NCERT Solutions for Class 9 Maths Chapter 5 - I'm Up and Down, and Round and Round

Get reliable Chapter-wise NCERT Solutions for Class 9 Maths in one place. Prepared in line with the latest NCERT syllabus and CBSE guidelines, these solutions provide step-by-step explanations, important questions, and FREE PDF downloads to support effective learning and exam preparation. You can also explore subject-wise resources for Maths, Physics, Chemistry, and Biology, thoughtfully curated by ALLEN Experts.

1.0Download NCERT Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round Solutions PDF

I'm Up, Down, and Round and Round introduces the fundamental concepts of circles and their different parts. It helps students understand the properties of circles through activities and practical examples. Download the NCERT Solutions Class 9 Maths Im up and down ,and round and round Solutions PDF with Detailed Answers for all the Exercise Probelms

NCERT Solutions Class 9 Maths Chapter 5

2.0Learning Outcomes – NCERT Class 9 Maths Chapter 5: I'm Up and Down, and Round and Round

Identify and describe the different parts of a circle, including the centre, radius, diameter, chord, and arc.
Understand the symmetry and unique properties of circles.
Construct the circumcircle of a triangle and locate its circumcentre.
Apply the properties of chords, arcs, and angles to solve geometric problems.
Use the relationships between chords and their distances from the centre of a circle.
Understand and apply the properties of cyclic quadrilaterals and concyclic points.
Solve theorem-based and application-oriented questions involving circles with confidence.
Develop logical reasoning and geometric proof skills through circle-based constructions and problems.

3.0Detailed Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round Solutions

Exercise : 5.1

1. Draw $△ ABC$ with $AB = 5 cm, ∠ A = 7 0^{\circ}$ and $∠ B = 6 0^{\circ}$ . Draw the circumcircle of $△ ABC$ . Is the centre inside or outside the triangle?

Sol.

(i) Draw the line segment

AB = 5 cm

. (ii) At point A , construct

∠ CAB = 7 0^{\circ}

. (iii) At point B , construct

∠ CBA = 6 0^{\circ}

. (iv) Let the two rays meet at C . Then

△ ABC

is formed. (v) Draw the perpendicular bisector of AB . (vi) Draw the perpendicular bisectors of BC and AC . (vii) Let the three perpendicular bisectors meet at 0 . This point 0 is the circumcenter. (viii) With centre O and radius OA/OB/OC, draw a circle passing through

A, B

and C . This is the circumcircle of

△ ABC

Now: $∠ C = 18 0^{\circ} - (7 0^{\circ} + 6 0^{\circ}) = 5 0^{\circ}$ Since all the angles of the triangle are less than $9 0^{\circ}$ , the triangle is acute-angled. In an acute angled triangle, the circumcenter lies inside the triangle. Therefore, the centre is inside the triangle. 2. Draw $△ ABC$ with $AB = 5 cm, ∠ A = 10 0^{\circ}$ , $AC = 4 cm$ . Draw the circumcircle of $△ ABC$ . Is the centre inside or outside the triangle?

Sol.

(i) Draw the line segment

AB = 5 cm

. (ii) At point A , construct

∠ BAC = 10 0^{\circ}

. (iii) On the ray AX , mark point C such that

AC = 4 cm

. (iv) Join C to B . Then

△ ABC

is formed. (v) Draw the perpendicular bisector of AB . (vi) Draw the perpendicular bisector of AC and BC . (vii) Let these bisectors meet at 0 . This point O is the circumcenter. (viii) With centre O and radius OA/OB/OC draw the circle through

A, B

and C . This is the circumcircle.

Now, $∠ A = 10 0^{\circ}$ , the triangle is obtuseangled.

In an obtuse-angled triangle, the circumcenter lies outside the triangle.

Therefore, the centre is outside the triangle. 3. Draw $△ ABC$ , with $AB = 6 cm, BC = 7 cm$ and $CA = 7 cm$ . Draw the circumcircle of $△ ABC$ . Let the circumcenter be O . Measure $OA, OB, OC$ .

Sol.

(i) Draw

AB = 6 cm

. (ii) From

A

and draw an arc of length 7 cm . (iii) Now from B draw another arc cutting the first arc at C of length 7 cm . (iv) Join AC and BC . Then

△ ABC

is formed. (v) Draw the perpendicular bisector of AB . (vi) Draw the perpendicular bisectors of

A C

and

BC

. (vii) Let them intersect at 0 . This point 0 is the circumcenter. (viii) With centre O and radius

OA / OB / OC

, draw the circumcircle.

Since 0 is the circumcenter it is equidistant from all three vertices. Therefore, $OA = OB = OC$ Approximate measurement: For the triangle with sides $7 cm, 7 cm, 6 cm$ , $OA = OB = OC \approx 4 cm$ (By ruler)

So, the three measurements are equal, each approximately 4 cm . 4. What is the least possible radius of a circle through two points A and B ?

Sol. For all circles passing through two fixed points A and B , the centre lies on the perpendicular bisector of AB .

The smallest circle occurs when the centre is the midpoint of AB .

In that case, AB becomes the diameter of the circle.

