.

We use the condition that the perpendicular from the centre of a circle divides the chord in two equal parts. Then we use the Pythagoras theorem to find the required distance.
The Pythagoras Theorem states that the square of hypotenuse is equal to sum of squares of other two sides that is for the triangle shown below

The Pythagoras theorem is given as \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\].

Complete step by step answer:
We are given that the radius of the circle is 13cm.
We know that the radius of a circle is the distance from the centre to one of the points on the circle.
By using this definition we can take the radius of circle from the figure that is
\[\Rightarrow OA=13cm\]
We are given that the length of chord of circle as 24cm
Let us assume that the give chord as
\[\Rightarrow AB=24cm\]
We know that the condition that the perpendicular from the centre of a circle divides the chord in two equal parts
By using the above condition to the figure we get
\[\begin{align}
  & \Rightarrow AC=\dfrac{AB}{2} \\
 & \Rightarrow AC=12cm \\
\end{align}\]
Now, let us assume that the distance of chord from the centre of the circle as
\[\Rightarrow OC=x\]
Now, let us consider the triangle \[\Delta OAC\]
We know that the Pythagoras Theorem states that the square of hypotenuse is equal to sum of squares of other two sides that is for the triangle shown below

The Pythagoras theorem is given as \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\].
By using the Pythagoras theorem to \[\Delta OAC\] we get
\[\begin{align}
  & \Rightarrow O{{A}^{2}}=O{{C}^{2}}+C{{A}^{2}} \\
 & \Rightarrow O{{C}^{2}}=O{{A}^{2}}-C{{A}^{2}} \\
\end{align}\]
Now by substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}={{13}^{2}}-{{12}^{2}} \\
 & \Rightarrow {{x}^{2}}=169-144=25 \\
\end{align}\]
Now, by applying the square root on both sides we get
\[\Rightarrow x=\pm 5\]
Here we get two values for \['x'\] one by taking positive sign and other by taking negative sign.
We know that the distance can never be negative.

Therefore we can conclude that the distance of the given chord from the centre of circle is 5cm.

Note: We have the direct formula for finding the distance of chord from the centre of the circle.

The formula for distance of chord from the centre having the radius \['r'\] and the chord length as \['d'\] is given as
\[\Rightarrow D=\dfrac{1}{2}\sqrt{4{{r}^{2}}-{{d}^{2}}}\]
Now, by using the above formula to given question we get
\[\begin{align}
  & \Rightarrow D=\dfrac{1}{2}\sqrt{4{{\left( 13 \right)}^{2}}-{{24}^{2}}} \\
 & \Rightarrow D=\dfrac{1}{2}\sqrt{676-576} \\
 & \Rightarrow D=\dfrac{10}{2}=5 \\
\end{align}\]