---
title: "NCERT Solutions Class 9 Maths I’m Up and Down, and Round and Round"
url: https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions
dateModified: 2026-10-07T15:44:30+00:00
---

# NCERT Solutions Class 9 Maths I’m Up and Down, and Round and Round

This chapter's questions cover fundamental properties of circles, including chords, angles subtended by arcs, perpendicular bisectors, concyclicity, and cyclic quadrilaterals.

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## Activity

### Question 1

*Activity*

List some objects from nature that resemble a circle.

**Solution**

1. Observe circular patterns and objects present in the natural world.
2. Examples include cross-sections of plant stems, the full moon, the sun, ripples formed when raindrops fall on water, and the inflorescence of a sunflower.

**Answer:** Objects from nature that resemble a circle include the full moon, the sun, water ripples, and the cross-section of a plant stem.

## Think and Reflect

### Question 1

*2 marks · Very short answer*

Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

**Solution**

1. Fold the circular paper into half along any diameter and make a crease.
2. Unfold the paper and fold it again along a different diameter, then the intersection of the two creases gives the centre of the circle.

**Answer:** Amina suggested folding the circular paper twice along two different diameters; their intersection point is the centre.

> Common mistake: Stating only a single fold, which gives a chord instead of the centre.

## Think and Reflect

### Question 1

*3 marks · Short answer*

What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?

**Solution**

1. A square has rotational symmetry of orders $4$ ($90^\circ$, $180^\circ$, $270^\circ$, $360^\circ$) and $4$ lines of reflection symmetry.
2. A regular pentagon has rotational symmetry of orders $5$ ($72^\circ$ and its multiples) and $5$ lines of reflection symmetry.
3. A regular hexagon has rotational symmetry of orders $6$ ($60^\circ$ and its multiples) and $6$ lines of reflection symmetry.

**Answer:** Square: 4 rotational symmetries, 4 lines of reflection symmetry; regular pentagon: 5, 5; regular hexagon: 6, 6.

> Common mistake: Confusing the number of rotational symmetry angles or orders with lines of symmetry.

### Question 2

*3 marks · Short answer*

What is the length of the longest chord in a circle of radius $5$ units? Is there a smallest chord?

**Solution**

1. The longest chord of a circle is its diameter, which passes through the centre and is equal to twice the radius.
2. For a circle of radius $5$ units, the length of the longest chord is $2 \times 5 = 10$ units.
3. There is no smallest chord because the length of a chord can be made arbitrarily close to zero as the two points on the circle get closer to each other.

**Answer:** The longest chord is $10$ units (diameter); there is no smallest chord.

> Common mistake: Stating that the radius is the smallest chord or giving a non-zero minimum chord length.

### Question 3

*3 marks · Short answer*

The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points? (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)

**Solution**

1. A point that is equidistant from two given points $A$ and $B$ lies on the perpendicular bisector of the line segment $AB$.
2. Conversely, every point lying on the perpendicular bisector of $AB$ is equidistant from $A$ and $B$.
3. Therefore, the locus of points equidistant from two given points is the perpendicular bisector of the line segment joining the two points.

**Answer:** The locus of points equidistant from two given points is the perpendicular bisector of the line segment joining them.

> Common mistake: Stating only that the midpoint is the locus instead of the entire perpendicular bisector line.

## Think and Reflect

### Question 1

*2 marks · Very short answer*

How many circles pass through two points on a plane?

**Solution**

1. Infinitely many circles can pass through two given points A and B on a plane.
2. The centres of all such circles lie on the perpendicular bisector of the line segment AB.

**Answer:** Infinitely many circles.

> Common mistake: Stating that only one or two circles can pass through two points.

### Question 2

*3 marks · Short answer*

Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?

**Solution**

1. Yes, there are circles of all possible radii greater than or equal to half the length of AB passing through A and B.
2. The smallest circle passing through A and B has its centre at the midpoint of AB, with its radius equal to half the length of AB (where AB is the diameter).
3. There is no largest circle, as the radius can grow indefinitely up to infinity (a straight line).

**Answer:** Radii of all lengths $\ge \frac{1}{2}AB$. Smallest radius is $\frac{1}{2}AB$ and largest radius is infinite.

> Common mistake: Forgetting that a straight line can be considered a circle of infinite radius.

### Question 3

*2 marks · Very short answer*

As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?

**Solution**

1. As you move away from segment AB along its perpendicular bisector, the distance from the centre to points A and B increases.
2. Therefore, the radii of the circles containing A and B increase.

**Answer:** The radii increase.

> Common mistake: Stating that the radii decrease as you move away.

### Question 4

*2 marks · Very short answer*

As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?

**Solution**

1. As the radius of the circle increases when moving away along the perpendicular bisector, the circle becomes larger.
2. Consequently, the circle drawn from that point through A and B will appear less curved.

**Answer:** The circle will appear less curved.

> Common mistake: Confusing larger radius with higher curvature.

### Question 5

*3 marks · Short answer*

You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?

**Part (i)**

1. Consider a square of side length less than or equal to the distance between A and B.
2. We can place the square such that A and B lie on its boundary in various positions.
3. By varying the size and position of the square, infinitely many squares can be drawn with A and B on the boundary.

Answer (i): Infinitely many squares.

**Part (ii)**

1. Given two points A and B, the distance between them fixes the side length or diagonal length of the square if they are adjacent or opposite corners.
2. For adjacent corners, the side length is equal to the distance AB, and two such squares can be drawn on either side of AB.
3. For opposite corners, AB is the diagonal, and infinitely many squares can be drawn having AB as a chord/diagonal by rotating the square about the midpoint of AB.

Answer (ii): Infinitely many squares.

**Answer:** Infinitely many squares can be drawn with A and B on the boundary, and infinitely many squares can be drawn with A and B as corners of the square.

> Common mistake: Confusing adjacent and opposite corners of a square with the given distance AB.

## Exercise Set 5.1

### Question 1

*3 marks · Short answer*

Draw $\Delta ABC$ with $AB = 5\text{ cm}$, $\angle A = 70^\circ$ and $\angle B = 60^\circ$. Draw the circumcircle of $\Delta ABC$. Is the centre inside or outside the triangle?

