NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.5#
Theorem 8: If $AB > DE$ are two chords of a circle, then $AB$ is closer to the centre than $DE$. The longer chord is nearer the centre.
For a chord at distance $d$ from the centre of a circle of radius $r$, the perpendicular from the centre, half the chord and a radius form a right triangle.
1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm. #
Solution
Let the chord be $AB$ with midpoint $M$, and let $O$ be the centre, so $OA = 7$ cm and $OM = 6$ cm. By Theorem 5, $OM \perp AB$ and $M$ is the midpoint of $AB$.
In right $\triangle OMA$, by the Baudhāyana–Pythagoras theorem:
$$AM^2 = OA^2 - OM^2$$$$AM^2 = 7^2 - 6^2$$
$$AM^2 = 49 - 36$$
$$AM^2 = 13$$
$$\Rightarrow AM = \sqrt{13} \text{ cm}$$
$M$ is the midpoint, so:
$$AB = 2AM$$$$AB = 2\sqrt{13} \text{ cm}$$
∴ The length of the chord is $2\sqrt{13}$ cm, which is about 7.21 cm.
2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is $2\sqrt{r^2 - d^2}$. #
Solution
Let $AB$ be the chord, $O$ the centre, and $M$ the foot of the perpendicular from $O$ to $AB$. Then $OM = d$ and $OA = r$.
By Theorem 5, the perpendicular from the centre bisects the chord, so $M$ is the midpoint of $AB$:
$$AB = 2AM$$$\triangle OMA$ is right-angled at $M$, with hypotenuse $OA$. By the Baudhāyana–Pythagoras theorem:
$$AM^2 + OM^2 = OA^2$$$$\Rightarrow AM^2 = r^2 - d^2$$
$$\Rightarrow AM = \sqrt{r^2 - d^2}$$
So:
$$AB = 2AM$$$$AB = 2\sqrt{r^2 - d^2}$$
∴ The chord length is $2\sqrt{r^2 - d^2}$. (When $d = 0$ the chord is a diameter of length $2r$, and when $d = r$ the chord shrinks to a single point.)
*3. 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. #
Solution
No, we cannot conclude that.
Let the radius be $r$, let $CD$ be at distance $d$, and $AB$ at distance $2d$. By Question 2:
$$CD = 2\sqrt{r^2 - d^2}$$$$AB = 2\sqrt{r^2 - 4d^2}$$
These are not in the ratio $2 : 1$ in general. For example, take $r = 5$ cm and $d = 2$ cm:
$$CD = 2\sqrt{25 - 4} = 2\sqrt{21} \approx 9.17 \text{ cm}$$$$AB = 2\sqrt{25 - 16} = 2\sqrt{9} = 6 \text{ cm}$$$$\Rightarrow 2AB = 12 \neq 9.17 \approx CD$$
∴ $CD \neq 2AB$. The chord length depends on $\sqrt{r^2 - d^2}$, not on $d$ itself, so doubling the distance does not halve the chord. All we can say (by Theorem 8) is that $CD$, being closer to the centre, is longer than $AB$.