Question

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, then find the radius of the circle?

Answer 1

Hint: We start solving the problem by drawing all the given information. We then use the Pythagoras theorem to the right-angled triangles formed due to radius, chords, and the line joining the chords. We then make the necessary calculations and arrangements to these right-angled triangles to get the required value of the radius of the circle.

Complete step-by-step solution:
According to the problem, we are given that two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its center. We need to find the radius of the circle if the distance between AB and CD is 6 cm.
Let us draw the given information to get a better view.

Let us assume ‘r’ be the radius of the circle.
From the figure, we can see that $EO+OF=EF$. Let us assume the length of the line segment EO is ‘x’ cm.
So, we get $x+OF=6\Leftrightarrow OF=\left( 6-x \right)cm$.
From the figure, we can see that the triangle OEC is a right-angled triangle.
We know that according to the Pythagoras theorem, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
So, we get $O{{E}^{2}}+E{{C}^{2}}=O{{C}^{2}}$.
$\Rightarrow {{x}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}={{r}^{2}}$, since $EC=\dfrac{CD}{2}$.
$\Rightarrow {{x}^{2}}+\dfrac{121}{4}={{r}^{2}}$ -(1).
From the figure, we can see that the triangle OFB is a right-angled triangle.
So, we get $O{{F}^{2}}+F{{B}^{2}}=O{{B}^{2}}$.
$\Rightarrow {{\left( 6-x \right)}^{2}}+{{\left( \dfrac{5}{2} \right)}^{2}}={{r}^{2}}$, since $FB=\dfrac{AB}{2}$.
From equation (1), we get
$\Rightarrow 36-12x+{{x}^{2}}+\dfrac{25}{4}={{x}^{2}}+\dfrac{121}{4}$.
$\Rightarrow 36-12x=\dfrac{121}{4}-\dfrac{25}{4}$.
$\Rightarrow 36-12x=\dfrac{96}{4}$.
$\Rightarrow 36-12x=24$.
$\Rightarrow 12x=12$.
$\Rightarrow x=1$. Let us substitute this in equation (1).
So, we get ${{1}^{2}}+\dfrac{121}{4}={{r}^{2}}$.
$\Rightarrow 1+\dfrac{121}{4}={{r}^{2}}$.
$\Rightarrow \dfrac{4+121}{4}={{r}^{2}}$.
$\Rightarrow \dfrac{125}{4}={{r}^{2}}$.
$\Rightarrow \dfrac{5\sqrt{5}}{2}=r$.
So, we have found the radius of the circle as $\dfrac{5\sqrt{5}}{2}cm$.
$\therefore$ The radius of the given circle is $\dfrac{5\sqrt{5}}{2}cm$.

Note: We should consider the perpendicular distance if the distance between two parallel lines is mentioned. Here OC and OB are radii of the circle as they are the line segments connecting the center and the point on the circumference of the circle. We should not stop solving the problem after finding the value of ‘x’ in cm. Similarly, we can expect problems to find the diameter of the circle.