**Solution**

1. Construct triangle $ABC$ with $AB = 5\text{ cm}$, $\angle A = 70^\circ$ and $\angle B = 60^\circ$.
2. Draw perpendicular bisectors of any two sides, say $AB$ and $BC$, to intersect at point $O$.
3. The circumcentre $O$ lies inside the triangle since $\Delta ABC$ is an acute-angled triangle.

**Answer:** The centre is inside the triangle.

> Common mistake: Drawing the bisectors of the angles instead of perpendicular bisectors of the sides.

### Question 2

*3 marks · Short answer*

Draw $\Delta ABC$ with $AB = 5\text{ cm}$, $\angle A = 100^\circ$, $AC = 4\text{ cm}$. Draw the circumcircle of $\Delta ABC$. Is the centre inside or outside the triangle?

**Solution**

1. Construct triangle $ABC$ with $AB = 5\text{ cm}$, $\angle A = 100^\circ$ and $AC = 4\text{ cm}$.
2. Draw perpendicular bisectors of sides $AB$ and $AC$ to intersect at point $O$.
3. The circumcentre $O$ lies outside the triangle since $\Delta ABC$ is an obtuse-angled triangle.

**Answer:** The centre is outside the triangle.

> Common mistake: Confusing acute and obtuse triangle circumcentre locations.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. Construct triangle $ABC$ with sides $AB = 6\text{ cm}$, $BC = 7\text{ cm}$ and $CA = 7\text{ cm}$.
2. Draw the perpendicular bisectors of the sides to locate the circumcentre $O$.
3. Measure the distances from $O$ to the vertices, obtaining $OA = OB = OC$ (approximately $3.9\text{ cm}$).

**Answer:** $OA = OB = OC \approx 3.9\text{ cm}$

> Common mistake: Inaccurate construction leading to unequal distances from the centre to the vertices.

### Question 4

*2 marks · Very short answer*

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

**Solution**

1. The smallest circle passing through two points $A$ and $B$ has the line segment $AB$ as its diameter.
2. The radius of this circle is half the length of $AB$.

**Answer:** Half the distance between A and B.

> Common mistake: Stating the radius equals the full length of AB instead of half.

## Think, Draw and Infer

### Question 1

*4 marks · Proof*

A, B and C are three collinear points. Can you find a point P such that $PA = PB = PC$? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?

**Solution**

1. Let A, B and C be three collinear points on a line.
2. Suppose there is a point P such that $PA = PB = PC$.
3. Since $PA = PB$, point P must lie on the perpendicular bisector of segment AB.
4. Since $PB = PC$, point P must lie on the perpendicular bisector of segment BC.
5. The perpendicular bisectors of AB and BC are parallel because A, B and C are collinear, so two parallel lines cannot intersect at P.
6. Thus, no such point P exists, the perpendicular bisectors of AB and BC are parallel, no circle can pass through three collinear points, and a straight line can intersect a circle in at most two distinct points.
7. Hence proved.

**Answer:** No such point P exists, the perpendicular bisectors are parallel, no circle can pass through collinear points, and a line cannot cut a circle in three distinct points.

> Common mistake: Assuming three collinear points can form a triangle or a circle.

### Question 2

*3 marks · Short answer*

The circumcircle of a given $\Delta ABC$ is drawn. Can there be other triangles congruent to $\Delta ABC$ that share the same circumcircle?

**Solution**

1. Yes, there can be other triangles congruent to $\Delta ABC$ that share the same circumcircle.
2. Since the circumcircle is fixed by the three vertices A, B and C, any triangle whose vertices lie on the same circle and whose side lengths are equal to the corresponding sides of $\Delta ABC$ will be congruent to $\Delta ABC$ by SSS congruence.
3. By rotating or reflecting the triangle's vertices along the circumference of the same circumcircle, we obtain infinitely many distinct triangles that are congruent to $\Delta ABC$ and share the same circumcircle.

**Answer:** Yes, other triangles congruent to $\Delta ABC$ can share the same circumcircle by positioning their vertices at different points along the circumference such that their side lengths remain equal.

> Common mistake: Thinking that fixing the circumcircle restricts the triangle's orientation to only one position.

## Exercise Set 5.2

### Question 1

*3 marks · Proof*

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

**Solution**

1. Let AB be a chord of a circle with centre C.
2. Join CA and CB to form ∃CAB.
3. Since CA and CB are both radii of the same circle, $CA = CB$.
4. Therefore, ∃CAB is an isosceles triangle.
5. Hence proved.

**Answer:** Hence proved that the triangle formed by a chord and the centre is isosceles.

> Common mistake: Forgetting to state that the two sides are radii of the same circle.

### Question 2

*4 marks · Proof*

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

**Solution**

1. Let $\Delta CAB$ and $\Delta C'A'B'$ be two triangles formed by chords $AB$ and $A'B'$ with their respective centres $C$ and $C'$.
2. We are given that the base lengths are equal, so $AB = A'B'$.
3. The other two sides of each triangle are radii of the circles, so $CA = CB = C'A' = C'B'$.
4. By SSS congruence, $\Delta CAB \cong \Delta C'A'B'$.
5. Hence proved.

**Answer:** Hence proved that the two isosceles triangles are congruent to each other.

> Common mistake: Assuming the circles have the same radius without justification if not specified, though base equality and radii properties apply.

## Exercise Set 5.3

### Question 1

*3 marks · Proof*

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? (Hint: Use Fig. 5.12. You are told that $\angle CMA = \angle CMB = 90^\circ$. You need to show that $AM = BM$.)

**Solution**

1. Given: A circle with centre C, chord AB, and CM $\perp$ AB such that $\angle CMA = \angle CMB = 90^\circ$.
2. To show: AM = BM.
3. In $\Delta CMA$ and $\Delta CMB$, CA = CB (radii of the circle).
4. Side $\text{CM} = \text{CM}$ (common side).
5. Angle $\angle CMA = \angle CMB = 90^\circ$ (given).
6. By RHS congruence, $\Delta CMA \cong \Delta CMB$.
7. Hence, AM = BM (corresponding parts of congruent triangles).
8. Therefore, the perpendicular from the centre to a chord bisects the chord. Hence proved.

