In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD Class 9

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD Class 9

Question 1.

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer. Class 9

Solution:

Let AB and CD be the two chords of circle with centre O.

ON = d and OM = 2d are the lengths of perpendicular drawn from centre O to chords AB and CD.

Join OB and OD.

Let r = radius of circle.

In right-angled ∆OMB,

OB² = OM² + BM² ...[By Baudhayana-Pythagoras theorem]

∴ BM² = OB² - OM²

= r² - (2d)²

∴ BM = $\sqrt{r^2-4 d^2}$

∵ AB = 2 BM ... [M is the midpoint of chord AB]

∴ AB = $2 \sqrt{r^2-4 d^2}$ ...(i)

In right-angled ∆OND,

OD² = ON² + ND² ...[By Baudhayana-Pythagoras theorem]

∴ ND² = OD² - ON²

= r² - d²

∴ ND = $\sqrt{r^2-d^2}$

∵ CD = 2 ND ... [N is the midpoint of CD]

∴ CD = 2$\sqrt{r^2-d^2}$ ...(ii)

Since, r² - 4d² < r² - d²

∴ $2 \sqrt{r^2-4 d^2}<\sqrt{r^2-d^2}$

∴ AB < CD ... [From (i) and (ii)]

∴ CD > AB

But, we cannot conclude that CD = 2 AB