Therefore, the least possible radius is: Radius $= \frac{AB}{2}$ Hence, the least possible radius of a circle through A and B is half of AB .

Exercise: 5.2

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

Sol.

Let AB be a chord of a circle with centre O. Join OA and OB.

In triangle OAB $OA = OB$ [Radii of the same circle] Therefore, triangle OAB is an isosceles triangle. Hence proved. 2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Sol.

Let AB and PQ be two equal chords of the same circle with centre 0 . Then triangles formed are $△ OAB$ and $△ OPQ$ .

Now, in $△ OAB$ and $△ OPQ$ $OA = OP$ [Radii of the same circle] $OB = OQ$ [Radii of the same circle] $AB = PQ$ [Given] So, by SSS congruence criterion $△ OAB ≅ △ OPQ$ . Hence, if two such isosceles triangles have equal base length, they are congruent to each other.

Hence proved.

Exercise : 5.3

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? (Hint: Use Fig. 5.12. You are told that $∠ CMA = ∠ CMB = 9 0^{\circ}$ . You need to show that $AM = BM$ .)

Sol. Theorem 4 : The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.

Let AB be a chord of a circle with centre C , and let $CM ⊥ AB$ .

To Prove: $AM = BM$ In $△ CMA$ and $△ CMB$ , $CA = CB$ [Radii of the same circle] CM = CM [Common to both triangles] $∠ CMA = ∠ CMB$ [Each $9 0^{\circ}$ , as $CM ⊥ AB$ ] Therefore, by RHS congruence $Δ CMA ≅ △ CMB$ Hence, $AM = BM [CPCT]$ So, $M$ is the midpoint of $A B$ . Therefore, the perpendicular from the centre of a circle to a chord bisects the chord. Hence proved.

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.

Sol.

△ ABC

is inscribed in a circle and

AB = AC

. Since

AB = AC

, triangle ABC is isosceles with base BC.

Let AD be the altitude from A to BC . Then $A D ⊥ BC$ In $△ ABD$ and $△ ACD$ , $AB = AC$ [Given] $AD = AD$ [Common to both triangles] $∠ ADB = ∠ ADC [$ Each $9 0^{\circ}$ , as $AD ⊥ BC]$ Therefore, by RHS congruence $△ ABD ≅ △ ACD$ Hence, $BD = DC [CPCT]$ So, AD is bisector of BC . Since AD is perpendicular to BC and also bisects $BC, AD$ is the perpendicular bisector of chord BC. Therefore, AD passes through the centre of the circle. Hence, the altitude from A to BC passes through the centre of the circle. Hence proved.

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 the chords.

Sol

Given: Radius of the circle $OA = OC = 5 cm$ Lengths of the two chords are $AB = 6 cm$ and $CD = 8 cm$

The perpendicular from the centre 0 to a chords AB and CD bisects the chords at M and N respectively.

So, half-lengths are: For the 6 cm chord: $AM = 3 cm$ For the 8 cm chord: $CN = 4 cm$ Let the distances of the chords from the centre be $d_{1}$ and $d_{2}$ .

In triangle OAM, using Pythagoras theorem: $OM^{2} + AM^{2} = OA^{2}$ $\Rightarrow d_{1}^{2} + 3^{2} = 5^{2}$ $\Rightarrow d_{1}^{2} + 9 = 25$ $\Rightarrow d_{1}^{2} = 16$ $\Rightarrow d_{1} = 4 cm$ Similarly, in triangle OCN, using Pythagoras theorem: $ON^{2} + CN^{2} = OC^{2}$ $\Rightarrow d_{2}^{2} + 4^{2} = 5^{2}$ $\Rightarrow d_{2}^{2} + 16 = 25$ $\Rightarrow d_{2}^{2} = 9$ $\Rightarrow d_{2} = 3 cm$ Since the chords are on opposite sides of the centre and their centers lies on same line. So, the distance between their midpoints $= d_{1} + d_{2} = 4 + 3 = 7 cm$

Therefore, the distance between the midpoints of the two chords is 7 cm .

Exercise: 5.4

1.Use the Baudhāyana-Pythagoras to show why Theorem 6 must be true.

Sol. Theorem 6: Chords of a circle having the same length are all at the same distance from the centre of the circle.

Let AB and CD be two equal chords of a circle with centre 0 .

Let $OM ⊥ AB$ and $ON ⊥ CD$ . We know that the perpendicular from the centre to a chord bisects the chord.

So, $AM = MB = \frac{AB}{2}$ and $CN = ND = \frac{CD}{2}$ Since $AB = CD$ , we get: $AM = CN$ Also, $OA = OC =$ radius of the circle. So by congruency OM ana ON are equal.

2. Consider Figure if CE is perpendicular to $AB, CH$ is perpendicular to GF , and $CE = CH$ , show that $AB = GF$ .

Sol. Given: $CE ⊥ AB, CH ⊥ GF$ and $CE = CH$ . To Prove: $AB = GF$ Proof: Since CE is perpendicular to chord AB , the perpendicular from the centre to a chord bisects the chord.

