Class 9 · MathsNCERT 2026 Revised Syllabus
Chapter 5: I’m Up and Down, and Round and Round (0904CH05)
Ganita Manjari Part I · 44 questions · 7 exercises
Chapter code: 0904ch05 · Book: Ganita Manjari Part I · Class 9 Maths
Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 5. Every exercise has a clear, step-by-step answer for chapter code 0904CH05.
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Exercise Set 5.1(4 questions)
1Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
School Level Solution
Construction: Draw AB = 5 cm. At A construct ∠A = 70° and at B construct ∠B = 60°. The two rays meet at C.∠C = 180° − 70° − 60° = 50°.All three angles are less than 90°, so ΔABC is an acute-angled triangle.Construct the perpendicular bisectors of two sides. Their intersection is the circumcentre O inside the triangle.Therefore, the circumcentre lies inside the triangle.
2Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
School Level Solution
Construction: Draw AB = 5 cm. At A construct ∠A = 100°. With A as centre and radius 4 cm, locate C on the ray and join BC.Since ∠A = 100° > 90°, ΔABC is an obtuse-angled triangle.Construct the perpendicular bisectors of two sides. Their intersection is the circumcentre O outside the triangle.Therefore, the circumcentre lies outside the triangle.
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.
School Level Solution
Construction: Draw AB = 6 cm. With A and B as centres and radius 7 cm, draw arcs intersecting at C. Join AC and BC.Since BC = CA = 7 cm, the triangle is isosceles.Construct the perpendicular bisectors of the sides. Their intersection gives the circumcentre O.Measuring the radii gives OA ≈ OB ≈ OC.OA ≈ OB ≈ OC ≈ 3.85 cm.
4What is the least possible radius of a circle through two points A and B?
School Level Solution
The least possible circle through two points A and B is obtained when AB is a diameter.Therefore, radius = AB2.∴ Least possible radius = AB2 Ans.
Exercise Set 5.2(2 questions)
1Show that the triangle formed by a chord and the centre of the circle is isosceles.
School Level Solution
Let AB be a chord and O be the centre of the circle.OA = OB = radius of the circle.Therefore, ΔOAB has two equal sides OA and OB.∴ ΔOAB is isosceles.
2Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
School Level Solution
Let the two triangles be OAB and OCD, with AB = CD.OA = OC and OB = OD, since all are radii.Also, AB = CD is given.Therefore, by SSS congruence, ΔOAB ≅ ΔOCD.
Exercise Set 5.3(3 questions)
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?(Hint) Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.
School Level Solution
Given, ∠CMA = ∠CMB = 90° and CA = CB.CM is common to both triangles.Therefore, ΔCMA and ΔCMB are congruent by RHS.∴ AM = BM.Hence, the perpendicular from the centre of a circle to a chord bisects the chord.
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.
School Level Solution
In isosceles ΔABC, AB = AC.The altitude from A to BC bisects the base BC.So it is the perpendicular bisector of chord BC.The perpendicular bisector of a chord passes through the centre of the circle.Therefore, the altitude from A passes through the centre O.
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.
School Level Solution
For chord of length 6 cm, half chord = 3 cm.Distance of the 6 cm chord from the centre = √(5² − 3²) = 4 cm.For chord of length 8 cm, half chord = 4 cm.Distance of the 8 cm chord from the centre = √(5² − 4²) = 3 cm.The chords are on opposite sides of the centre.Distance between their midpoints = 4 + 3 = 7 cm Ans.
Exercise Set 5.4(3 questions)
1Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
School Level Solution
Let the equal chords be AB and FG, and let CE and CH be their perpendicular distances from the centre C.Since E and H are midpoints of the chords, AE = FH.By Baudhāyana–Pythagoras,CA² = CE² + AE²CF² = CH² + FH²But CA = CF, AE = FH.Therefore CE² = CH², so CE = CH — hence Theorem 6 is true.
2Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
School Level Solution
Fig. 5.15
3Solve the previous question using the Baudhāyana–Pythagoras theorem.
School Level Solution
Using Baudhāyana–Pythagoras,CA² = CE² + AE²CF² = CH² + FH²CA = CF and CE = CH.Therefore AE² = FH² ⇒ AE = FH.∴ AB = 2AE = 2FH = GF.
Exercise Set 5.5(3 questions)
1Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
School Level Solution
Given, radius r = 7 cm and perpendicular distance d = 6 cm.Half of chord = √(r² − d²)= √(7² − 6²) = √13 cm.Therefore, length of chord = 2√13 = 2√13 cm Ans.
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 2√(r² − d²).
