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I'm Up and Down, and Round and Round

CBSE · Class 9 · Mathematics

NCERT Solutions for I'm Up and Down, and Round and Round — CBSE Class 9 Mathematics.

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EXERCISE SET 5.1

1Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?Show solution

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2Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?Show solution

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3Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.Show solution

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4What is the least possible radius of a circle through two points A and B?Show solution
For a circle through two points A and B, the smallest possible circle is when AB itself is a diameter. Then the centre is the midpoint of AB, so the radius is half of AB. Any smaller radius cannot pass through both points.

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1Show that the triangle formed by a chord and the centre of the circle is isosceles.Show solution
Let C be the centre of the circle and AB be a chord. Join CA and CB.

Since CA and CB are radii of the same circle, CA = CB.

So triangle CAB has two equal sides, hence it is an isosceles triangle.

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2Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.Show solution
If two such triangles have the same base length, then each has two equal sides that are radii of the same circle, and their bases are equal by given information.

So the two triangles are congruent by the SSS congruence rule.

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Think, Draw and Infer

1A, 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?Show solution

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2The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?Show solution
Yes. For a given triangle ABC, there is a unique circumcircle through its three non-collinear vertices. But other triangles can be drawn that are congruent to ΔABC and still share the same circumcircle, by choosing other sets of three points on the same circle that give a congruent triangle. So, more than one congruent triangle can share the same circumcircle.

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Think and Reflect

1What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?Show solution
A square has rotational symmetry through angles of 90∘90^\circ, 180∘180^\circ, 270∘270^\circ, and 360∘360^\circ. It has 4 lines of reflection symmetry.

A regular pentagon has rotational symmetry through multiples of 72∘72^\circ and has 5 lines of reflection symmetry.

A regular hexagon has rotational symmetry through multiples of 60∘60^\circ and has 6 lines of reflection symmetry.

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2What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?Show solution
The longest chord of a circle is the diameter. For radius 55 units, diameter =2×5=10= 2\times 5 = 10 units.

There is no smallest chord in the strict sense, because a chord can be made as short as we like, approaching length 00.

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3The 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?Show solution
The locus of points equidistant from two given points is the perpendicular bisector of the line segment joining those two points.

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1How many circles pass through two points on a plane?Show solution
Through two points on a plane, we can draw infinitely many circles. Their centres lie on the perpendicular bisector of the segment joining the two points, and each point on that bisector gives a different circle.

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2Are 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?Show solution
Yes, there are circles of all possible radii passing through A and B.

- The smallest circle has AB as diameter, so its radius is **AB/2AB/2.
- There is no largest radius**, because the centre can be taken farther and farther away on the perpendicular bisector, making the radius larger without bound.

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3As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?Show solution
As you move away from segment AB along its perpendicular bisector, the centres get farther from the segment, so the circles must have larger radii. Therefore, the radii increase.

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4As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?Show solution
As the centre moves farther away along the perpendicular bisector, the circle through A and B becomes larger and looks less curved. A larger circle has a gentler curvature.

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5You 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?Show solution
If A and B are only required to be on the boundary of a square, then there are infinitely many squares.

If A and B are required to be corners of the square, then there are 2 possible squares, one on each side of segment AB.

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EXERCISE SET 5.3

1Can 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?Show solution
Let C be the centre of the circle and AB be a chord. Join CA and CB.

Since CA = CB as they are radii, triangle ABC is isosceles.

Now let CM be the perpendicular from the centre to the chord AB. In triangles CMA and CMB:
- CA = CB,
- CM is common,
- and the angle at M is 90∘90^\circ in both triangles.

So the two triangles are congruent, which gives AM = MB. Therefore, the perpendicular from the centre bisects the chord.

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2An 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.Show solution
Let M be the foot of the altitude from A to BC. Since AB = AC, triangle ABC is isosceles.

In an isosceles triangle, the altitude from the vertex A to the base BC is also the perpendicular bisector of BC.

The centre of the circumcircle lies on the perpendicular bisectors of the sides. Hence the altitude from A passes through the centre of the circle.

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3Two 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.Show solution
For a chord of length ll at distance dd from the centre in a circle of radius rr,

(l2)2+d2=r2 \left(\frac{l}{2}\right)^2 + d^2 = r^2

Here l=6l=6 cm and r=5r=5 cm, dd is required. So

32+d2=52 3^2 + d^2 = 5^2
9+d2=25 9 + d^2 = 25
d2=16 d^2 = 16
d=4 cm d = 4\text{ cm}

Since the chords are on opposite sides of the centre, the distance between their midpoints is 4+?4+?

But the given chord lengths are 6 cm and 8 cm. Their half-lengths are 3 cm and 4 cm.
For the 6 cm chord:
32+d12=25⇒d1=4 3^2 + d_1^2 = 25 \Rightarrow d_1 = 4
For the 8 cm chord:
42+d22=25⇒d2=3 4^2 + d_2^2 = 25 \Rightarrow d_2 = 3
They are on opposite sides, so distance between midpoints = d1+d2=4+3=7d_1 + d_2 = 4 + 3 = 7 cm.

The correct result is 7 cm.

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EXERCISE SET 5.4

1Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.Show solution
Let CE and CH be the perpendiculars from the centre to two equal chords AB and FG.

Then E and H are the midpoints of the chords, so:
- AE=12ABAE = \tfrac12 AB,
- FH=12FGFH = \tfrac12 FG.

