Question

In a circle, two parallel chords of lengths 4 cm and 6 cm are 5 cm apart. Then the radius of the circle is

• A. $\sqrt{13}$ cm
• B. $2\sqrt{13}$ cm
• C. $\frac{1}{\sqrt{13}}$ cm
• D. $\frac{\sqrt{13}}{2}$ cm

Answer 1

The correct option is A

$\sqrt{13}$ cm

Let AB be a chord of length 4 cm and CD be a chord of length 6 cm.

Given, AB and CD are 5 cm apart.

Let O be the centre of the circle.

Let the distance of chord AB from the centre of the circle be $x$ cm. Then the distance of chord CD from the centre of the circle is $5-x$ cm.

Let the radius of the circle be $r$.

We know, in a circle, the square of half the length of a chord is the difference of the squares of the radius and the perpendicular distance of the chord from the centre of the circle.

So, we must have,

$(\frac{1}{2}\times 4{)}^2=r^2-x^2$ and $(\frac{1}{2}\times 6{)}^2=r^2-(5-x{)}^2$

I.e., $4=r^2-x^2$ and $9=r^2-(5-x{)}^2$

Now, $9=r^2-(5-x{)}^2⟹9=r^2-(25+x^2-10x)⟹9=r^2-x^2+10x-25$

From $4=r^2-x^2$ and $9=r^2-x^2+10x-25$, we get, $x=3$ cm.

Now, $4=r^2-x^2$ and $x=3$ cm implies $r^2=13$

So, $r=\sqrt{13}$ cm.