School Level Solution
Let M be the midpoint of the chord.The perpendicular from the centre to the chord bisects it.So, OM = d and half chord = AB2.By Baudhāyana–Pythagoras,r² = d² + AB²4⇒ AB²4 = r² − d²⇒ AB = 2√(r² − d²) Ans.
★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.
School Level Solution
No, we cannot conclude that CD = 2AB.For any chord, length = 2√(r² − d²).Thus chord length depends on the radius and the distance from the centre; it is not directly proportional to the distance.For example, taking r = 5 cm, d_AB = 2 cm and d_CD = 1 cm,AB = 2√21 cm while CD = 4√6 cm.Therefore, CD ≠ 2AB.
Exercise Set 5.6(3 questions)
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?
School Level Solution
Given, ∠AOB = 60° and radius OA = OB = 12 cm.Join O to the midpoint M of chord AB.In right ΔAOM, ∠AOM = 30°.AM = 12 sin 30° = 6 cm.Therefore, AB = 2AM = 12 cm Ans.
2Let 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 ∠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?
School Level Solution
(i) No. Points on the same segment of chord AB subtend equal angles at AB.
(ii) No. Equal subtended angles do not by themselves determine which side of AB the points lie on; points on opposite sides can subtend equal angles in special cases.
(iii) No, not without the same-side condition. The converse theorem used in the chapter requires the two points to be on the same side of AB. Equal angles alone are not sufficient to conclude that X and Y lie on one circle in all configurations.
3Find x in Fig. 5.26.
School Level Solution
Fig. 5.26
End-of-Chapter Exercises(26 questions)
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?
School Level Solution
Given, radius r = 13 cm and distance of chord from centre d = 5 cm.Half chord = √(13² − 5²) = √144 = 12 cm.Therefore, chord length = 24 cm Ans.
2An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
School Level Solution
Angle at centre = 70°.Angle subtended by the same arc at a point on the circle = 70°2 = 35° Ans.
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.
School Level Solution
Diameter = 26 cm ⇒ radius = 13 cm.Half chord = 242 = 12 cm.Distance from centre = √(13² − 12²) = √25 = 5 cm Ans.
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?
School Level Solution
Given, r = 15 cm and d = 9 cm.Half chord = √(15² − 9²) = √144 = 12 cm.Therefore, chord length = 24 cm Ans.
5Prove that the perpendicular bisector of a chord passes through the centre of the circle.
School Level Solution
Let AB be a chord and O be the centre.Let M be the midpoint of AB.Then OA = OB (radii), AM = BM (M is midpoint) and OM is common.Therefore, ΔOMA ≅ ΔOMB by SSS.∴ ∠OMA = ∠OMB.They form a linear pair, so each = 90°.Hence OM is the perpendicular bisector of AB and therefore passes through O.
6The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
School Level Solution
AB is a diameter.Therefore, the arc AB is a semicircle and the angle subtended by it at C is half of 180°.∠ACB = 90° Ans.
7ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?
School Level Solution
Opposite angles of a cyclic quadrilateral are supplementary.∠C = 180° − 75° = 105°.∠D = 180° − 110° = 70°.
8Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.
School Level Solution
∠P + ∠R = 180°.(2x + 10) + (3x − 20) = 180⇒ 5x − 10 = 180⇒ x = 38.∠P = 2(38)+10 = 86°.∠R = 3(38)−20 = 94°.
9The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
School Level Solution
Half chord = 162 = 8 cm.Given distance d = 6 cm.r² = 6² + 8² = 36 + 64 = 100.∴ r = 10 cm Ans.
10A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
School Level Solution
Sides of the cyclic quadrilateral are 5, 5, 12 and 12 units.Semiperimeter, s = 5+5+12+122 = 17.By Brahmagupta’s formula,Area = √[(s−5)(s−5)(s−12)(s−12)]= √(12×12×5×5)= 60 square units Ans.
★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?
School Level Solution
Construct the perpendicular bisectors of two sides of the cyclic quadrilateral.Their point of intersection is the centre of the circumcircle.If this point lies inside the quadrilateral, the circumcentre is inside; otherwise it is outside.This is the best method because the perpendicular bisectors locate the circumcentre directly.
★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.
School Level Solution
Let equal chords AB and CD intersect at P.Given, AB = CD.So PA + PB = PC + PD.Also, by the intersecting-chords theorem, PA×PB = PC×PD.Thus PA and PB are the two roots of the same quadratic as PC and PD.Therefore, the corresponding segments are equal: PA = PC and PB = PD.
★13Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
School Level Solution
Half of the required chord = 62 = 3 cm and distance from centre = 3 cm.Join the centre O to one endpoint A and the midpoint M of the chord.OA² = OM² + AM² = 3² + 3² = 18.Therefore, radius = 3√2 cm.So draw a circle of radius 3√2 cm with the chord 6 cm long at distance 3 cm from its centre.
