Question

In a circle, two parallel chords of lengths 6 cm and 8 cm are 6 cm apart. Then the distance of chord of length 6 cm from the centre is

• A. $\frac{43}{12}$ cm
• B. 4 cm
• C. $\frac{52}{12}$ cm
• D. 6 cm

Answer 1

The correct option is A

$\frac{43}{12}$ cm

Let AB be a chord of length 6 cm and CD be a chord of length 8 cm.

Given, AB and CD are 6 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 $6-x$ 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,

$(\frac{1}{2}\times 6{)}^2=r^2-x^2$ and $(\frac{1}{2}\times 8{)}^2=r^2-(6-x{)}^2$

I.e., $9=r^2-x^2$ and $16=r^2-(6-x{)}^2$

Now, $16=r^2-(6-x{)}^2⟹16=r^2-(36+x^2-12x)⟹16=r^2-x^2+12x-36$

From $9=r^2-x^2$ and $16=r^2-x^2+12x-36$, we get, $x=\frac{43}{12}$ cm.

Therefore the distance of chord of length 6 cm from the centre is $\frac{43}{12}$ cm.