**Answer:** Hence proved.

> Common mistake: Using SAS instead of RHS congruence.

### Question 2

*4 marks · Proof*

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**

1. Given: An isosceles triangle ABC inscribed in a circle with AB = AC, and AD as the altitude from A to BC.
2. To show: The altitude AD passes through the centre of the circle.
3. Since AB = AC, $\Delta ABC$ is isosceles with base BC.
4. The perpendicular bisector of the base of an isosceles triangle passes through the opposite vertex.
5. Therefore, the perpendicular bisector of BC passes through A and is the altitude AD.
6. The centre of the circumcircle of $\Delta ABC$ lies on the perpendicular bisector of BC.
7. Since AD is the perpendicular bisector of BC, the centre of the circle must lie on AD.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Assuming the centre lies on the altitude without stating that the altitude of an isosceles triangle is also the perpendicular bisector of the base.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. Given: Radius $r = 5\text{ cm}$, lengths of parallel chords $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$ on opposite sides of the centre.
2. Let O be the centre. Draw perpendiculars from O to both chords, meeting them at midpoints M and N.
3. Since the perpendicular from the centre bisects the chord, $\text{AM} = \frac{1}{2}AB = 3\text{ cm}$ and $\text{CN} = \frac{1}{2}CD = 4\text{ cm}$.
4. In right-angled triangle $\Delta OMA$, $\text{OM} = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ cm}$.
5. In right-angled triangle $\Delta ONC$, $\text{ON} = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3\text{ cm}$.
6. Since the chords are on opposite sides of the centre, the distance between their midpoints is $\text{OM} + \text{ON} = 4\text{ cm} + 3\text{ cm} = 7\text{ cm}$.

**Answer:** $7\text{ cm}$

> Common mistake: Subtracting the distances instead of adding them when chords are on opposite sides of the centre.

## Exercise Set 5.4

### Question 1

*4 marks · Proof*

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

**Solution**

1. Let a circle have centre C and chords AB and FG of equal length, so that $AB = FG$.
2. Let E and H be the midpoints of AB and FG respectively, so that $CE$ and $CH$ are perpendicular to AB and FG.
3. By Theorem 5, $AE = \frac{1}{2}AB$ and $FH = \frac{1}{2}FG$. Since $AB = FG$, we have $AE = FH$.
4. Consider right-angled triangles $\Delta CEA$ and $\Delta CHF$. By the Baudhāyana–Pythagoras theorem, $AC^2 = CE^2 + AE^2$ and $CF^2 = CH^2 + FH^2$.
5. Since $AC = CF$ (both are radii) and $AE = FH$, it follows that $CE^2 = CH^2$, giving $CE = CH$. Hence, chords of equal length are equidistant from the centre. Thus proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to state that radii of the same circle are equal.

### Question 2

*4 marks · Proof*

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

**Solution**

1. Given a circle with centre C, $CE \perp AB$, $CH \perp GH$ (where $GH$ is denoted as $FG$ in the theorem statement), and $CE = CH$.
2. From Theorem 5, the perpendicular from the centre bisects the chord, so $AE = \frac{1}{2}AB$ and $FH = \frac{1}{2}FG$.
3. Join CA and CF. Here $CA = CF$ since both are radii of the circle.
4. Consider right-angled triangles $\Delta CEA$ and $\Delta CHF$. We have $\angle CEA = \angle CHF = 90^\circ$, hypotenuse $CA = CF$, and side $CE = CH$.
5. By RHS congruence, $\Delta CEA \cong \Delta CHF$. Therefore, $AE = FH$. Since $AB = 2AE$ and $FG = 2FH$, it follows that $AB = FG$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Using SSS instead of RHS congruence when the hypotenuse and one side are known.

### Question 3

*4 marks · Proof*

Solve the previous question using the Baudhāyana–Pythagoras theorem.

**Solution**

1. Let the circle have centre C and radius $r$. We are given $CE = CH = d$, where $CE \perp AB$ and $CH \perp FG$.
2. By the Baudhāyana–Pythagoras theorem in right-angled triangle $\Delta CEA$, $AC^2 = CE^2 + AE^2$, which gives $r^2 = d^2 + AE^2$.
3. Similarly, in right-angled triangle $\Delta CHF$, $CF^2 = CH^2 + FH^2$, which gives $r^2 = d^2 + FH^2$.
4. Equating the expressions for $r^2$, we get $d^2 + AE^2 = d^2 + FH^2$, so $AE^2 = FH^2$, which means $AE = FH$.
5. Since $AB = 2AE$ and $FG = 2FH$ (perpendicular from centre bisects the chord), we obtain $AB = FG$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not explicitly stating that the perpendicular from the centre bisects the chord to relate $AE$ to $AB$.

## Exercise Set 5.5

### Question 1

*3 marks · Short answer*

Find the length of the chord of a circle where the radius is $7\text{ cm}$ and perpendicular distance is $6\text{ cm}$.

**Solution**

1. Given: Radius $r = 7\text{ cm}$, perpendicular distance from centre $d = 6\text{ cm}$.
2. By the Baudhāyana--Pythagoras theorem applied to half the chord, radius, and perpendicular distance: half chord length $= \sqrt{r^2 - d^2} = \sqrt{7^2 - 6^2} = \sqrt{49 - 36} = \sqrt{13}\text{ cm}$.
3. The total length of the chord is twice the length of half the chord.
4. Result: $2\sqrt{13}\text{ cm}$.

**Answer:** $2\sqrt{13}\text{ cm}$

> Common mistake: Forgetting to multiply the half-chord length by 2 to get the full chord length.

### Question 2

*3 marks · Proof*

Explain why the following statement is true: If the perpendicular distance of a chord from the centre is $d$ and the radius is $r$, then the chord length is $2\sqrt{r^2 - d^2}$.