Therefore, $AE = EB$ Similarly, since CH is perpendicular to chord GF. So, GH = HF In $△ CEA$ and $△ CHG$ , we have $CA = CG$ [Radii of the same circle] $CE = CH$ [Given] $∠ CEA = ∠ CHG [$ Each $9 0^{\circ}]$ Therefore, by RHS congruence $Δ CEA ≅ CHG$ So, $AE = GH [CPCT]$ $\Rightarrow 2 AE = 2 GH$ $\Rightarrow AB = GF [AB = 2 AE$ and $GF = 2 GH]$ Hence proved.

3. Solve the previous question using the Baudhāyana-Pythagoras theorem. Sol.

Given: $CE ⊥ AB, CH ⊥ GF$ and $CE = CH$ . To Prove: $AB = GF$ Proof: Since CE is perpendicular to chord AB , it bisects AB . So, E be the midpoint of AB . $\Rightarrow AE = \frac{AB}{2}$ Similarly, since CH is perpendicular to chord GF, it bisects GF. So, H be the midpoint of GF . $\Rightarrow GH = \frac{GF}{2}$

Now apply the Baudhāyana-Pythagoras theorem in right triangles $△ CEA$ and $Δ CHG$ .

In $△ CEA : CA^{2} = CE^{2} + AE^{2}$ In $△ CHG : CG^{2} = CH^{2} + GH^{2}$ But, CA = CG [Radii of same circle] $CE = CH$ [Given] Therefore, $CE^{2} + AE^{2} = CH^{2} + GH^{2}$ $\Rightarrow AE^{2} = GH^{2}$ [Since CE = CH] Hence, $AE = GH$ So, $\frac{AB}{2} = \frac{GF}{2}$ Therefore, $AB = GF$

Exercise: 5.5

1.Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm .

Sol.

Let the chord be AB and O be the centre of the circle.

Let OM be the perpendicular from O to chord AB .

Given: Radius, $OA = OB = 7 cm$ Perpendicular distance, $OM = 6 cm$ We know that the perpendicular from the centre to a chord bisects the chord.

So, $AM = MB$

In right triangle OMA, $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow 7^{2} = 6^{2} + AM^{2}$ $\Rightarrow 49 = 36 + AM^{2}$ $\Rightarrow AM^{2} = 13$ $\Rightarrow AM = 13$ Therefore, chord AB $= 2 \times AM = 213 cm$ Hence, the length of the chord is $213 cm$ .

2. 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 $2 r^{2} - d^{2}$ . Sol.

Let AB be a chord of a circle with centre 0 . Let OM be the perpendicular from O to AB . Given: Radius $= r$ Let perpendicular distance from centre to chord = d Since the perpendicular from the centre to the chord bisects the chord, so M is the midpoint of AB . So, $AM = MB =$ half of the chord length. In right triangle OMA, By the Baudhāyana-Pythagoras theorem: $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow r^{2} = d^{2} + AM^{2}$ $\Rightarrow A M^{2} = r^{2} - d^{2}$ $\Rightarrow A M = (r^{2} - d^{2})$ But chord length $AB = 2 \times AM$ So, $AB = 2 (r^{2} - d^{2})$ Hence, the chord length is $2 (r^{2} - d^{2})$

3. 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. Sol.

The length of a chord depends on the formula:

Chord length $= 2 (r^{2} - d^{2})$ [Proved] This relation is not directly proportional to the distance from the centre.

Let the distance of chord CD from the centre be d .

Then the distance of chord AB from the centre is 2d.

So, $AB = 2 (r^{2} - (2 d)^{2} = 2 (r^{2} - 4 d^{2})$ $CD = 2 (r^{2} - d^{2})$ Clearly, CD is not equal to 2 AB in general. Therefore, the statement is false. Hence, we cannot conclude that $CD = 2 AB$ .

Exercise: 5.6

1. In a circle with centre $O$ , the central angle AOB is $6 0^{\circ}$ . If the radius of the circle is 12 cm , what is the length of the chord AB ? Sol.

Given: Radius $OA = OB = 12 cm$ $∠ AOB = 6 0^{\circ}$ In $△ AOB$ : $OA = OB$ So, $∠ OAB = ∠ OBA$ [Angles opposite to equal sides of triangle] Let $∠ OAB = ∠ OBA = x$ In $△ OAB$ , $∠ AOB + ∠ OAB + ∠ OBA = 18 0^{\circ}$ [Angles sum property of the triangle] $\Rightarrow 6 0^{\circ} + x + x = 18 0^{\circ}$ [Since $∠ AOB = 6 0^{\circ}$ ] $\Rightarrow 2 x = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ $\Rightarrow x = \frac{12 0 ^{\circ}}{2} = 6 0^{\circ}$ $\Rightarrow ∠ OAB = ∠ OBA = 6 0^{\circ}$ Thus, triangle AOB is equilateral. Therefore, $AB = OA = 12 cm$ Hence, the length of chord AB is 12 cm .

