Question

Find the length of a chord which is at a distance of $ 5\;cm $ from the centre of a circle of radius $ 13\;cm $ .

Answer 1

Hint: To find length of the chord of circle of radius $ 13\;cm $ which is at a distance of $ 5\;cm $ from centre. We first draw perpendicular from centre to chord and then applying Pythagoras theorem in the right angle triangle so formed and then substituting values in it to get the third side of the triangle and then using it length of required chord.
Formulas used: Pythagoras theorem $ {\left( {Hyp.} \right)^2} = {\left( {Perp} \right)^2} + {\left( {Base} \right)^2} $

Complete step-by-step answer:
Given,
There is a circle of radius $ 13\;cm $ .

Let O be the centre of this circle. Such that $ OA = OB = 13\;cm $
Where AB is a chord of a circle. Which is at a distance of $ 5\;cm $ from the centre of the circle.
Such that $ OM = 5\;cm $
 $ \Delta OAM $ is a right angle triangle.
Therefore, applying Pythagoras theorem. We have
 $ {\left( {OA} \right)^2} = {\left( {OM} \right)^2} + {\left( {AM} \right)^2} $
Substituting values in above we have
 $
  {\left( {13} \right)^2} = {\left( 5 \right)^2} + {\left( {AM} \right)^2} \\
   \Rightarrow 169 = 25 + {\left( {AM} \right)^2} \\
   \Rightarrow 169 - 25 = {\left( {AM} \right)^2} \\
   \Rightarrow 144 = {\left( {AM} \right)^2} \\
   \Rightarrow {\left( {AM} \right)^2} = {\left( {12} \right)^2} \\
   \Rightarrow AM = 12 \;
  $
Therefore from above we see that length of AM is $ 12\;cm $
Hence, length of chord AB will be given as: $ 2 \times AM $
Length of chord AB = $ 2 \times 12 $
Length of chord AB = $ 24\;cm $ .
Hence, the required length of the chord of a circle which is at a distance of $ 5\;cm $ from centre is $ 24\;cm. $
So, the correct answer is “24 cm”.

Note: In this type of problem we use Pythagoras theorem to form an equation by substituting values and then solving equations so formed to get the value of the third side of right angle triangle and then using the length of the chord asked in the given problem.