Question

In a circle, two parallel chords of lengths 4 cm and 10 cm are 5 cm apart. Then the distance of the longer chord from the centre is ________.

• A. 0.1 cm
• B. 0.2 cm
• C. 0.3 cm
• D. 0.4 cm

Answer 1

The correct option is D

0.4 cm

Let AB be a chord of length 4 cm and CD be a chord of length 10 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 $(\frac{1}{2}\times 10{)}^2=r^2-(5-x{)}^2$

I.e., $4=r^2-x^2$ and $25=r^2-(5-x{)}^2$

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

From $4=r^2-x^2$ and $25=r^2-x^2+10x-25$, we get, $x=4.6$ cm.

Therefore $5-x=5-4.6=0.4$ cm.

Hence the distance of the longer chord from the centre is 0.4 cm