NCERT Solutions Class 9 Maths Exercise 5.6 — 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 5.6. All 3 questions solved.

The chapter now turns from lengths to angles. Three results carry this exercise:

• The central angle is twice the inscribed angle. An arc subtends at the centre twice the angle it subtends at any point on the remaining circle.
• Angles in the same segment are equal. Any two points on the same side of a chord see it at the same angle — an immediate consequence, since both are half the same central angle.
• Opposite angles of a cyclic quadrilateral add to $180°$.

Key insight. “Angles in the same segment are equal” has a powerful converse, and Question 2 is about it: if a segment $AB$ subtends equal angles at two points on the same side of it, those two points and $A$ and $B$ all lie on one circle. That converse is what lets you prove four points are concyclic without ever drawing the circle.

Question 1

In a circle with centre $O$, the central angle $AOB$ is $60°$. If the radius of the circle is $12$ cm, what is the length of the chord $AB$?

Solution. Join $OA$ and $OB$ to form $\triangle OAB$.

Step 1 — the triangle is isosceles. $OA = OB = 12$ cm, both radii.

Step 2 — find the base angles. The angles of a triangle sum to $180°$, and the two base angles of an isosceles triangle are equal:

$$\angle OAB = \angle OBA = \frac{180° – 60°}{2} = \frac{120°}{2} = 60°$$

Step 3 — recognise the triangle. All three angles are $60°$, so $\triangle OAB$ is equilateral. Hence

$$AB = OA = OB = 12 \text{ cm}$$

A second route, using the general formula. Dropping the perpendicular from $O$ to $AB$ bisects both the chord and the central angle, giving a right-angled triangle with hypotenuse $r$ and an angle of $30°$ at $O$:

$$AB = 2r \sin\left( \frac{\theta}{2} \right) = 2(12)\sin 30° = 24 \times \frac{1}{2} = 12 \text{ cm}$$

The agreement is a useful check, and the first route explains why the answer is exactly the radius: a $60°$ central angle always produces a chord equal to the radius, whatever the circle.

Question 2

Let $A$ and $B$ be two points on a circle with centre $O$.

Solution.

(i) Are there points $X$, $Y$ on the circle, on the same side of $AB$, such that $\angle AXB$ is different from $\angle AYB$?

No. The chord $AB$ divides the circle into two arcs. Any point on the same arc sees $AB$ at the same angle, because each such angle is half the central angle $\angle AOB$:

$$\angle AXB = \frac{1}{2}\angle AOB = \angle AYB$$

The central angle does not depend on where $X$ or $Y$ is, so neither does the inscribed angle. This is the “angles in the same segment” theorem.

(Points on opposite sides of $AB$ do see it differently: those two angles add to $180°$, which is the cyclic-quadrilateral result.)

(ii) Is it true that if $\angle AXB = \angle AYB$, then $X$ and $Y$ lie on the same side of the circle?

Not necessarily. Equal angles usually do mean the same side — but there is one case where points on opposite sides also give equal angles.

Points on opposite arcs give angles that add to $180°$. Those two angles are equal exactly when each is $90°$, which happens precisely when $AB$ is a diameter.

So:

• if $\angle AXB = \angle AYB \neq 90°$, then $X$ and $Y$ are on the same arc;
• if $\angle AXB = \angle AYB = 90°$, then $AB$ is a diameter and $X$, $Y$ may be on either arc — the angle in a semicircle is $90°$ on both sides.

(iii) If $\angle AXB = \angle AYB$, and $X$ and $Y$ do not lie on the circle, does the circle through $A$, $B$ and $X$ also pass through $Y$?

Yes, provided $X$ and $Y$ are on the same side of $AB$. This is the converse of the same-segment theorem, stated in the chapter as Theorem 10.

Why. $A$, $B$, $X$ are three points not in a straight line, so exactly one circle passes through them. Suppose $Y$ were not on it — then $Y$ lies either inside or outside that circle. Extend $AY$ to meet the circle at a point $Z$. Then $\angle AZB$ is an angle in the same segment as $\angle AXB$, so they are equal; but $\angle AYB$ and $\angle AZB$ are angles of a triangle in which one is exterior to the other, so they cannot be equal. That contradicts $\angle AYB = \angle AXB$. Hence $Y$ must lie on the circle.

This is the tool for proving concyclicity: four points lie on one circle whenever a segment joining two of them subtends equal angles at the other two, on the same side.

Question 3

Find $x$ in Fig. 5.26.

Solution. Fig. 5.26 shows a quadrilateral $ABCD$ with all four vertices on the circle — a cyclic quadrilateral. The angle at $D$ is marked $100°$ and the angle at $B$ is marked $x$.

Going round the quadrilateral $A \to B \to C \to D$, the vertices $B$ and $D$ are opposite each other. In a cyclic quadrilateral opposite angles are supplementary:

$$\angle B + \angle D = 180°$$ $$x + 100° = 180°$$ $$x = 80°$$

Why opposite angles add to $180°$. The two angles stand on the same chord $AC$ but from opposite arcs. Each is half its own central angle, and the two central angles together make a full turn of $360°$ — so the two inscribed angles together make half of that, $180°$.

Common mistakes

• Question 1, halving the central angle to get the chord. The angle halves when the perpendicular is dropped; the chord does not. Here the chord is $12$ cm, not $6$ cm.
• Question 1, assuming the answer is a coincidence. A $60°$ central angle always gives a chord equal to the radius, in any circle — the triangle is equilateral every time.
• Question 2(i), thinking the angle depends on the position of $X$. It does not, so long as $X$ stays on the same arc. Moving $X$ along the arc changes the triangle’s shape but never the angle at $X$.
• Question 2(ii), answering a flat “yes”. The diameter case is the exception, and it is the point of the question: two right angles on opposite arcs are also equal.
• Question 2(iii), thinking a second circle is needed. Three non-collinear points determine exactly one circle, so there is no second circle available for $Y$ to be on.
• Question 3, adding adjacent angles. Only opposite angles of a cyclic quadrilateral sum to $180°$. Adjacent ones have no fixed relationship. Reading the vertex order $ABCD$ carefully is what identifies which pair is which.
• Question 3, halving or doubling the $100°$. The doubling rule relates a central angle to an inscribed one. Both $x$ and $100°$ are inscribed angles here, so the supplementary rule applies instead.

Practise next

• The End-of-Chapter Exercises, Questions 2, 6, 7 and 8 — the central–inscribed angle rule, the angle in a semicircle and two cyclic-quadrilateral calculations.
• The End-of-Chapter Exercises, Question 24 — asks you to justify the angle in a semicircle from a figure, which is Question 2(ii)’s special case proved properly.