Question

In a circle, two parallel chords of lengths 16 cm and 24 cm are 20 cm apart. Then the radius of the circle is

• A. $\sqrt{224}$ cm
• B. 15 cm
• C. $\sqrt{108}$ cm
• D. $\sqrt{208}$ cm

Answer 1

The correct option is D

$\sqrt{208}$ cm

Let AB be a chord of length 16 cm and CD be a chord of length 24 cm.

Given, AB and CD are 20 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 $20-x$ 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,

$(\frac{1}{2}\times 16{)}^2=r^2-x^2$ and $(\frac{1}{2}\times 24{)}^2=r^2-(20-x{)}^2$

I.e., $64=r^2-x^2$ and $144=r^2-(20-x{)}^2$

Now, $144=r^2-(20-x{)}^2⟹144=r^2-(400+x^2-40x)⟹144=r^2-x^2+40x-400$

From $64=r^2-x^2$ and $144=r^2-x^2+40x-400$, we get, $x=12$ cm.

Now, $64=r^2-x^2$ and $x=12$ cm implies $r^2=208$

So, $r=\sqrt{208}$ cm.