2. Let A and B be two points on a circle with centre 0 . (i) Are there points $X, Y$ on the circle, on the same side of AB , such that $∠ AXB$ is different from $∠ AYB$ ? (ii) Is it true that if $∠ AXB = ∠ AYB$ , then X and Y lie on the same side of the circle? (iii) If $∠ AXB = ∠ AYB$ , and X and Y do not lie on the circle, does the circle through $A, B$ and X also pass through Y ?

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

Answer: No. Reason: If X and Y lie on the same side of chord AB , then they lie on the same $arc AB$ . Angles subtended by the same chord in the same segment of a circle are equal.

Therefore, $∠ AXB = ∠ AYB$ So, there are no such points X and Y on the same side of AB for which the angles are different. (ii) Is it true that if $∠ AXB = ∠ AYB$ , then X and Y lie on the same side of the circle?

Answer: No, this is not always true. Reason: Equal angles can also be subtended by the same chord AB at points on opposite arcs.

So, even if $∠ AXB = ∠ AYB, X$ and Y need not lie on the same side of $A B$ .

Therefore, the statement is false. (iii) If $∠ AXB = ∠ AYB$ , and X and Y do not lie on the circle, does the circle through $A, B$ and X also pass through Y ?

Answer: Yes. Reason: If $∠ AXB = ∠ AYB$ , then the line segment AB subtends equal angles at X and Y.

By the converse theorem of concyclicity: If a line segment subtends equal angles at two points on the same side, then the four points are concyclic. Hence, $A, B, X$ and Y lie on the same circle. Therefore, the circle through $A, B$ and X also passes through Y.

3. Find x in figure.

Sol. In the figure $A, D, C$ and B are points on the same circle.

So, quadrilateral ADCB is a cyclic quadrilateral.

Given: $∠ ADC = 10 0^{\circ}$ In a cyclic quadrilateral, opposite angles are supplementary.

Therefore, $∠ ABC + ∠ ADC = 18 0^{\circ}$ $\Rightarrow x + 10 0^{\circ} = 18 0^{\circ}$ $\Rightarrow x = 8 0^{\circ}$ Hence, $x = 8 0^{\circ}$ .

End of Chapter Exercise

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm , what is the length of the chord?

Sol.

Let AB be the chord and O be the centre. Let OM be the perpendicular from O to AB . Given: $OM = 5 cm$ $OA = 13 cm$ Since the perpendicular from the centre to a chord bisects the chord, so, $AM = MB$ .

In right triangle OMA, $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow 1 3^{2} = 5^{2} + AM^{2}$ $\Rightarrow 169 = 25 + AM^{2}$ $\Rightarrow AM^{2} = 144$ $\Rightarrow AM = 12 cm$ Therefore, $AB = 2 \times AM$

$= 2 \times 12 = 24 cm$

Hence, the length of the chord is 24 cm .

2. An arc of a circle subtends an angle of $7 0^{\circ}$ at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

Sol. We know: Angle subtended by an arc at the centre $= 2 \times$ angle subtended by the same arc at a point on the circle

Given: Angle at the centre $= 7 0^{\circ}$ So, angle at the circle $= \frac{7 0 ^{\circ}}{2} = 3 5^{\circ}$ Hence, the required angle is $3 5^{\circ}$ .

3. The diameter of a circle is 26 cm . A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.

Sol.

Diameter $= 26 cm$ So, radius $= 13 cm$ Chord AB length $= 24 cm$ Half of chord AM = 12 cm Let OM be the perpendicular distance from the centre 0 to the chord AB .

Then M is the midpoint of the chord. In right triangle OMA, $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow 1 3^{2} = OM^{2} + 1 2^{2}$ $\Rightarrow 169 = OM^{2} + 144$ $\Rightarrow OM^{2} = 25$ $\Rightarrow OM = 5 cm$ Hence, the distance from the centre to the chord is 5 cm .

4. A circle has a radius of 15 cm . A chord is drawn. The distance from the centre of the circle to the chord is 9 cm . What is the length of the chord?

Sol.

Let AB be the chord and O be the centre. Let OM be the perpendicular from 0 to AB . Given: $OA = 15 cm$ $OM = 9 cm$ Since the perpendicular from the centre to a chord bisects the chord, $AM = MB$ In right triangle OMA, $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow 1 5^{2} = 9^{2} + AM^{2}$ $\Rightarrow 225 = 81 + AM^{2}$ $\Rightarrow AM^{2} = 144$ $\Rightarrow AM = 12 cm$ Therefore, $AB = 2 \times AM = 24 cm$ Hence, the length of the chord is 24 cm .

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

Sol.

Let AB be a chord of a circle with centre 0 . Let M be the midpoint of AB .

We have to show that OM is perpendicular to AB , which means the perpendicular bisector of AB passes through 0.

So, in triangles OMA and OMB: $OA = OB$ [Radii of the same circle] $AM = MB [M$ is the midpoint of AB $]$ $OM = OM$ [Common] Therefore, by SSS congruence $△ OMA ≅ △ OMB$ Hence, $∠ OMA = ∠ OMB$ [CPCT]

But these two angles form a linear pair, so $∠ OMA + ∠ OMB = 18 0^{\circ}$

Since they are equal, $∠ OMA = ∠ OMB = 9 0^{\circ}$ Therefore, $OM ⊥ AB$

So, the line through O and M is the perpendicular bisector of chord AB . Hence proved.

