NCERT Solutions Class 9 Maths Exercise 5.3 — 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 5.3. All 3 questions solved.

This exercise is about the relationship between a chord and the perpendicular from the centre to it. Two statements, each the converse of the other:

• Theorem 4. The line from the centre to the midpoint of a chord is perpendicular to that chord.
• Its converse. The perpendicular from the centre to a chord bisects that chord.

Both come from the isosceles triangle of Exercise Set 5.2, and once they are available, any question about a chord becomes a right-angled triangle problem with sides

$$\text{radius}, \qquad \text{half the chord}, \qquad \text{distance from the centre}$$

related by $r^2 = \left( \tfrac{\ell}{2} \right)^2 + d^2$.

Key insight. Dropping the perpendicular from the centre to a chord does two jobs at once: it creates a right angle and it cuts the chord exactly in half. That is why one construction turns every chord question into Pythagoras.

Question 1

Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?

Solution. Let the circle have centre $O$ and let $AB$ be a chord. Drop the perpendicular from $O$ to $AB$, meeting it at $M$, so $\angle OMA = \angle OMB = 90°$.

To show. $AM = MB$.

Proof. Compare $\triangle OMA$ and $\triangle OMB$. Both are right-angled at $M$.

$$OA = OB \qquad \text{(radii of the same circle — the hypotenuses)}$$ $$OM = OM \qquad \text{(the same segment, common to both)}$$ $$\angle OMA = \angle OMB = 90° \qquad \text{(by construction)}$$

Two right-angled triangles with equal hypotenuses and one pair of equal legs are congruent by the RHS criterion:

$$\triangle OMA \cong \triangle OMB$$

Corresponding parts of congruent triangles are equal, so

$$AM = MB$$

That is, $M$ is the midpoint of $AB$, and the perpendicular bisects the chord. $\blacksquare$

Using the hint instead. The same conclusion follows from the Baudhāyana–Pythagoras theorem without congruence. In the two right-angled triangles,

$$AM^2 = OA^2 – OM^2 \qquad \text{and} \qquad MB^2 = OB^2 – OM^2$$

The right-hand sides are equal, since $OA = OB$. So $AM^2 = MB^2$, and as both are lengths, $AM = MB$.

Question 2

An isosceles triangle $ABC$ is inscribed in a circle, with $AB = AC$. Show that the altitude from $A$ to $BC$ passes through the centre of the circle.

Solution. Let the circle have centre $O$. Let $AD$ be the altitude from $A$, meeting $BC$ at $D$, so $AD \perp BC$.

The key idea: show that both $A$ and $O$ lie on the perpendicular bisector of $BC$; then the line through them is that bisector, which is also the altitude.

Step 1 — $O$ lies on the perpendicular bisector of $BC$. $BC$ is a chord of the circle, and $OB = OC$ since both are radii. A point equidistant from $B$ and $C$ lies on the perpendicular bisector of $BC$.

Step 2 — $A$ lies on it too. We are given $AB = AC$, so $A$ is also equidistant from $B$ and $C$, and therefore also lies on the perpendicular bisector of $BC$.

Step 3 — conclude. Two distinct points determine exactly one line, so the line $AO$ is the perpendicular bisector of $BC$. Being perpendicular to $BC$ and passing through $A$, it is exactly the altitude from $A$.

Hence the altitude from $A$ passes through the centre $O$. $\blacksquare$

The same argument shows the altitude also bisects $BC$ and bisects $\angle BAC$ — all four lines (altitude, median, angle bisector from $A$, and perpendicular bisector of $BC$) coincide in an isosceles triangle, and here they all pass through the circumcentre.

Question 3

Two parallel chords of lengths $6$ cm and $8$ cm are on opposite sides of the centre of a circle. If the radius of the circle is $5$ cm, find the distance between the midpoints of these chords.

Solution. Let the centre be $O$. Drop perpendiculars from $O$ to each chord; by Question 1 each perpendicular meets its chord at the midpoint, so the feet of these perpendiculars are exactly the midpoints we want.

Distance to the $6$ cm chord. Half the chord is $3$ cm, and the radius is $5$ cm. By the Baudhāyana–Pythagoras theorem:

$$d_1 = \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = \sqrt{16} = 4 \text{ cm}$$

Distance to the $8$ cm chord. Half the chord is $4$ cm:

$$d_2 = \sqrt{5^2 – 4^2} = \sqrt{25 – 16} = \sqrt{9} = 3 \text{ cm}$$

Combine them. Both chords are parallel, so both perpendiculars lie along the same line through $O$. The chords are on opposite sides of the centre, so $O$ lies between the two midpoints and the distances add:

$$d_1 + d_2 = 4 + 3 = 7 \text{ cm}$$

Note the check on Theorem 8: the longer chord ($8$ cm) is the nearer one ($3$ cm from the centre), exactly as it should be.

If the chords were on the same side, $O$ would lie outside the gap and the distances would subtract, giving $4 – 3 = 1$ cm. Question 18 of the End-of-Chapter Exercises is that version, and it is worth comparing the two.

Common mistakes

• Question 1, using SAS or SSS. Neither applies: the angle known is not between the two known sides, and only two sides are known. RHS is the criterion designed for exactly this configuration.
• Question 1, assuming what is to be proved. Writing “$AM = MB$ since the perpendicular bisects the chord” is circular — that is the statement being established.
• Question 2, proving only that $AD$ bisects $BC$. That shows $AD$ is a median. What is needed is that the centre lies on it, which comes from $OB = OC$.
• Question 3, forgetting to halve the chord. The right-angled triangle has half the chord as a leg, not the whole chord. Using $6$ instead of $3$ makes $5^2 – 6^2$ negative, which is itself the signal that something is wrong.
• Question 3, subtracting the distances. They subtract only when the chords are on the same side of the centre. Here they are on opposite sides, so the centre lies between them and the distances add.
• Question 3, pairing the wrong distance with the wrong chord. The longer chord is nearer the centre. If a calculation puts the $8$ cm chord further away than the $6$ cm one, the halves have been swapped.

Practise next

• Exercise Set 5.4 — equal chords are equidistant from the centre, which is this exercise’s right-angled triangle used in the opposite direction.
• The End-of-Chapter Exercises, Question 18 — the same configuration as Question 3 with the chords on the same side of the centre, and with the radius to be found rather than given.