**Solution**

1. Let the circle have centre $C$, radius $r$, and let AB be a chord.
2. Let $M$ be the midpoint of AB, so that CM is perpendicular to AB with length $d$.
3. Join CA to form right-angled triangle CMA, where $\angle CMA = 90^\circ$ and hypotenuse $CA = r$.
4. By the Baudhāyana--Pythagoras Theorem in $\Delta CMA$, $AM^2 + CM^2 = CA^2$.
5. Substitute $CM = d$ and $CA = r$: $AM^2 + d^2 = r^2$, which gives $AM = \sqrt{r^2 - d^2}$.
6. Since the perpendicular from the centre bisects the chord, $AB = 2 \times AM = 2\sqrt{r^2 - d^2}$.
7. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not stating that the perpendicular from the centre bisects the chord.

### Question 3

*3 marks · Short answer*

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that $CD = 2 AB$? Give reasons for your answer.

**Solution**

1. Let the distance of chord AB from the centre be $d_1$ and the distance of chord CD be $d_2$. We are given that $d_1 = 2d_2$.
2. From the formula for chord length, $AB = 2\sqrt{r^2 - d_1^2}$ and $CD = 2\sqrt{r^2 - d_2^2}$.
3. Substituting $d_1 = 2d_2$, we get $AB = 2\sqrt{r^2 - 4d_2^2}$, whereas $2CD = 4\sqrt{r^2 - d_2^2}$.
4. Since $\sqrt{r^2 - 4d_2^2} \neq 2\sqrt{r^2 - d_2^2}$, we cannot conclude that $CD = 2AB$.

**Answer:** No, we cannot conclude that $CD = 2AB$.

> Common mistake: Assuming linear proportionality between chord length and its distance from the centre.

## Exercise

### Question 1

*3 marks · Short answer*

A circle with centre O is drawn, and A, B, C, D are points on the circle (see Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than $180^\circ$, it is a minor arc. If the angle at the centre is greater than $180^\circ$, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

**Solution**

1. An arc is a minor arc if the angle it subtends at the centre is less than $180^\circ$.
2. An arc is a major arc if the angle it subtends at the centre is greater than $180^\circ$.
3. Based on Fig. 5.19, arc AKB subtends an angle less than $180^\circ$ at the centre O, so arc AKB is a minor arc, and arc CLD subtends an angle greater than $180^\circ$ at the centre O, so arc CLD is a major arc.

**Answer:** Arc AKB is a minor arc and arc CLD is a major arc.

> Common mistake: Confusing minor and major arcs by looking at the arc length instead of the central angle.

## Exercise Set 5.6

### Question 1

*3 marks · Short answer*

In a circle with centre O, the central angle AOB is $60^\circ$. If the radius of the circle is $12\text{ cm}$, what is the length of the chord AB?

**Solution**

1. Given: radius $OA = OB = 12\text{ cm}$, central angle $\angle AOB = 60^\circ$.
2. In $\triangle OAB$, $OA = OB$, so $\angle OAB = \angle OBA$.
3. Since $\angle AOB = 60^\circ$, the sum of the remaining two angles is $180^\circ - 60^\circ = 120^\circ$.
4. Thus, $\angle OAB = \angle OBA = \frac{120^\circ}{2} = 60^\circ$, making $\triangle OAB$ an equilateral triangle.
5. Therefore, the length of the chord $AB = OA = OB = 12\text{ cm}$.

**Answer:** 12 cm

> Common mistake: Assuming chord length is equal to radius without proving the triangle is equilateral.

### Question 2

*3 marks · Case-based*

Let A and B be two points on a circle with centre O. (i) Are there points X, Y on the circle, on the same side of AB, such that $\angle AXB$ is different from $\angle AYB$? (ii) Is it true that if $\angle AXB = \angle AYB$, then X and Y lie on the same side of the circle? (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?

**Part (i)**

1. Angles in the same segment of a circle are equal.
2. Therefore, for any points X and Y on the circle on the same side of AB, $\angle AXB = \angle AYB$.
3. Thus, there are no such points where the angles are different.

Answer (i): No

**Part (ii)**

1. If $\angle AXB = \angle AYB$, then X and Y subtend equal angles at the circumference from the chord AB.
2. By the properties of arcs and angles, X and Y must lie on the same arc segment of the circle.
3. Therefore, it is true that they lie on the same side of the circle.

Answer (ii): Yes

**Part (iii)**

1. If X and Y do not lie on the circle, equality of angles $\angle AXB = \angle AYB$ only implies that A, B, X, Y are concyclic.
2. The circle passing through A, B and X is unique by Theorem 1.
3. Since Y does not lie on the circle initially, the circle through A, B and X does not necessarily pass through Y unless Y is on that circle.

Answer (iii): No

**Answer:** (i) No, (ii) Yes, (iii) No

> Common mistake: Confusing points on the circle with points outside the circle.

### Question 3

*3 marks · Short answer*

Find $x$ in Fig. 5.26.

**Solution**

1. Consider the cyclic quadrilateral formed by the four points on the circle in Fig. 5.26.
2. The angle opposite to the angle measuring $100^\circ$ is $x$.
3. The sum of opposite angles of a cyclic quadrilateral is $180^\circ$, so $x + 100^\circ = 180^\circ$.

**Answer:** $80^\circ$

> Common mistake: Confusing the opposite angles of a cyclic quadrilateral.

## Exercise

### Question 1

*3 marks · Short answer*

A cyclic quadrilateral has angles measuring $\angle A = 80^\circ$, $\angle B = 110^\circ$, $\angle C = 100^\circ$, and $\angle D = 70^\circ$. Can such a quadrilateral be drawn? Explain why or why not.

**Solution**

1. In a cyclic quadrilateral, the sum of opposite pairs of angles must be equal to $180^\circ$.
2. Check the sum of opposite angles $\angle A$ and $\angle C$: $\angle A + \angle C = 80^\circ + 100^\circ = 180^\circ$.
3. Check the sum of opposite angles $\angle B$ and $\angle D$: $\angle B + \angle D = 110^\circ + 70^\circ = 180^\circ$.
4. Since both pairs of opposite angles add up to $180^\circ$, such a cyclic quadrilateral can be drawn.

**Answer:** Yes, such a quadrilateral can be drawn because the sum of each pair of opposite angles is $180^\circ$.

> Common mistake: Forgetting that both pairs of opposite angles must sum to $180^\circ$, not just one pair.