6. The diameter of a circle is $A B$ . Point $C$ is on the circumference. What is the measure of the $∠ A CB$ ? Explain your reasoning.

Sol.

AB is a diameter. The angle subtended by a diameter at any point on the circle is $9 0^{\circ}$ .

Therefore, $∠ ACB = 9 0^{\circ}$ Hence, the measure of $∠ ACB$ is $9 0^{\circ}$ .

7. ABCD is a cyclic quadrilateral inscribed in a circle. If $∠ A$ measures $7 5^{\circ}$ , what is the measure of $∠ C$ ? If $∠ B$ measures $11 0^{\circ}$ , what is the measure of $∠ D$ ?

Sol.

In a cyclic quadrilateral, opposite angles are supplementary.

So, $∠ A + ∠ C = 18 0^{\circ}$ $\Rightarrow 7 5^{\circ} + ∠ C = 18 0^{\circ}$ $\Rightarrow ∠ C = 10 5^{\circ}$ Also, $∠ B + ∠ D = 18 0^{\circ}$ $\Rightarrow 11 0^{\circ} + ∠ D = 18 0^{\circ}$ $\Rightarrow ∠ D = 7 0^{\circ}$ Hence, $∠ C = 10 5^{\circ}$ and $∠ D = 7 0^{\circ}$ .

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

Sol.

Since PQRS is a cyclic quadrilateral, opposite angles are supplementary. So, $∠ P + ∠ R = 18 0^{\circ}$ $\Rightarrow (2 x + 10) + (3 x - 20) = 180$ $\Rightarrow 5 x - 10 = 180$ $\Rightarrow 5 x = 190$ $\Rightarrow x = 38$ Now, $∠ P = 2 x + 10 = 2 (38) + 10 = 8 6^{\circ}$ $∠ R = 3 x - 20 = 3 (38) - 20 = 9 4^{\circ}$ Hence, $x = 38, ∠ P = 8 6^{\circ}$ and $∠ R = 9 4^{\circ}$ .

9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm . Find the radius of the circle. Sol.

Chord length $AB = 16 cm$ Let $OM ⊥ AB$ Half chord $AM = MB = 8 cm$ Distance from centre to chord OM = 6 cm Let $r$ be the radius. In the right triangle OAM, $OA^{2} = AM^{2} + OM^{2}$ $\Rightarrow r^{2} = 8^{2} + 6^{2}$ $\Rightarrow r^{2} = 64 + 36$ $\Rightarrow r^{2} = 100$ $\Rightarrow r = 10 cm$ Hence, the radius of the circle is 10 cm .

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

Sol. Since the cyclic quadrilateral has sides $5, 5, 12, 12$ , its semi perimeter is $s = \frac{( 5 + 5 + 12 + 12 )}{2}$ $= \frac{34}{2} = 17$ For a cyclic quadrilateral, area is given by Brahmagupta's formula:

Area $= [(s - a) (s - b) (s - c) (s - d)]$ Substituting the values, we have: Area $= (17 - 5) (17 - 5) (17 - 12) (17 - 12)$ $= [(12) (12) (5) (5)]$ $= 3600 = 60$ Hence, the area of the cyclic quadrilateral is 60 square units.

11. 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?

Sol. In a cyclic quadrilateral, we know that opposite angles are supplementary (their sum is $18 0^{\circ}$ .

To find whether the centre of the circumcircle (circumcenter) lies inside or outside, we can observe the angles of the quadrilateral: If all angles are less than $9 0^{\circ}$ (acute), then the circumcenter lies inside the quadrilateral.

If any one angle is greater than $9 0^{\circ}$ (obtuse), then the circumcenter lies outside the quadrilateral.

Best way: Check the angles of the quadrilateral.

This method is easy and does not require drawing the circumcircle.

So, by just looking at whether the quadrilateral has an obtuse angle or not, we can decide the position of the circumcenter.

12. 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.

Sol.

Let chords AB and CD intersect at point P inside the circle, and 0 be the centre. From the figure, $ON ⊥ CD$ and $OM ⊥ AB$ .

Given: $AB = CD$

To Prove: $PB = PD$ and $AP = CP$ . Equal chord equal distances from centre Proof : So, the perpendicular distances from the centre are equal. So, $OM = ON$

In $△ OPM$ and $△ OPN$ , we have OP = OP [common side] $OM = ON$ [Proved above] $∠ OMP = ∠ ONP = 9 0^{\circ}$ [Perpendiculars] So, by RHS congruency $△ OPM ≅ △ OPN$ So, $PM = PN [CPCT]$

Now, $OM ⊥ AB$ , So, $AM = MB$ [Perpendicular from the centre bisect the chord]

Similarly, $ON ⊥ CD$ , So, $CN = ND$ Since, $AB = CD$ [From (1)] Therefore, $BM = DN$ [ $BM = \frac{1}{2} AB$ and $DN = \frac{1}{2} CD$ ]

Adding (2) and (3), we get $PM + BM = PN + DN$ $\Rightarrow PB = PD$

Subtracting (4) from (1), we get $AB - PB = CD - PD$ $\Rightarrow AP = CP$ Hence proved.