If AB=FGAB = FG, then AE=FHAE = FH.

Now in right triangles CEA and CHF:
- CA=CFCA = CF because both are radii,
- AE=FHAE = FH,
- ∠CEA=∠CHF=90∘\angle CEA = \angle CHF = 90^\circ.

So the triangles are congruent by RHS. Hence CE=CHCE = CH, which means equal chords are at equal distances from the centre.

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2Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.Show solution
Let E and H be the feet of the perpendiculars from the centre C to the chords AB and GF.

Because a perpendicular from the centre bisects a chord, E and H are the midpoints of the chords.

Given CE = CH and the radii are equal, the right triangles formed with the half-chords are congruent by RHS. Therefore the half-chords are equal, so the whole chords are equal:

AB=GF. AB = GF.

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3Solve the previous question using the Baudhāyana–Pythagoras theorem.Show solution
Use the Baudhāyana–Pythagoras theorem on the two right triangles formed by the perpendiculars from the centre to the chords.

If the distances from the centre are equal, say CE = CH, and the radii are equal, then:

CA2=CE2+AE2,CF2=CH2+FH2 CA^2 = CE^2 + AE^2, \qquad CF^2 = CH^2 + FH^2

Since CA=CFCA = CF and CE=CHCE = CH, we get

AE2=FH2⇒AE=FH. AE^2 = FH^2 \Rightarrow AE = FH.

Because the perpendicular from the centre bisects each chord, the full chords are equal:

AB=GF. AB = GF.

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EXERCISE SET 5.5

1Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.Show solution
The radius is 77 cm and the perpendicular distance from the centre to the chord is 66 cm.

Using the right triangle formed by half the chord:
(l2)2+62=72 \left(\frac{l}{2}\right)^2 + 6^2 = 7^2
(l2)2=49−36=13 \left(\frac{l}{2}\right)^2 = 49 - 36 = 13
l2=13 \frac{l}{2} = \sqrt{13}
So,
l=213 cm l = 2\sqrt{13}\text{ cm}
This is the exact answer. Since the computed value is not among the printed options, the correct length is **2132\sqrt{13} cm**.

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2Explain 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 2r2−d22\sqrt{r^2 - d^2}.Show solution
Drop a perpendicular from the centre to the chord. It bisects the chord, so half the chord, the distance from the centre to the chord, and the radius form a right triangle.

If the radius is rr and the perpendicular distance is dd, then by the Baudhāyana–Pythagoras theorem:

(l2)2+d2=r2 \left(\frac{l}{2}\right)^2 + d^2 = r^2

So,

(l2)2=r2−d2 \left(\frac{l}{2}\right)^2 = r^2 - d^2

Taking square roots:

l2=r2−d2 \frac{l}{2} = \sqrt{r^2 - d^2}

Therefore the chord length is

l=2r2−d2. l = 2\sqrt{r^2 - d^2}.

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3In 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.Show solution
No, we cannot conclude that CD = 2AB.

From the chapter, if a chord is farther from the centre, it is shorter. So if the distance of AB from the centre is twice the distance of CD, then AB is farther away, so AB is shorter than CD.

Thus the chord lengths do not have a simple ratio of 2:1; in fact, the longer chord is the one closer to the centre.

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Exercise

1A 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°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.Show solution

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EXERCISE SET 5.6

1In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?Show solution
The central angle is 60∘60^\circ and radius r=12r=12 cm.

In triangle formed by the two radii and the chord, the chord length is
AB=2rsin⁡(60∘2)=2⋅12⋅sin⁡30∘=24⋅12=12 cm. AB = 2r\sin\left(\frac{60^\circ}{2}\right)=2\cdot 12\cdot \sin 30^\circ =24\cdot \frac12 = 12\text{ cm}.
So the chord length is 12 cm.

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2(i)Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?Show solution
On the same side of chord AB, all angles subtended by AB at points on that same arc segment are equal. So there are no points X, Y on the same side of AB on the circle for which ∠AXB\angle AXB is different from ∠AYB\angle AYB.

So the answer is No.

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2(ii)Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?Show solution
If ∠AXB=∠AYB\angle AXB = \angle AYB, then by the result on angles in the same arc segment, X and Y must lie on the same side of AB on the circle, not on opposite sides.

So the statement “they lie on the same side of the circle” is not the correct conclusion as written. The correct idea is that they lie on the same segment/side of AB.

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2(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?
2Let A and B be two points on a circle with centre O.

END-OF-CHAPTER EXERCISES

1In 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?
3The 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.
4A 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?
5Prove that the perpendicular bisector of a chord passes through the centre of the circle.
9The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
10A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
11Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies
12When 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.
13Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
14Show that rectangle is the only parallelogram that can be inscribed in a circle.
15Show 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.
16Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
17In 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 ∠BAC”.
18Two 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.
19A 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.
20A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
21Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
22“There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
23Let 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.
24How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
25In 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.
26How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
6The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB ? Explain your reasoning.
7ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75∘75^\circ , what is the measure of ∠C ? If ∠B measures 110∘110^\circ , what is the measure of ∠D ?
8Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)^\circand∠R=(3x−20)∘ and ∠R = (3x - 20)^\circ , find the value of xx and the measures of ∠P and ∠R .
11Consider 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?

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I'm Up and Down, and Round and Round covers several key topics that are frequently asked in CBSE Class 9 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in I'm Up and Down, and Round and Round — CBSE Class 9 Mathematics?
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