Class 9 Mathematics · Ganita Manjari Part 1

Class 9 Maths Chapter 5 – Exercise Set 5.5 Solutions

Given: $r=7$ cm and perpendicular distance $d=6$ cm.

The perpendicular from the centre bisects the chord. If half the chord is $x$:

$$7^2=6^2+x^2.$$ $$49=36+x^2\Rightarrow x^2=13\Rightarrow x=\sqrt{13}.$$

Therefore the complete chord is

$$2x=2\sqrt{13}\text{ cm}\approx7.21\text{ cm}.$$

Answer: $\boxed{2\sqrt{13}\text{ cm}}$.

Let $r$ be the radius and $d$ the perpendicular distance from the centre to chord $AB$. If $M$ is the midpoint of $AB$, then $AM=AB/2$.

Using Baudhāyana–Pythagoras in right triangle $OMA$:

$$r^2=d^2+AM^2.$$

Hence

$$AM=\sqrt{r^2-d^2}.$$

Therefore

$$\boxed{AB=2\sqrt{r^2-d^2}}.$$

This formula also immediately shows that the chord becomes shorter as its distance from the centre increases.

Answer: No. Chord length is not directly proportional to distance from the centre.

If chord $CD$ is at distance $d$ and $AB$ at distance $2d$, then

$$CD=2\sqrt{r^2-d^2},$$ $$AB=2\sqrt{r^2-4d^2}.$$

There is no general relation $CD=2AB$. For example, with $r=5$ cm and $d=2$ cm:

$$CD=2\sqrt{21},\qquad AB=6\text{ cm}.$$

But $2AB=12$ cm, which is not $2\sqrt{21}$ cm.

Conclusion: The proposed doubling statement is false.