NCERT Solutions Class 9 Maths Exercise 5.5 — I’m Up and Down, and Round and Round

NCERT Class 9 Mathematics — Ganita Manjari Part I, I’m Up and Down, and Round and Round, Exercise Set 5.5. All 3 questions solved.

Everything in this exercise is the single right-angled triangle formed by the radius, half a chord, and the perpendicular distance from the centre to that chord:

$$r^2 = \left( \frac{\ell}{2} \right)^2 + d^2$$

Rearranged for the chord length, this gives the formula the exercise is built around:

$$\ell = 2\sqrt{r^2 – d^2}$$

Key insight. That formula is not linear in $d$ — the $d$ sits under a square root and inside a subtraction. So doubling the distance does not halve the chord, and it is Question 3 that makes this point. Any time a question asks “if this doubles, does that halve?”, the shape of the formula is what answers it.

Question 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 centre be $O$, the chord $AB$, and let $M$ be the foot of the perpendicular from $O$. Then $M$ is the midpoint of $AB$, and $\triangle OMA$ is right-angled at $M$ with

$$OA = 7 \text{ cm (radius)}, \qquad OM = 6 \text{ cm (distance)}$$

By the Baudhāyana–Pythagoras theorem:

$$AM^2 = OA^2 – OM^2 = 7^2 – 6^2 = 49 – 36 = 13$$ $$AM = \sqrt{13} \text{ cm}$$

The full chord is twice this:

$$AB = 2\sqrt{13} \text{ cm} \approx 7.21 \text{ cm}$$

A sanity check: the chord must be shorter than the diameter, $14$ cm, and it is. And $6$ cm is a fairly large distance for a $7$ cm radius, so a chord noticeably shorter than the diameter is what one would expect.

Question 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 $O$ be the centre and $AB$ the chord. Drop the perpendicular from $O$ to $AB$, meeting it at $M$, so $OM = d$.

Step 1 — the perpendicular bisects the chord. By the converse of Theorem 4 (Exercise Set 5.3, Question 1), $M$ is the midpoint of $AB$, so

$$AB = 2 \times AM$$

Step 2 — apply Pythagoras. $\triangle OMA$ has a right angle at $M$, hypotenuse $OA = r$ (a radius), and legs $OM = d$ and $AM$:

$$OA^2 = OM^2 + AM^2$$ $$r^2 = d^2 + AM^2$$ $$AM^2 = r^2 – d^2$$ $$AM = \sqrt{r^2 – d^2}$$

Step 3 — double it.

$$AB = 2\sqrt{r^2 – d^2} \qquad \blacksquare$$

Two things the formula tells you at a glance.

• When $d = 0$, the chord passes through the centre and $\ell = 2r$ — the diameter, the longest possible chord.
• When $d = r$, the formula gives $\ell = 0$: the “chord” has shrunk to a single point where the line just touches the circle. For $d > r$ the line misses the circle altogether, and indeed $r^2 – d^2$ would be negative, with no real square root — the algebra refusing to describe something that does not exist.

Question 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 $CD = 2AB$? Give reasons for your answer.

Solution. No. The chord length is not proportional to the distance, so doubling one does not halve the other.

Why not, from the formula. With $CD$ at distance $d$ and $AB$ at distance $2d$:

$$CD = 2\sqrt{r^2 – d^2} \qquad AB = 2\sqrt{r^2 – 4d^2}$$

For $CD = 2AB$ we would need

$$2\sqrt{r^2 – d^2} = 4\sqrt{r^2 – 4d^2}$$

which is a special condition on $r$ and $d$, not something that holds in general. The distances enter squared and subtracted from $r^2$, so nothing proportional can come out.

A counterexample settles it. Take a circle of radius $r = 5$ cm, with $AB$ at distance $4$ cm and $CD$ at distance $2$ cm — so the distance of $AB$ is twice that of $CD$, as required.

$$AB = 2\sqrt{25 – 16} = 2\sqrt{9} = 6 \text{ cm}$$ $$CD = 2\sqrt{25 – 4} = 2\sqrt{21} \approx 9.17 \text{ cm}$$

If the claim were true, $CD$ would be $2 \times 6 = 12$ cm. It is about $9.17$ cm, so the claim fails.

What is true. The nearer chord is always the longer one — that is Theorem 8, and it holds here: $CD$ at $2$ cm is longer than $AB$ at $4$ cm. But “longer” is all that can be said; the amount by which it is longer depends on the radius as well.

Common mistakes

• Question 1, forgetting to double. $\sqrt{13}$ is half the chord. Stopping there gives about $3.6$ cm, which is roughly a quarter of the diameter and clearly too short for a chord only $6$ cm from the centre of a $7$ cm circle.
• Question 1, giving a rounded answer only. $2\sqrt{13}$ cm is exact; $7.21$ cm is a rounding of it. Give the exact form first.
• Question 1, adding instead of subtracting. $7^2 + 6^2 = 85$ would give a chord of $2\sqrt{85} \approx 18.4$ cm, longer than the $14$ cm diameter — impossible, and a useful check. The radius is the hypotenuse, so the other two sides are subtracted from it.
• Question 2, assuming the chord is bisected without saying why. It is the converse of Theorem 4, and it is the step that lets half the chord be treated as one leg.
• Question 3, answering “yes” by analogy. The relation between chord and distance is not linear, so intuitions about proportion do not apply. A single counterexample is enough to settle it, and constructing one is quicker than arguing.
• Question 3, getting the direction backwards. The chord further from the centre is the shorter one. Here $AB$, at twice the distance, is the shorter chord — so even the inequality in the claim points the wrong way.

Practise next

• Exercise Set 5.6 — angles subtended by chords and arcs, which is the chapter’s other main theme.
• The End-of-Chapter Exercises, Questions 1, 3, 4 and 9 — the same formula solved for each of its three quantities in turn: chord, distance and radius.