## End-of-Chapter Exercises

### Question 1

*3 marks · Short answer*

In a circle, a chord is $5\text{ cm}$ away from the centre. If the radius of the circle is $13\text{ cm}$, what is the length of the chord?

**Solution**

1. Given: Radius $r = 13\text{ cm}$, perpendicular distance from centre $d = 5\text{ cm}$.
2. The perpendicular from the centre to a chord bisects the chord, forming a right-angled triangle with the radius as hypotenuse.
3. Half-chord length $x = \sqrt{r^2 - d^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$.
4. Length of the chord $= 2 \times 12 = 24\text{ cm}$.

**Answer:** 24 cm

> Common mistake: Forgeting to multiply half the chord length by 2 to get the total length of the chord.

### Question 2

*3 marks · Short answer*

An arc of a circle subtends an angle of $70^\circ$ at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

**Solution**

1. Given: Central angle $= 70^\circ$.
2. By Theorem 9, the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the circle outside the arc.
3. Angle at the point on the circle $= \frac{70^\circ}{2} = 35^\circ$.

**Answer:** 35°

> Common mistake: Multiplying the central angle by 2 instead of dividing.

### Question 3

*3 marks · Short answer*

The diameter of a circle is $26\text{ cm}$. A chord of length $24\text{ cm}$ is drawn in the circle. Find the distance from the centre of the circle to the chord.

**Solution**

1. Given: Diameter $= 26\text{ cm}$, so radius $r = 13\text{ cm}$. Chord length $= 24\text{ cm}$, so half-chord length $x = 12\text{ cm}$.
2. The perpendicular from the centre bisects the chord and forms a right-angled triangle with the radius and distance $d$.
3. Using the Baudhāyana-Pythagoras theorem: $d^2 = r^2 - x^2 = 13^2 - 12^2 = 169 - 144 = 25$.
4. Distance from the centre $d = \sqrt{25} = 5\text{ cm}$.

**Answer:** 5 cm

> Common mistake: Using the diameter instead of the radius in the Pythagoras theorem calculation.

### Question 4

*3 marks · Short answer*

A circle has a radius of $15\text{ cm}$. A chord is drawn. The distance from the centre of the circle to the chord is $9\text{ cm}$. What is the length of the chord?

**Solution**

1. Given: Radius $r = 15\text{ cm}$, distance from centre $d = 9\text{ cm}$.
2. Let half the chord length be $x$. By the Baudhāyana-Pythagoras theorem, $x^2 = r^2 - d^2$.
3. $x^2 = 15^2 - 9^2 = 225 - 81 = 144$, so $x = 12\text{ cm}$.
4. Length of the chord $= 2x = 2 \times 12 = 24\text{ cm}$.

**Answer:** 24 cm

> Common mistake: Forgetting to double the half-chord length.

### Question 5

*4 marks · Proof*

Prove that the perpendicular bisector of a chord passes through the centre of the circle.

**Solution**

1. Given: A circle with centre O, chord AB, and line $l$ which is the perpendicular bisector of AB meeting AB at M.
2. To prove: Line $l$ passes through the centre O.
3. Join OA and OB.
4. In $\triangle OMA$ and $\triangle OMB$, $AM = BM$ (since $l$ bisects AB), $\angle OMA = \angle OMB = 90^\circ$ (since $l$ is perpendicular to AB), and $OM = OM$ (common side).
5. By SAS congruence, $\triangle OMA \cong \triangle OMB$.
6. Therefore, $OA = OB$.
7. Since the centre is the unique point equidistant from the endpoints of chord AB, and O is the centre, the perpendicular bisector passing through M must pass through O. Hence proved.

**Answer:** Hence proved.

> Common mistake: Assuming O lies on the line beforehand instead of proving triangle congruence first.

### Question 6

*3 marks · Short answer*

The diameter of a circle is AB. Point C is on the circumference. What is the measure of the $\angle ACB$? Explain your reasoning.

**Solution**

1. Given: Diameter AB of a circle with centre O, and point C on the circumference.
2. Reasoning: The diameter AB subtends a straight angle ($180^\circ$) at the centre O.
3. By the corollary of Theorem 9, the angle subtended by a diameter at any point on the circle is half of the central angle.
4. Therefore, $\angle ACB = \frac{1}{2} \times 180^\circ = 90^\circ$.

**Answer:** 90°

> Common mistake: Confusing the central angle of a semicircle with the inscribed angle.

### Question 7

*3 marks · Short answer*

ABCD is a cyclic quadrilateral inscribed in a circle. If $\angle A$ measures $75^\circ$, what is the measure of $\angle C$? If $\angle B$ measures $110^\circ$, what is the measure of $\angle D$?

**Part (i)**

1. We know that the sum of opposite angles of a cyclic quadrilateral is 180 degrees.
2. So, angle A + angle C = 180 degrees.
3. 75 degrees + angle C = 180 degrees, which gives angle C = 105 degrees.

Answer (i): 105^\circ

**Part (ii)**

1. We know that the sum of opposite angles of a cyclic quadrilateral is 180 degrees.
2. So, angle B + angle D = 180 degrees.
3. 110 degrees + angle D = 180 degrees, which gives angle D = 70 degrees.

Answer (ii): 70^\circ

**Answer:** angle C = 105 degrees and angle D = 70 degrees

> Common mistake: Confusing adjacent angles with opposite angles in a cyclic quadrilateral.

### Question 8

*3 marks · Short answer*

Quadrilateral PQRS is inscribed in a circle. If $\angle P = (2x + 10)^\circ$ and $\angle R = (3x - 20)^\circ$, find the value of $x$ and the measures of $\angle P$ and $\angle R$.

**Solution**

1. Since quadrilateral PQRS is inscribed in a circle, it is a cyclic quadrilateral, so the sum of its opposite angles is $180^\circ$.
2. Therefore, $\angle P + \angle R = 180^\circ$, which gives $(2x + 10)^\circ + (3x - 20)^\circ = 180^\circ$.
3. Solving for $x$: $5x - 10 = 180 \implies 5x = 190 \implies x = 38$.
4. Substituting $x = 38$, $\angle P = 2(38) + 10 = 86^\circ$ and $\angle R = 3(38) - 20 = 94^\circ$.