★14Show that rectangle is the only parallelogram that can be inscribed in a circle.
School Level Solution
Let ABCD be a parallelogram inscribed in a circle.Opposite angles of a parallelogram are equal, while opposite angles of a cyclic quadrilateral are supplementary.Hence each pair of opposite angles is both equal and supplementary.Therefore each angle = 90°.∴ the parallelogram is a rectangle.
★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.
School Level Solution
In a rectangle, diagonals are equal and bisect each other.Let O be the intersection of its diagonals.Then OA = OB = OC = OD.Therefore O is equidistant from all four vertices.Hence O is the centre of the circumcircle.
★16Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
School Level Solution
Let a chord have fixed length l in a circle of radius R.The perpendicular from O to the chord bisects it, so half the chord is l2.Hence the distance d of its midpoint from O is √(R² − (l2)²), which is constant.Therefore, the midpoints of all such chords form a circle with centre O.
★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”.
School Level Solution
Given, chords AB and AC are congruent.Consider ΔOAB and ΔOAC.OA = OC and OB = OA are radii, while AB = AC is given.Therefore, ΔOAB ≅ ΔOAC by SSS.∴ ∠BAO = ∠OAC.Hence O 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.
School Level Solution
Half of chord 10 cm = 5 cm and half of chord 24 cm = 12 cm.Let their distances from centre be d₁ and d₂ respectively.Since they are on the same side and are 7 cm apart, d₁ − d₂ = 7.Also r² = d₁² + 5² = d₂² + 12².So d₁² − d₂² = 119.(d₁−d₂)(d₁+d₂)=119 ⇒ 7(d₁+d₂)=119 ⇒ d₁+d₂=17.Thus d₁=12, d₂=5.r²=12²+5²=169.∴ r = 13 cm Ans.
★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.
School Level Solution
Join the centre to two adjacent vertices of the regular hexagon.The central angle = 360°/6 = 60°.The triangle formed is equilateral.Therefore, each side of the hexagon = r.Distance of each side from centre = r cos30° = √3r/2.
20A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
School Level Solution
MN is a diameter.∠MOP is the central angle subtended by chord MP, while ∠MNP is the angle subtended by the same chord at the circle.Therefore, by the theorem of angles subtended by an arc,∠MOP = 2∠MNP.
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).
School Level Solution
In cyclic quadrilateral ABCD, ∠B + ∠D = 180°.The exterior angle at D = 180° − ∠D.Therefore, exterior angle at D = ∠B.Similarly, the exterior angle at any vertex equals the interior opposite angle.
★22“There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
School Level Solution
Let AB be any chord and O the centre.If OM is perpendicular to AB, then M is the midpoint of AB.So AB = 2√(r² − OM²) ≤ 2r.But 2r is the diameter.Therefore, no chord can be longer than the diameter.
★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.
School Level Solution
Let a chord through A have perpendicular distance d from O.Its length is 2√(R² − d²).For a line through A, the greatest possible perpendicular distance from O is OA, and this occurs when the chord is perpendicular to OA.A larger d gives a smaller chord.Therefore, the shortest chord through A is the one perpendicular to OA.
24How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
School Level Solution
In Fig. 5.30, let the endpoints of the diameter be L and R and let A be the point on the semicircle. Since OA=OL=OR, triangles AOL and AOR are isosceles. If the base angles at L and R are a and b, then ∠AOL = 180°−2a and ∠AOR = 180°−2b. Because LOR is a straight line, ∠AOL+∠AOR=180°. Hence 2a+2b=180°, so a+b=90°. But ∠LAR=a+b, therefore ∠LAR=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.
School Level Solution
Because AB is a diameter through the centre O and each chord CC′ and DD′ is perpendicular to AB, the endpoints C,C′ are reflections of each other in AB, and likewise D,D′. Reflection in AB sends chord CD to chord C′D′. Therefore it sends the midpoint M of CD to the midpoint M′ of C′D′. The segment joining a point to its reflection is perpendicular to the mirror line, so MM′ ⟂ 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°?
School Level Solution
In Fig. 5.31, OA=OB=OC=OD, so each of the four triangles AOB, BOC, COD and DOA is isosceles. Their base angles are p,q,u,v as marked. Thus the four central angles are 180°−(p+q), 180°−(q+u), 180°−(u+v), and 180°−(v+p). These four central angles make one complete turn around O, so their sum is 360°. Therefore 720°−2(p+q+u+v)=360°, giving p+q+u+v=180°. The opposite interior angles are ∠A=p+v and ∠C=q+u, hence ∠A+∠C=180°; similarly ∠B+∠D=180°.