Question

Two parallel chords, 10 cm apart, of lengths 8 cm and 12 cm, respectively are there in a circle. Find the distance of the shorter chord from the centre.

Answer 1

Let AB be a chord of length 8 cm and CD be a chord of length 12 cm.

Given, AB and CD are 10 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 $10-x$ 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,

$(\frac{1}{2}\times 8{)}^2=r^2-x^2$ and $(\frac{1}{2}\times 12{)}^2=r^2-(10-x{)}^2$

I.e., $16=r^2-x^2$ and $36=r^2-(10-x{)}^2$

Now, $36=r^2-(10-x{)}^2⟹36=r^2-(100+x^2-20x)⟹36=r^2-x^2+20x-100$

From $16=r^2-x^2$ and $36=r^2-x^2+20x-100$, we get, $x=6$ cm.