**Answer:** $x = 38$, $\angle P = 86^\circ$, $\angle R = 94^\circ$

> Common mistake: Adding adjacent angles instead of opposite angles to $180^\circ$.

### Question 9

*3 marks · Short answer*

The distance of a chord of length $16\text{ cm}$ from the centre of a circle is $6\text{ cm}$. Find the radius of the circle.

**Solution**

1. Let $AB$ be the chord of length $16\text{ cm}$ and $OM$ be the perpendicular from the centre $O$ to $AB$, so $OM = 6\text{ cm}$.
2. The perpendicular from the centre to a chord bisects the chord, so $AM = \frac{16}{2} = 8\text{ cm}$.
3. In right-angled triangle $OMA$, using the Baudhāyana-Pythagoras theorem, $OA^2 = OM^2 + AM^2$.
4. Substituting the values, $OA^2 = 6^2 + 8^2 = 36 + 64 = 100$, so $OA = \sqrt{100} = 10\text{ cm}$.

**Answer:** $10\text{ cm}$

> Common mistake: Using the full chord length instead of half the chord length in the Pythagorean theorem.

### Question 10

*3 marks · Short answer*

A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

**Solution**

1. A cyclic quadrilateral with side lengths $a, b, c, d$ can be inscribed in a circle, and its area is given by Brahmagupta's formula $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ is the semi-perimeter.
2. Here the sides are $5, 5, 12, 12$, so the semi-perimeter $s = \frac{5 + 5 + 12 + 12}{2} = 17$.
3. Calculating the area: $\text{Area} = \sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5}$.
4. Simplifying the square root gives $\text{Area} = 12 \times 5 = 60\text{ square units}$.

**Answer:** $60\text{ square units}$

> Common mistake: Using standard rectangle or parallelogram area formulas instead of Brahmagupta's formula for a cyclic quadrilateral.

### Question 11

*3 marks · Short answer*

Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?

**Solution**

1. Draw the perpendicular bisectors of any two adjacent sides or chords formed by the vertices of the cyclic quadrilateral.
2. The intersection point of these perpendicular bisectors gives the centre of the circumcircle.
3. By observing whether this intersection point lies inside the region bounded by the quadrilateral's sides or outside, we can determine its location without drawing the full circle.

**Answer:** By finding the intersection of the perpendicular bisectors of the sides.

> Common mistake: Trying to draw the circle blindly without geometric constructions.

### Question 12

*4 marks · Proof*

When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

**Solution**

1. Given: Two intersecting chords AB and CD of equal length in a circle with centre O.
2. To show: The line segments of one chord are equal to the corresponding line segments of the other chord.
3. Draw perpendiculars from the centre O to the two chords AB and CD, meeting them at M and N respectively.
4. Since the chords are equal in length (AB = CD), by Theorem 6 they are equidistant from the centre, so OM = ON.
5. Consider the right triangles formed by joining the centre O to the point of intersection of the chords and using the perpendicular distances.
6. Using congruence of triangles and the perpendicular bisector properties, the segments from the intersection point to the endpoints of equal chords are corresponding and equal. Hence proved.

**Answer:** Hence proved.

> Common mistake: Failing to drop perpendiculars from the centre to establish distances.

### Question 13

*3 marks · Short answer*

Draw a circle in which a chord of $6\text{ cm}$ length stands at a distance of $3\text{ cm}$ from the centre. (Hint: Is it a circumcircle of a suitable triangle?)

**Solution**

1. Given: A circle with a chord of length $6\text{ cm}$ at a distance of $3\text{ cm}$ from the centre.
2. Let AB be the chord of length $6\text{ cm}$ and let C be the centre, with perpendicular distance $CM = 3\text{ cm}$.
3. The perpendicular from the centre bisects the chord, so $AM = MB = \frac{6}{2} = 3\text{ cm}$.
4. Using the Baudhāyana-Pythagoras theorem in right-angled triangle CMA, $AC = \sqrt{AM^2 + CM^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\text{ cm}$.
5. Draw a circle of radius $3\text{\sqrt{2}}\,\text{cm}$, and mark a chord of length $6\text{ cm}$ at distance $3\text{ cm}$ from the centre.

**Answer:** A circle of radius $3\sqrt{2}\text{ cm}$ containing the given chord.

> Common mistake: Confusing the chord length with the segment length AM.

### Question 14

*4 marks · Proof*

Show that rectangle is the only parallelogram that can be inscribed in a circle.

**Solution**

1. Given: A parallelogram ABCD inscribed in a circle.
2. To show: ABCD is a rectangle.
3. Since ABCD is a cyclic parallelogram, the opposite angles are equal ($\angle A = \lfloor C$ and $\angle B = \lfloor D$) and their sum is $180^\circ$.
4. Therefore, $\angle A + \angle C = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ$.
5. A parallelogram with one interior angle equal to $90^\circ$ is a rectangle.
6. Hence proved.

**Answer:** A parallelogram inscribed in a circle must be a rectangle.

> Common mistake: Forgetting to state that opposite angles of a cyclic quadrilateral add up to $180^\circ$.

### Question 15

*4 marks · Proof*

Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

**Solution**

1. Given: A rectangle ABCD inscribed in a circle.
2. To show: The point of intersection of its diagonals lies at the centre of the circle.
3. The diagonals of a rectangle are equal in length and bisect each other at the point of intersection, say O.
4. Since the rectangle is inscribed in the circle, its vertices are equidistant from the centre of the circumcircle.
5. The circumcentre of a right-angled triangle or rectangle lies at the midpoint of its hypotenuse or diagonal.
6. Thus, the intersection point of the diagonals coincides with the centre of the circle. Hence proved.

**Answer:** The intersection of the diagonals of an inscribed rectangle lies at the centre.

> Common mistake: Assuming the centre is the intersection without referencing the circumcentre of triangles.

### Question 16

*3 marks · Short answer*

Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

**Solution**

1. Consider all chords of a fixed length in a given circle.
2. By Theorem 6, chords of equal length are all at the same distance from the centre of the circle.
3. Since the distance from the centre to each chord is constant, the locus of the midpoints of these equal chords is another circle concentric with the original circle.

**Answer:** A concentric circle.

> Common mistake: Stating that the shape is a straight line instead of a concentric circle.

