Question

In the figure \[O\] is the centre of the circle. \[OB = 5{\text{ }}cm\] , Distance from \[O\] to chord \[AB\] is \[3{\text{ }}cm\] .Find the length of \[AB\]

Answer 1

Hint: First we will input the given data into the diagram so that it will be easy to understand. When we will input the data, a triangle \[OPB\] will be formed. Then we will apply the Pythagoras theorem in the triangle \[OPB\] to get the value of \[PB\] and hence multiply it by \[2\] to get the value of \[AB\] because we know that the perpendicular from the centre of circle bisects the chord.

Complete step by step answer:
A figure is given in which \[O\] is the centre of the circle and the value of \[OB\] is given as \[5{\text{ }}cm\] and the value of distance from \[O\] to chord \[AB\] i.e., \[OP\] is given as \[3{\text{ }}cm\] and we have to find the length of chord \[AB\]

As it can be seen in the figure that \[OP\] is drawn perpendicular to \[O\] on the line \[AB\]
So, \[\angle P = {90^ \circ }\]
Now we will apply the Pythagoras theorem in the triangle \[OPB\] to get the value of \[PB\] because we have the values of the other two sides of the triangle and also \[\angle P = {90^ \circ }\]
So, according to Pythagoras theorem,
\[{H^2} = {B^2} + {P^2}\]
where \[H\] represents the hypotenuse of a triangle, \[B\] represents the base of the triangle and \[P\] represents the perpendicular of the triangle.
In triangle \[OPB\] , by Pythagoras theorem
\[O{B^2} = O{P^2} + P{B^2}\]
Now, substitute the values of \[OB\] and \[OP\] in the above expression to find the value of \[PB\]
\[ \Rightarrow {5^2} = {3^2} + P{B^2}\]
\[ \Rightarrow 25 = 9 + P{B^2}\]
On simplifying it, we get
\[ \Rightarrow P{B^2} = 25 - 9\]
\[ \Rightarrow P{B^2} = 16\]
Taking square roots both sides, we get
\[PB = 4\]
Hence the value of \[PB\] is \[4{\text{ }}cm\]
Now, we have the value of \[PB\] and as we know that the perpendicular from the centre of the circle bisects the chord into two equal parts
\[\therefore AB = 2 \times PB\]
On Substituting the value of \[PB\] we get
\[AB = 2 \times 4\]
\[ \Rightarrow AB = 8\]
Hence, the length of \[AB\] is \[{\text{8 }}cm\]

Note:
Remember that the perpendicular drawn from the centre of the circle to the chord always bisects the chord in two equal parts, thus we can conclude that \[AP = BP\] in the given problem because it is the base point of the whole solution. Also, while taking square roots do not consider the negative root as \[OP\] denotes the distance of length and length can never be negative. The length of \[AB\] is itself a distance we require.