NCERT Solutions Class 9 Maths Exercise 5.1 — 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 5.1. All 4 questions solved.

A circle is the set of all points at a fixed distance — the radius — from a fixed point, the centre. This first exercise is about what it takes to pin a circle down.

• Through one point: infinitely many circles.
• Through two points $A$ and $B$: still infinitely many, but their centres all lie on the perpendicular bisector of $AB$ — since the centre must be equidistant from both.
• Through three points not in a straight line: exactly one circle, the circumcircle, whose centre is where the perpendicular bisectors of the sides meet.

To construct a circumcircle: draw the perpendicular bisectors of any two sides, mark their intersection $O$, and draw the circle of radius $OA$ centred at $O$.

Key insight. The circumcentre is equidistant from the three vertices, but it need not be inside the triangle. For an acute triangle it lies inside; for a right-angled triangle it lies exactly at the midpoint of the hypotenuse; for an obtuse triangle it falls outside. Questions 1 and 2 are the first two of these cases, and noticing which kind of triangle you have predicts the answer before you draw anything.

Question 1

Draw $\triangle ABC$ with $AB = 5$ cm, $\angle A = 70°$ and $\angle B = 60°$. Draw the circumcircle of $\triangle ABC$. Is the centre inside or outside the triangle?

Solution. Construction.

1. Draw $AB = 5$ cm.
2. At $A$, construct an angle of $70°$; at $B$, construct an angle of $60°$. Where the two arms meet is $C$.
3. Construct the perpendicular bisector of $AB$ and of $BC$.
4. Mark their intersection $O$ — the circumcentre.
5. With centre $O$ and radius $OA$, draw the circle. It passes through $B$ and $C$ as well.

Where is the centre? Work out the third angle first:

$$\angle C = 180° – 70° – 60° = 50°$$

All three angles — $70°$, $60°$, $50°$ — are less than $90°$, so the triangle is acute. The circumcentre of an acute triangle lies inside it, and that is what the drawing shows.

Question 2

Draw $\triangle ABC$ with $AB = 5$ cm, $\angle A = 100°$, $AC = 4$ cm. Draw the circumcircle of $\triangle ABC$. Is the centre inside or outside the triangle?

Solution. Construction.

1. Draw $AB = 5$ cm.
2. At $A$, construct an angle of $100°$.
3. Along that arm, mark $C$ with $AC = 4$ cm.
4. Join $BC$. Construct the perpendicular bisectors of two of the sides and mark their intersection $O$.
5. Draw the circle with centre $O$ and radius $OA$.

Where is the centre? The angle at $A$ is $100°$, which is more than $90°$, so the triangle is obtuse. The circumcentre of an obtuse triangle lies outside it — on the far side of the side opposite the obtuse angle, here beyond $BC$.

The reason is worth seeing: the circumcentre lies on the perpendicular bisector of $BC$, and the obtuse angle at $A$ pushes $A$ so close to the chord $BC$ that the centre ends up on the other side of it.

Question 3

Draw $\triangle ABC$ with $AB = 6$ cm, $BC = 7$ cm and $CA = 7$ cm. Draw the circumcircle of $\triangle ABC$. Let the circumcentre be $O$. Measure $OA$, $OB$, $OC$.

Solution. Construction.

1. Draw $AB = 6$ cm.
2. With centre $A$ and radius $7$ cm, draw an arc; with centre $B$ and radius $7$ cm, draw another. They meet at $C$.
3. Construct the perpendicular bisectors of two sides; they meet at $O$.
4. Draw the circle with centre $O$ and radius $OA$.

Measuring $OA$, $OB$ and $OC$. They come out equal — about $3.9$ cm for this triangle.

They must be equal, and no measurement is really needed to know it: $O$ is the centre of a circle passing through $A$, $B$ and $C$, so each of $OA$, $OB$, $OC$ is a radius of that circle. Measuring is a check on the drawing, not on the mathematics.

The triangle is isosceles ($CA = CB = 7$ cm), so the perpendicular bisector of $AB$ also passes through $C$ — which makes the construction quicker, since that bisector is one of the two lines needed.

Question 4

What is the least possible radius of a circle through two points $A$ and $B$?

Solution. Infinitely many circles pass through $A$ and $B$, with centres all along the perpendicular bisector of $AB$. The question is which of them is smallest.

For any such circle, $AB$ is a chord. No chord of a circle is longer than the diameter, so

$$AB \leq 2r \qquad \text{that is} \qquad r \geq \frac{AB}{2}$$

So no circle through $A$ and $B$ can have radius less than half of $AB$.

Is that value achieved? Yes. Take the centre at the midpoint $M$ of $AB$ and the radius as $\frac{AB}{2}$. Then $MA = MB = \frac{AB}{2}$, so both points lie on the circle, and $AB$ is a diameter of it.

So the smallest possible radius is exactly $\frac{AB}{2}$, attained by the unique circle having $AB$ as a diameter.

Common mistakes

• Questions 1 to 3, using the angle bisectors. Those meet at the incentre, the centre of the circle that touches all three sides. The circumcentre needs the perpendicular bisectors of the sides, because it must be equidistant from the three vertices.
• Question 1, drawing the angles at the wrong vertices. $\angle A = 70°$ is at $A$ and $\angle B = 60°$ at $B$. Swapping them gives a different triangle, though in this case still an acute one.
• Question 2, expecting the centre inside. Obtuse triangles are exactly the case where it is not. Computing the largest angle before drawing predicts the answer.
• Question 3, expecting $OA$, $OB$ and $OC$ to differ. They cannot: all three are radii. If a measurement disagrees, the construction has drifted, not the geometry.
• Question 4, answering $AB$. That is the diameter of the smallest circle, so the radius is half of it.
• Question 4, saying there is no smallest. There is. The centres run along the whole perpendicular bisector, and the radius is smallest at the point of that line closest to $A$ and $B$ — the midpoint of $AB$ itself.

Practise next

• Exercise Set 5.2 — the isosceles triangle formed by a chord and the centre, which is the idea behind almost every proof in the rest of the chapter.
• The End-of-Chapter Exercises, Question 13 — asks you to draw a circle with a chord of given length at a given distance from the centre, which is this construction run backwards.