### Question 17

*4 marks · Proof*

In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of $\angle BAC$”.

**Solution**

1. Given: A circle with centre O, and congruent chords AB and AC.
2. To show: The centre O lies on the angle bisector of $\angle BAC$.
3. Join OB, OC, and draw the line segment from O to the midpoint of BC or use triangle congruence.
4. Consider $\triangle OAB$ and $\triangle OAC$: $OA = OA$ (common), $OB = OC$ (radii), and $AB = AC$ (given).
5. By SSS congruence, $\triangle OAB \cong \triangle OAC$.
6. Therefore, $\angle BAO = \\angle CAO$ (corresponding parts of congruent triangles).
7. Thus, AO is the angle bisector of $\angle BAC$, meaning the centre O lies on the angle bisector. Hence proved.

**Answer:** The centre O lies on the angle bisector of $\angle BAC$.

> Common mistake: Not explicitly stating the SSS congruence conditions for triangles OAB and OAC.

### Question 18

*3 marks · Short answer*

Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ are on the same side of the centre of a circle. The distance between the chords is $7\text{ cm}$. Find the radius of the circle.

**Solution**

1. Given: Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ on the same side of the centre, with distance between them equal to $7\text{ cm}$.
2. Let the radius of the circle be $r$. Let the perpendicular distance from the centre to the chord of length $10\text{ cm}$ be $x$, so the distance to the chord of length $24\text{ cm}$ is $x + 7$.
3. Half the lengths of the chords are $5\text{ cm}$ and $12\text{ cm}$ respectively.
4. Using the Baudhāyana-Pythagoras theorem for both chords: $r^2 = x^2 + 5^2$ and $r^2 = (x + 7)^2 + 12^2$.
5. Equating the two expressions for $r^2$: $x^2 + 25 = (x + 7)^2 + 144 \implies x^2 + 25 = x^2 + 14x + 49 + 144$.
6. Simplifying: $14x = 25 - 193 = -168$, giving $x = 5\text{ cm}$ (taking positive side or setting distances correctly: distance from centre to longer chord is smaller, so let distance to $24\text{ cm}$ chord be $x$, and to $10\text{ cm}$ chord be $x+7$).
7. Re-assigning: distance to $24\text{ cm}$ chord is $x$, distance to $10\text{ cm}$ chord is $x+7$. Then $r^2 = x^2 + 12^2$ and $r^2 = (x+7)^2 + 5^2$.
8. Expanding: $x^2 + 144 = x^2 + 14x + 49 + 25 \implies 144 = 14x + 74 \implies 14x = 70 \implies x = 5\text{ cm}$.
9. Substitute $x = 5$ into $r^2 = x^2 + 12^2 = 5^2 + 12^2 = 25 + 144 = 169$.
10. Result: $r = \sqrt{169} = 13\text{ cm}$.

**Answer:** $13\text{ cm}$

> Common mistake: Placing the shorter chord closer to the centre instead of the longer chord.

### Question 19

*3 marks · Short answer*

A regular hexagon is inscribed in a circle of radius $r$. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

**Solution**

1. Given: A regular hexagon inscribed in a circle of radius $r$.
2. The vertices of a regular hexagon divide the circle into 6 equal arcs, subtending an angle of $\frac{360^\circ}{6} = 60^\circ$ at the centre.
3. In the triangle formed by the centre and two adjacent vertices of the hexagon, two sides are radii $r$ and the included angle is $60^\circ$, making it an equilateral triangle. Thus, the length of each side of the hexagon is equal to the radius $r$.
4. The distance of each side from the centre is the perpendicular bisector length from the centre to the side, which forms a right-angled triangle with hypotenuse $r$ and base $\frac{r}{2}$.
5. By the Baudhāyana--Pythagoras theorem, the distance is $\sqrt{r^2 - \left(\frac{r}{2}\right)^2} = \frac{\sqrt{3}}{2} r$.

**Answer:** Side length = $r$, Distance from centre = $\frac{\sqrt{3}}{2} r$

> Common mistake: Confusing the side length of the hexagon with the radius or miscalculating the perpendicular distance.

### Question 20

*3 marks · Short answer*

A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about $\angle MOP$ and $\angle MNP$? Explain your reasoning.

**Solution**

1. Given: Quadrilateral MNOP inscribed in a circle with MN as a diameter.
2. Since MN is a diameter, it subtends a right angle at any point on the circle. Point P lies on the circle, so $\angle MPN = 90^\circ$.
3. The angle $\angle MOP$ is the central angle subtended by the arc MP, while $\angle MNP$ is the angle subtended by the same arc at the circumference, so $\angle MOP = 2\angle MNP$.
4. Alternatively, $\triangle MOP$ is isosceles with two radii as sides, and $\angle MOP$ depends on the position of P.

**Answer:** $\angle MOP = 2\angle MNP$ and $\angle MPN = 90^\circ$

> Common mistake: Forgetting that the angle subtended by a diameter at the circumference is $90^\circ$.

### Question 21

*4 marks · Proof*

Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle CDE = \angle ABC$, where E is a point on the extension of side CD).

**Solution**

1. Given: ABCD is a cyclic quadrilateral and E is a point on the extension of side CD.
2. To show: $\angle CDE = \angle ABC$.
3. Proof: The opposite angles of a cyclic quadrilateral add up to $180^\circ$. So, $\angle ADC + \angle ABC = 180^\circ$.
4. Also, the angles on a straight line CD--E sum to $180^\circ$. So, $\angle ADC + \angle CDE = 180^\circ$.
5. Comparing the two equations, $\angle ADC + \angle ABC = \angle ADC + \angle CDE$.
6. Subtracting $\angle ADC$ from both sides gives $\angle CDE = \angle ABC$.
7. Hence proved.

**Answer:** $\angle CDE = \angle ABC$

> Common mistake: Not stating the linear pair property clearly for the exterior angle.

### Question 22

*3 marks · Proof*

“There is no chord of a circle that is longer than its diameter.” How do you justify this statement?