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

Given: Chord length $= 6 cm$ Distance from centre to chord $= 3 cm$ Steps of Construction:

Draw a line segment $AB = 6 cm$
Find the midpoint M of AB .
Draw a perpendicular line to AB at M .
On this perpendicular, mark a point O such that $OM = 3 cm$ .
Join OA or OB .
With centre O and radius $\frac{OA}{OB}$ , draw a circle.

Since OM is perpendicular to AB and passes through its midpoint, AB is a chord of the circle.

Also, $AM = 3 cm$ and $OM = 3 cm$ So, $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow OA^{2} = 3^{2} + 3^{2} = 18$ $\Rightarrow OA = 32 cm$ Therefore, the required circle has centre 0 and radius $32$ .

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

Sol.

Let ABCD be a parallelogram inscribed in a circle.

Since it is a cyclic quadrilateral: $∠ A + ∠ C = 18 0^{\circ}$ But in a parallelogram, opposite angles are equal: $∠ A = ∠ C$

Therefore, $∠ A + ∠ A = 18 0^{\circ}$ $\Rightarrow 2∠ A = 18 0^{\circ}$ $\Rightarrow ∠ A = 9 0^{\circ}$ Similarly, all angles are $9 0^{\circ}$ . Hence, the parallelogram is a rectangle. Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.

15. 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.

Sol.

Let ABCD be a rectangle inscribed in a circle.

Let its diagonals AC and BD intersect at 0 . In a rectangle, diagonals are equal and bisect each other.

So, $OA = OC$ and $OB = OD$ Also, $AC = BD$ Therefore, $OA = OB = OC = OD$ This means 0 is equidistant from all four vertices A, B, C and D. The point equidistant from all points on a circle is the centre.

Hence, the point of intersection of the diagonals is the centre of the circle.

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

Sol. Let the radius of the circle be r . Let each chord have fixed length x . The perpendicular from the centre to a chord bisects the chord.

So, for every chord, half chord $= \frac{x}{2}$ . Let d be the distance of the midpoint of the chord from the centre.

Using Pythagoras theorem: $r^{2} = d^{2} + (\frac{x}{2})^{2}$ So, $d^{2} = r^{2} - (\frac{x}{2})^{2}$ Since r and x are fixed, d is also fixed. Therefore, every midpoint is at the same distance from the centre.

Hence, the midpoints form a circle with the same centre as the original circle.

17. In a circle with centre 0 , chords $A B$ and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of $∠ BAC$ .

Sol.

Given: $AB = AC$ Join $OA, OB$ and OC . In $△ AOB$ and $△ AOC$ , we have Now, $OB = OC$ [Both are radii] $OA = OA$ [Common] $AB = AC$ [Given] Therefore, by SSS congruence $△ AOB ≅ AOC$ So, $∠ BAO = ∠ OAC$ Hence, AO bisects $∠ BAC$ . Therefore, the centre 0 lies on the angle bisection of $∠ BAC$ .

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

Sol.

Let the radius be r . So, $OA = OC = r$ . For chord AB of length 10 cm Half of chord $AB = 5 cm$ Let its distance from centre $0 M = d_{1}$ For chord CD of length 24 cm : Half of chord CD = 12 cm Let its distance from centre $ON = d_{2}$ Since the longer chord is closer to the centre: $d_{1} - d_{2} = 7$

In triangle OAM, using Pythagoras theorem, we have $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow r^{2} = d_{1}^{2} + 5^{2}$ $\Rightarrow r^{2} = d_{1}^{2} + 25$ In triangle OCN, using Pythagoras theorem, we have $OC^{2} = ON^{2} + CN^{2}$ $\Rightarrow r^{2} = d_{2}^{2} + 1 2^{2}$ $\Rightarrow r^{2} = d_{2}^{2} + 144$ So, $d_{1}^{2} + 25 = d_{2}^{2} + 144$ $\Rightarrow d_{1}^{2} - d_{2}^{2} = 119$ $\Rightarrow (d_{1} - d_{2}) (d_{1} + d_{2}) = 119$ $\Rightarrow 7 (d_{1} + d_{2}) = 119$ [Since $d_{1} - d_{2} = 7$ ]

$\Rightarrow d_{1} + d_{2} = 17$

Now solving equations (1) and (2), we have $2 d_{1} = 24$ $\Rightarrow d_{1} = 12$ Then $d_{2} = 5$ Now, $r^{2} = d_{1}^{2} + 5^{2}$ $= 1 2^{2} + 5^{2}$ $= 144 + 25$ = 169 $\Rightarrow r = 13 cm$ Therefore, the radius of the circle is 13 cm .

19. 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.

Sol.

A regular hexagon divides the circle into 6 equal central angles. Each central angle $= \frac{36 0 ^{\circ}}{6} = 6 0^{\circ}$ Let AB be one side of the hexagon and 0 be the centre.

