Question

The length of two parallel chords of a circle are $6\;{\rm{cm}}$ and $8\;{\rm{cm}}$. If the smaller chord is at distance $4\;{\rm{cm}}$ from the center, what is the distance of the other chord from the center?

Answer 1

Hint: In the solution we will use Pythagoras theorem to find out the distance of the chord from the center. Pythagoras theorem shows the relation between hypotenuse, base and perpendicular of a right triangle.

Complete Step-by-step Solution
Given:
The length of the first chord of the circle is ${l_1} = 6\;{\rm{cm}}$.
The length of the second chord of the circle is ${l_2} = 8\;{\rm{cm}}$.
The distance of the smaller chord from the center is $d = 4\;{\rm{cm}}$.
 The following is the schematic diagram of the circle having two parallel chords.

The chord of the circle is divided by the perpendicular line into two parts. The radius of the circle is,
$R = \sqrt {{{\left( {\frac{{{l_1}}}{2}} \right)}^2} + {{\left( {\frac{{{l_2}}}{2}} \right)}^2}} $
Here, $R$ is the radius of the circle.
Substitute the values in the above equation to get the radius of the circle.
$\begin{array}{l}
R = \sqrt {{{\left( {\frac{{6\;{\rm{cm}}}}{2}} \right)}^2} + {{\left( {\frac{{8\;{\rm{cm}}}}{2}} \right)}^2}} \\
R = \sqrt {{{\left( {3\;{\rm{cm}}} \right)}^2} + {{\left( {4\;{\rm{cm}}} \right)}^2}} \\
R = 5\;{\rm{cm}}
\end{array}$

Since, the radius of the circle is $5\;{\rm{cm}}$. Now, the distance of the other chord from the center is,
\[D = \sqrt {{R^2} - {d^2}} \]
Here, \[D\] is the distance of the other chord from the center.
Substitute the values in the above equation.
\[\begin{array}{l}
D = \sqrt {{{\left( {5\;{\rm{cm}}} \right)}^2} - {{\left( {4\;{\rm{cm}}} \right)}^2}} \\
D = 3\;{\rm{cm}}
\end{array}\]
Therefore, the distance of the chord from the center of the circle is \[3\;{\rm{cm}}\].

Note: In such types of problems, the theorem of chords of a circle which states that the chord of the circle is divided by the perpendicular line into two equal parts is used to find the radius of the circle.