**Solution**

1. Given: A circle with centre O and any chord AB.
2. To show: $\text{Chord } AB \le \text{Diameter}$.
3. Proof: Join the centre O to the endpoints A and B to form $\triangle OAB$.
4. By the triangle inequality in $\triangle OAB$, the sum of any two sides is greater than the third side: $OA + OB > AB$.
5. Since $OA$ and $OB$ are both radii of the circle, their sum $OA + OB = 2r$, which is the diameter of the circle.
6. Therefore, $AB < 2r$ for any chord not passing through the centre, and $AB = 2r$ if the chord passes through the centre (diameter).
7. Hence, no chord can be longer than the diameter. Hence proved.

**Answer:** Chord length is always less than or equal to the diameter.

> Common mistake: Failing to cite the triangle inequality theorem properly.

### Question 23

*4 marks · Proof*

Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.

**Solution**

1. Given: A circle with centre O and a point A within the circle.
2. To show: The shortest chord passing through A is perpendicular to OA.
3. Proof: Let PQ be any chord passing through A, and let OC be the perpendicular from the centre O to PQ.
4. In right-angled triangle OCA, $OA$ is the hypotenuse, so $OC \le OA$, with $OC = OA$ only when point A coincides with the midpoint of the chord (making PQ perpendicular to OA).
5. We know that the length of a chord is inversely proportional to its distance from the centre: shorter distance means a longer chord, and a larger distance means a shorter chord.
6. Since the perpendicular distance from the centre to PQ is the minimum possible distance when A is the foot of the perpendicular from O, the chord perpendicular to OA is the shortest chord. Hence proved.

**Answer:** The shortest chord through an interior point A is perpendicular to OA.

> Common mistake: Confusing shortest chord with longest chord (diameter).

### Question 24

*4 marks · Proof*

How would you use the following figure to justify the statement that the angle in a semicircle is $90^\circ$?

**Solution**

1. Given: A circle with centre O, diameter AB, and a point C on the circumference forming $\triangle ABC$ (referring to Fig. 5.30).
2. To show: $\angle ACB = 90^\circ$.
3. Proof: The straight angle subtended by the diameter AB at the centre O is $180^\circ$.
4. By Theorem 9, the angle subtended by an arc at any point on the remaining part of the circle is half of the angle subtended at the centre.
5. Therefore, $\angle ACB = \frac{1}{2} \times 180^\circ = 90^\circ$.
6. Hence proved.

**Answer:** $\angle ACB = 90^\circ$

> Common mistake: Not mentioning Theorem 9 or the straight angle measure of $180^\circ$.

### Question 25

*5 marks · Proof*

In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C' D' is perpendicular to AB.

**Solution**

1. Let O be the centre of the circle and AB be the diameter.
2. Chords CC' and DD' are perpendicular to the diameter AB.
3. Let M be the midpoint of chord CD and M' be the midpoint of chord C'D'.
4. The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord, so OM passes through M and is along AB.
5. Similarly, OM' is along the diameter AB.
6. Since both M and M' lie on the diameter AB, the segment MM' lies entirely along the diameter AB.
7. As AB is perpendicular to chords CC' and DD', any segment lying along AB is perpendicular to lines parallel to the chords, hence MM' is perpendicular to AB.

**Answer:** Hence proved that the segment MM' is perpendicular to AB.

> Common mistake: Assuming M and M' are off the diameter without using the perpendicular bisector property from the centre.

### Question 26

*5 marks · Proof*

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$?

**Solution**

1. Consider the cyclic quadrilateral ABCD inscribed in a circle with centre O (Fig. 5.31).
2. Let the angles subtended at the centre O by the arcs be expressed using the angles marked at the centre as $p, q, u, v$.
3. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
4. Therefore, $\angle BCD = \frac{1}{2}(p + q)$ and $\angle BAD = \frac{1}{2}(u + v)$.
5. The sum of all angles around the centre O is a complete angle of $360^\circ$, so $(p + q + u + v) = 360^\circ$.
6. Adding the two opposite angles gives $\angle BCD + \angle BAD = \frac{1}{2}(p + q + u + v) = \frac{1}{2}(360^\circ) = 180^\circ$.
7. Hence proved that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$.

**Answer:** Hence proved that opposite angles add up to $180^\circ$.

> Common mistake: Confusing central angles with inscribed angles or forgetting that a complete rotation is $360^\circ$.

## Frequently asked questions

### How many exercises and questions are there in Class 9 Maths Chapter 5?

This chapter follows the new NCERT book for the 2026-27 session and contains multiple sections including Activity, Think and Reflect, six Exercise Sets, and 26 End-of-Chapter Exercises. You can find step-by-step solutions for all these questions in SwaVid's free PDF available on this page.

### Which mathematical topics are covered in these NCERT solutions?

The solutions cover various concepts such as properties of circles in nature, finding the centre of a circle, chords, circumcentres of acute and obtuse-angled triangles, and cyclic quadrilaterals. SwaVid provides detailed answers for all these topics on this page.

### Which question types are the most challenging in this chapter and how should I approach them?

Proof-based questions involving congruent triangles, perpendicular bisectors, and cyclic quadrilateral properties are often considered the hardest. To approach them, clearly state the given parameters, apply relevant circle theorems step by step, and structure your logical reasoning properly as shown in SwaVid's free PDF on this page.

### How can I write answers to score full marks in Class 9 Maths examinations?

To secure full marks, you should write clean mathematical statements, draw neat diagrams for circle constructions, and quote exact theorems like $...$ when using them in proofs. Referring to SwaVid's expert-verified solutions on this page will help you understand the ideal presentation format.

### Is the free PDF for this chapter available for download?

Yes, the complete chapter-wise free PDF and detailed solutions are available right here on this SwaVid page for the 2026-27 academic session. You can easily access these resources to prepare thoroughly for your Class 9 exams.

## Related pages

- [Exercise 5.1 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-1)
- [Exercise 5.2 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-2)
- [Exercise 5.3 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-3)
- [Exercise 5.4 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-4)
- [Exercise 5.5 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-5)
- [Exercise 5.6 solutions](https://www.swavid.com/maths/class/9/chapter/i-m-up-and-down-and-round-and-round/ncert-solutions/exercise-5-6)
- [Class 9 Maths chapters](https://www.swavid.com/maths/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student