Then $OA = OB = r$ and $∠ AOB = 6 0^{\circ}$ So, $△ AOB$ is an equilateral triangle. Therefore, $AB = r$ Thus, each side of the regular hexagon is r . Now find the distance of each side from the centre. Let OM be perpendicular to AB . Since $△ AOB$ is equilateral: $AM = \frac{AB}{2} = \frac{r}{2}$ In right triangle OMA $OA^{2} = OM^{2} + AM^{2}$ $\Rightarrow r^{2} = OM^{2} + (\frac{r}{2})^{2}$ $\Rightarrow OM^{2} = r^{2} - \frac{r ^{2}}{4}$ $\Rightarrow OM^{2} = \frac{3 r ^{2}}{4}$ ⇒ $OM = (\frac{3}{2}) r$ Therefore, Side length of hexagon $= r$ Distance of each side from centre $= (\frac{3}{2}) r$

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

Using theorem the angle subtended by an arc at the centre is double the angle subtended at the circle. $2∠ MOP = ∠ MCP$ $2∠ MNP = ∠ MCP$ So, $2∠ MOP = ∠ MNP$

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

In cyclic quadrilateral ABCD : $∠ ABC + ∠ ADC = 18 0^{\circ}$

Since E lies on the extension of CD , $∠ ADC$ and $∠ ADE$ form a linear pair. So, $∠ ADC + ∠ ADE = 18 0^{\circ}$

From (1) and (2), we have $∠ ABC + ∠ ADC = ∠ ADC + ∠ ADE$ Subtracting $∠ ADC$ from both sides, we get $∠ ABC = ∠ ADE$ Therefore, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Hence proved.

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

Sol. A chord is a line segment joining two points on a circle.

The longest chord passes through the centre of the circle.

Such a chord is called the diameter. For any other chord, its perpendicular distance from the centre is greater than 0 .

If radius $= r$ and distance of chord from centre $= d$ , then:

Chord length $= 2 (r^{2} - d^{2})$ Since $d > 0$ for any chord not passing through the centre, $r^{2} - d^{2} < r^{2}$ So, $2 (r^{2} - d^{2}) < 2 r$ But diameter $= 2 r$ Therefore, every chord other than the diameter is shorter than the diameter.

Hence, no chord of a circle is longer than its diameter.

23. Let A be any point within a given circle with centre 0 . Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA .

Sol.

Let the circle have centre 0 and let A be a point inside it.

Any chord passing through A will have some perpendicular distance from the centre 0.

We know that the farther a chord is from the centre, the shorter it is.

So, the shortest chord through A will be the chord whose distance from 0 is maximum.

Among all lines passing through A , the perpendicular distance from 0 to the line is greatest when the line is perpendicular to OA .

Therefore, the chord through A that is perpendicular to OA is farthest from the centre. Hence, it is the shortest chord passing through A.

Therefore, the shortest chord through A is the one perpendicular to OA . Hence proved.

24. How would you use the following figure to justify the statement that the angle in a semicircle is $9 0^{\circ}$ ?

Sol. Let the diameter be BC and O be the centre of the circle.

Point A lies on the semicircle. Join OA. Since 0 is the centre. $OA = OB = OC$ because all are radii of the same circle. So, $△ AOB$ is isosceles and $△ AOC$ is isosceles.

Let: $∠ ABO = a$ and $∠ ACO = b$ Then: $∠ BAO = a$ and $∠ OAC = b$ [Since $∠ ABO = ∠ BAO$ and $∠ ACO = ∠ OAC$ ] Therefore, $∠ BAC = a + b$ Now, in triangle ABC : $∠ ABC + ∠ BAC + ∠ ACB = 18 0^{\circ}$ So, $a + (a + b) + b = 18 0^{\circ}$ $\Rightarrow 2 a + 2 b = 18 0^{\circ}$ $\Rightarrow 2 (a + b) = 18 0^{\circ}$ $\Rightarrow a + b = 9 0^{\circ}$ But, $∠ BAC = a + b$ Therefore, $∠ BAC = 9 0^{\circ}$ Hence, the angle in a semicircle is $9 0^{\circ}$ .

25. 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 .

Sol. Let AB be the diameter of the circle. Since chords CC' and DD' are perpendicular to AB , the points C and $C^{'}$ are symmetric about AB . Similarly, D and $D^{'}$ are also symmetric about AB .

Now, let M be the midpoint of CD and $M^{'}$ be the midpoint of C'D'.

Since C corresponds to $C^{'}$ and D corresponds to $D^{'}$ by reflection in AB , the midpoint of CD corresponds to the midpoint of $C^{'} D^{'}$ .

Therefore, M and $M^{'}$ are symmetric about AB . The line joining two points that are symmetric about a line is perpendicular to that line.

Hence, $MM^{'} ⊥ AB$ Therefore, the segment joining the midpoints of CD and $C^{'} D^{'}$ is perpendicular to AB . Hence proved.

26. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $18 0^{\circ}$ ?

Sol. Let ABCD be a cyclic quadrilateral with centre 0 .

Join $OA, OB, OC$ and OD . Since $OA, OB, OC$ and OD are radii of the same circle: $OA = OB = OC = OD$

So, the triangles formed with the centre are isosceles triangles.

