NCERT Solutions Class 9 Maths Exercise 5.4 — 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 5.4. All 3 questions solved.

This exercise establishes the pair of statements that link a chord’s length to its distance from the centre:

• Theorem 6. Chords of equal length are equidistant from the centre.
• Theorem 7. Chords equidistant from the centre have equal length.

Questions 2 and 3 prove the second of these, by two different routes — and the exercise is built so that you meet both. Every argument rests on the same right-angled triangle, with sides radius, half-chord and distance:

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

Key insight. That one equation contains both theorems. With $r$ fixed, $\ell$ and $d$ determine each other: fix the chord length and the distance is forced; fix the distance and the length is forced. Everything in this exercise is that relation read in one direction or the other.

Question 1

Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

Solution. Theorem 6 says: in a circle with centre $C$, two chords of equal length are equidistant from the centre.

Set-up. Let $AB$ and $GF$ be chords with $AB = GF$. Drop perpendiculars from $C$: let $CE \perp AB$ with $E$ on $AB$, and $CH \perp GF$ with $H$ on $GF$. By the converse of Theorem 4, $E$ and $H$ are the midpoints of their chords.

To show. $CE = CH$.

Proof. Apply the Baudhāyana–Pythagoras theorem in the right-angled triangle $CEA$:

$$CA^2 = CE^2 + EA^2 \qquad \Longrightarrow \qquad CE^2 = CA^2 – EA^2$$

and in $\triangle CHG$:

$$CH^2 = CG^2 – HG^2$$

Now compare the two right-hand sides.

• $CA = CG$, since both are radii of the same circle.
• $EA = \frac{1}{2}AB$ and $HG = \frac{1}{2}GF$, since $E$ and $H$ are midpoints — and $AB = GF$ is given, so $EA = HG$.

Both terms being equal, $CE^2 = CH^2$, and since distances are positive,

$$CE = CH \qquad \blacksquare$$

So equal chords are the same distance from the centre.

Question 2

Consider Fig. 5.15. If $CE$ is perpendicular to $AB$, $CH$ is perpendicular to $GF$, and $CE = CH$, show that $AB = GF$.

Solution. This is the converse of Question 1: the distances are now given equal and the chord lengths are to be shown equal. Here it is proved by congruence.

Given. $CE \perp AB$, $CH \perp GF$, and $CE = CH$, where $C$ is the centre.

To show. $AB = GF$.

Proof. Compare $\triangle CEA$ and $\triangle CHG$.

$$\angle CEA = \angle CHG = 90° \qquad \text{(given perpendiculars)}$$ $$CA = CG \qquad \text{(radii of the same circle — the hypotenuses)}$$ $$CE = CH \qquad \text{(given)}$$

By the RHS criterion,

$$\triangle CEA \cong \triangle CHG$$

so corresponding parts are equal, giving

$$EA = HG$$

Now use the converse of Theorem 4: a perpendicular from the centre bisects its chord, so $E$ is the midpoint of $AB$ and $H$ the midpoint of $GF$. Hence

$$AB = 2 \times EA \qquad \text{and} \qquad GF = 2 \times HG$$

Since $EA = HG$, doubling both gives

$$AB = GF \qquad \blacksquare$$

Question 3

Solve the previous question using the Baudhāyana–Pythagoras theorem.

Solution. Same statement, different route — this time with no congruence at all.

Given. $CE \perp AB$, $CH \perp GF$, $CE = CH$.

Proof. In the right-angled triangle $CEA$:

$$EA^2 = CA^2 – CE^2$$

In the right-angled triangle $CHG$:

$$HG^2 = CG^2 – CH^2$$

Compare the right-hand sides. $CA = CG$ because both are radii, and $CE = CH$ is given, so the two expressions are identical. Hence

$$EA^2 = HG^2 \qquad \Longrightarrow \qquad EA = HG$$

taking positive square roots since these are lengths.

Since the perpendiculars from the centre bisect their chords, $AB = 2EA$ and $GF = 2HG$, so

$$AB = GF \qquad \blacksquare$$

Comparing the two proofs. Question 2 works with whole triangles and concludes that everything about them matches; Question 3 works with lengths only and concludes just what is needed. The second is shorter, and it generalises more easily — it is the same calculation used in Question 1, read in the opposite direction.

Together, Questions 1 and 2 establish both halves of the relationship, which the book states as Theorem 7: chords of a circle that are equidistant from the centre have equal length.

Common mistakes

• Questions 1 to 3, forgetting that the perpendicular bisects the chord. Without it, $EA$ is just “some part of $AB$” and no conclusion about the whole chord follows. It is the step that turns a half-chord result into a chord result.
• Question 2, using SAS. The known angle is at $E$ and $H$, not between the two known sides $CA$ and $CE$. RHS is the criterion for right-angled triangles with equal hypotenuses.
• Question 2, concluding $AB = GF$ directly from the congruence. The congruent triangles contain only the halves, $EA$ and $HG$. The doubling step must be written.
• Question 3, repeating the congruence proof. The question explicitly asks for the Baudhāyana–Pythagoras route, which is a different argument — lengths only, no triangle comparison.
• Confusing the two directions. Question 1 assumes equal chords and derives equal distances; Questions 2 and 3 assume equal distances and derive equal chords. Both are true, but each has to be proved separately, and using one to prove the other is circular.

Practise next

• Exercise Set 5.5 — the same right-angled triangle used numerically, to compute chord lengths from the radius and distance.
• The End-of-Chapter Exercises, Questions 3, 4 and 9 — three numerical versions of this relationship, each solving for a different one of the three quantities.