Question

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5cm.
A) 2 cm
B) 6 cm
C) 8 cm
D) 10 cm

Answer 1

Hint: First, we will draw the diagram from the given condition. Then, apply the properties of the circle and hence obtain a right triangle. Then, use Pythagoras theorem in the right triangle formed to find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5cm.

Complete step by step solution: 

We are given a circle of radius 5 cm.

 We will first draw the diagram from the given conditions.

Let PQ be the radius of the circle and is of 5 cm.

Also, OR is the perpendicular distance from centre to the chord PQ and is of 3 cm.

As we know, perpendicular drawn from the centre bisects the chord.

Then, \[{\text{PR = RQ}}\], let ${\text{PR}} = a$, then the length of the required chord will be $2a$

Since, ${\text{OR}} \bot {\text{PQ}}$, we can apply Pythagoras theorem in $\Delta {\text{PRO}}$

Pythagoras theorem states that, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{Perpendicula}}{{\text{r}}^{\text{2}}}{\text{ + bas}}{{\text{e}}^{\text{2}}}$

Therefore, we get,

${\text{P}}{{\text{O}}^2} = {\text{O}}{{\text{R}}^{\text{2}}}{\text{ + P}}{{\text{R}}^{\text{2}}}$

Substitute the values of PO and OR to find the value of PR.

${\left( 5 \right)^2} = {\left( 3 \right)^{\text{2}}}{\text{ + P}}{{\text{R}}^{\text{2}}}$

On solving the expression, we get,

 25 = 9 + ${\text{P}}{{\text{R}}^{\text{2}}}$

$  \Rightarrow {\text{P}}{{\text{R}}^{\text{2}}} = 25 - 9 $

$   \Rightarrow {\text{P}}{{\text{R}}^{\text{2}}} = 16 $

Taking square on both sides,

\[{\text{PR}} = \pm 4\]

But length cannot be negative, hence, \[{\text{PR}} = 4{\text{cm}}\]

And, we have to calculate the length of PQ which is twice of PR.

Hence, \[{\text{PQ}} = 8{\text{cm}}\]

Therefore, option C is the correct answer.

Note: A chord of a circle is a line segment whose endpoints both lie on the circle. Also, the diameter is the longest chord of the circle. Moreover, if we draw a perpendicular line from the centre to the chord, it bisects the chord in two equal parts. And, Pythagoras theorem can be applied only in a right triangle.