From the figure: Let $∠ OAB = p = ∠ OBA$ and $∠ OBC = q = ∠ OCB$ . Let $∠ ODA = v = ∠ OAD$ and $∠ OCD = u = ∠ ODC$ . $∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$ (Angle sum property of quadrilateral) $2 p + 2 q + 2 u + 2 v = 36 0^{\circ}$ $p + q + u + v = 18 0^{\circ}$ Now $∠ A + ∠ C = p + q + u + v = 18 0^{\circ}$ $∠ B + ∠ D = p + q + u + v = 18 0^{\circ}$ Therefore, the sum of opposite angles of a cyclic quadrilateral is $18 0^{\circ}$ .

Hence proved.

4.0Important Key Concepts of Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round

Topic	What Students Learn
Definitions	Formal introduction to a circle, its centre, radius, chord, diameter, and the concept of locus
Symmetries of a Circle	Exploration of rotational symmetry and reflection symmetry across diameters, and why these make circles geometrically unique
How Many Circles?	Proof that infinitely many circles pass through two points, and that a unique circle passes through any three non-collinear points
Circumcircle Construction	Step-by-step construction of a triangle's circumcircle using perpendicular bisectors, and locating the circumcentre
Chords and Their Properties	Understanding how a perpendicular from the centre bisects a chord, and how equal chords relate to their distance from the centre
Angles Subtended by Arcs	Relationship between the angle an arc subtends at the centre and the angle it subtends at a point on the circle
Cyclic Quadrilaterals	Properties of quadrilaterals inscribed in a circle, including the sum of opposite angles
Concyclic Points	Conditions under which a set of points lie on the same circle, and how to prove concyclicity
Theorem-Based Problem Solving	Application of the chapter's theorems to solve numerical and proof-based geometry problems using triangle congruence and the Baudhāyana–Pythagoras theorem

5.0Exercise-wise NCERT Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round

Chapter 5 has six exercises plus an End-of-Chapter set, taking students from basic circle constructions through chord properties, arc angles, and cyclic quadrilaterals, with each exercise building on the theorems covered before it.

Exercise	Topics Covered
Exercise 5.1	Construction of circumcircles, circumcentre, and circles passing through given points.
Exercise 5.2	Chords, radii, isosceles triangles, and congruence of triangles formed by equal chords.
Exercise 5.3	Perpendicular bisectors of chords, midpoint properties, and related circle theorems.
Exercise 5.4	Distance of chords from the centre and relationships between equal and unequal chords.
Exercise 5.5	Finding chord lengths using the radius and perpendicular distance from the centre.
Exercise 5.6	Angles subtended by arcs, angle in a semicircle, concyclic points, and cyclic quadrilaterals.
End-of-Chapter Exercises	Mixed questions covering constructions, circle theorems, chords, arcs, angles, cyclic quadrilaterals, and application-based problems from the chapter.

6.0Quick Revision on Class 9 Maths Chapter 5 - I'm Up and Down, and Round and Round

7.0Related Study Materials Class 9 Maths

Enhance your knowledge with ALLEN’s Class 9 Maths study materials for all chapters of the latest NCERT syllabus. You also get updated syllabus, NCERT textbooks, revision notes, sample papers and previous years question papers. These resources help you revise concepts systematically, and prepare effectively for school and CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 5 Maths Class 9 NCERT Solutions

Strengthens Integer Concepts: Builds a clear understanding of positive and negative numbers.
Simplifies Integer Operations: Explains addition, subtraction, multiplication, and division with step-by-step solutions.
Improves Number Line Skills: Helps students accurately compare, order, and represent integers.
Connects Maths to Real Life: Demonstrates the use of integers in everyday situations like temperature, banking, and elevation.
Enhances Accuracy: Reduces errors by clearly explaining the rules for operations involving integers.
Builds a Strong Foundation: Prepares students for advanced topics such as algebra and coordinate geometry.

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Why should I use the NCERT Solutions for Class 9 Maths Chapter 5?

Do these NCERT Solutions include detailed explanations for every exercise?

What is the significance of the circumcircle in Chapter 5?

How do students learn about chords in this chapter?

What role do angles play in Chapter 5?

How does Chapter 5 help in solving geometry problems?

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NCERT Solutions for Class 9 Maths Chapter 5 - I'm Up and Down, and Round and Round

1.0Download NCERT Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round Solutions PDF

2.0Learning Outcomes – NCERT Class 9 Maths Chapter 5: I'm Up and Down, and Round and Round

3.0Detailed Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round Solutions

Exercise : 5.1

Exercise: 5.2

Exercise : 5.3

Exercise: 5.4

Exercise: 5.5

Exercise: 5.6

End of Chapter Exercise

4.0Important Key Concepts of Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round

5.0Exercise-wise NCERT Class 9 Maths Chapter 5 I'm Up, Down, and Round and Round

6.0Quick Revision on Class 9 Maths Chapter 5 - I'm Up and Down, and Round and Round

7.0Related Study Materials Class 9 Maths

8.0Advantages of Chapter 5 Maths Class 9 NCERT Solutions