Question

Given in the following figure O is the center of the circle, OP =12cm and OB=13cm. Find the value of AB.

(a)8cm
(b)10cm
(c)12cm
(d)13cm

Answer 1

Hint: To solve this question we will use Pythagoras in triangle OPB. Pythagoras theorem is stated as “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. We will apply this to obtain the length of PB and then proceed to obtain the length of AB.

Complete step-by-step answer:
We are given in the following circle C, O is the Centre of the circle, OB = radius of the circle =13cm, and OP = 12cm.

Through O we have drawn a line to P then, OP is the perpendicular bisector of line AB, hence we have \[\angle OPB=\angle OPA=90\] and line AP= PB.
Now, we consider the triangle OPB, as \[\angle OPB=90\]then, triangle OPB is a right angled triangle, therefore we can apply Pythagoras theorem to it stated as- ‘In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides’.
Given that OB = radius of the circle =13cm, and OP = 12cm.
Then applying Pythagoras to triangle OPB we have,
\[O{{B}^{2}}=O{{P}^{2}}+P{{B}^{2}}\]
Substituting the values of OB=13cm and OP=12cm we get,
\[\begin{align}
  & {{13}^{2}}={{12}^{2}}+P{{B}^{2}} \\
 & \Rightarrow P{{B}^{2}}=169-144 \\
 & \Rightarrow P{{B}^{2}}=25 \\
 & \Rightarrow PB=5cm \\
\end{align}\]
So, we obtained the value of PB=5cm. Also, because OP is the perpendicular bisector of line AB then line AP= PB.
\[\begin{align}
  & AB=AP+PB \\
 & \Rightarrow AB=PB+PB \\
 & \Rightarrow AB=2PB=2(5) \\
 & \Rightarrow AB=10cm \\
\end{align}\]
Hence, we obtain the length of AB=10cm, which is option (b).

Note: The possibility of error in this question is that you can calculate the value of PB=5 and not evaluate the value of AB after that, which is wrong because we need to multiply by 2 to get the value of AB after calculating PB.