NCERT Solutions Class 9 Maths Exercise 5.2 — 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 5.2. Both questions solved.

Two short proofs, and between them they set up almost everything that follows in the chapter.

A chord is a line segment joining two points of a circle. Join its two ends to the centre and a triangle appears — and because two of its sides are radii, that triangle is always isosceles. Nearly every theorem in this chapter is that observation, pushed a little further.

The congruence criteria used here are the ones from earlier grades: SSS (three sides), SAS (two sides and the angle between them), ASA, and RHS for right-angled triangles.

Key insight. In a single circle, every radius has the same length. So any two triangles built from the centre and a chord already agree on two pairs of sides before anything is assumed — which is why one extra piece of information, the base, is enough to force congruence.

Question 1

Show that the triangle formed by a chord and the centre of the circle is isosceles.

Solution. Let the circle have centre $O$, and let $AB$ be any chord of it. Join $OA$ and $OB$ to form $\triangle OAB$.

Given. $A$ and $B$ lie on the circle with centre $O$.

To show. $\triangle OAB$ is isosceles.

Proof. By the definition of a circle, every point on it is at the same distance from the centre — that distance being the radius $r$. Both $A$ and $B$ lie on the circle, so

$$OA = r \qquad \text{and} \qquad OB = r$$

and therefore

$$OA = OB$$

A triangle with two equal sides is isosceles, so $\triangle OAB$ is isosceles, with the two radii as its equal sides and the chord $AB$ as its base. $\blacksquare$

A consequence worth recording, since the rest of the chapter uses it constantly: the angles opposite equal sides of a triangle are equal, so

$$\angle OAB = \angle OBA$$

— the two base angles a chord makes with the radii are always equal.

Notice how little the proof needs. It does not matter how long the chord is, or where on the circle it sits; the argument is just the definition of a circle.

Question 2

Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Solution. Let $AB$ and $CD$ be two chords of the same circle, with centre $O$, and suppose $AB = CD$. Consider $\triangle OAB$ and $\triangle OCD$.

Given. $AB = CD$, and all of $A$, $B$, $C$, $D$ lie on the circle with centre $O$.

To show. $\triangle OAB \cong \triangle OCD$.

Proof. Compare the three pairs of sides.

$$OA = OC \qquad \text{(both radii)}$$ $$OB = OD \qquad \text{(both radii)}$$ $$AB = CD \qquad \text{(given)}$$

All three pairs of corresponding sides are equal, so by the SSS criterion

$$\triangle OAB \cong \triangle OCD \qquad \blacksquare$$

What follows from it. Corresponding parts of congruent triangles are equal, so in particular

$$\angle AOB = \angle COD$$

— equal chords subtend equal angles at the centre, which is one of the named theorems of the chapter. The converse holds too, by the same triangles with SAS instead of SSS: if the angles at the centre are equal, the two radii and the included angle match, so the triangles are congruent and the chords are equal.

The same congruence also gives $\angle OAB = \angle OCD$, the base angles, which is used later when proving results about equidistant chords.

Common mistakes

• Question 1, assuming the triangle is equilateral. Only two of its sides are radii; the third is the chord, whose length depends on the chord chosen. It is equilateral only in the special case where the chord happens to equal the radius.
• Question 1, arguing from a drawing. A picture in which $OA$ and $OB$ look equal proves nothing. The reason they are equal is the definition of a circle, and saying so is the proof.
• Question 2, using SAS without justification. SAS needs the angle between the two known sides — here $\angle AOB$ and $\angle COD$ — and their equality is what the question leads to, not something given. Assuming it is circular reasoning. SSS avoids the problem entirely.
• Question 2, forgetting to say the radii are equal. It is the load-bearing step: without “all four are radii of the same circle” there are no equal sides at all, and the congruence fails.
• Question 2, applying it across two different circles. The argument needs $O$ to be the centre of one circle containing all four points. Two equal chords in circles of different sizes make triangles that are not congruent.

Practise next

• Exercise Set 5.3 — the perpendicular from the centre to a chord, proved with the same isosceles triangle and one more congruence.
• The End-of-Chapter Exercises, Question 17 — uses the congruence of Question 2 to show that the centre lies on the bisector of the angle between two equal chords.