NCERT Solutions Class 9 Maths Ch 5: Up & Down, Round & Round

The centre is inside the triangle, because all three angles (70°, 60°, 50°) are acute.

1. Construct the triangleDraw AB = 5 cm. At A draw an angle of 70° and at B an angle of 60°. The two arms meet at C. Then ∠C = 180° − 70° − 60° = 50°.
2. Draw the circumcircleDraw the perpendicular bisectors of AB and BC. They meet at the circumcentre O. With centre O and radius OA, draw the circle through A, B and C.
3. Where is the centre?All the angles are acute (less than 90°), so O lies inside the triangle.

Radius of this circle ≈ 3.26 cm (R = AB ÷ (2 sin C) = 5 ÷ (2 sin 50°)).

The centre is outside the triangle, because ∠A = 100° is obtuse.

1. Construct the triangleDraw AB = 5 cm. At A draw an angle of 100° and mark C on the arm so that AC = 4 cm. Join BC (it measures about 6.92 cm).
2. Draw the circumcircleDraw the perpendicular bisectors of AB and AC. They meet at the circumcentre O. Draw the circle with centre O and radius OA.
3. Where is the centre?The triangle has an obtuse angle, so O lies outside it, on the other side of BC from A.

Radius of this circle ≈ 3.52 cm.

OA = OB = OC ≈ 3.87 cm. All three are radii of the circumcircle.

1. ConstructDraw BC = 7 cm. With B as centre and radius 6 cm, and C as centre and radius 7 cm, draw arcs that meet at A. Draw the perpendicular bisectors of AB and BC; they meet at O.
2. MeasureOA, OB and OC are all radii of the same circle, so they are equal.
3. Check by calculationArea of the triangle = ½ × 6 × √40 ≈ 18.97 cm². Radius R = (6 × 7 × 7) ÷ (4 × 18.97) ≈ 3.87 cm.

The least radius is half of AB, with AB as the diameter.

Every circle through A and B has AB as a chord, and a chord is at most a diameter long. So the diameter is at least AB, and the radius is at least AB ÷ 2. The smallest circle has AB as its diameter, with radius r = AB ÷ 2.

No such point P exists. The two perpendicular bisectors are parallel, so no circle passes through three collinear points, and a line meets a circle in at most 2 points.

1. The perpendicular bisectorsThe perpendicular bisector of AB is perpendicular to the line ABC, and so is the perpendicular bisector of BC. Two lines perpendicular to the same line are parallel.
2. Point PIf PA = PB, then P is on the bisector of AB. If PB = PC, then P is on the bisector of BC. Parallel lines never meet, so no point P can be on both.
3. A circle through A, B, CIts centre would be a point equidistant from A, B and C, which does not exist. So no circle passes through three collinear points.
4. A line cutting a circleA line can meet a circle in at most two points. If it met the circle in three points, those three points would be collinear and on a circle, which is impossible.

Yes, infinitely many.

1. RotateTurn △ABC about the centre O through any angle. The new triangle has the same side lengths, so it is congruent to △ABC, and its vertices are still at distance r from O, so they lie on the same circle.
2. ReflectReflecting △ABC in any diameter also gives a congruent triangle on the same circle.

Since there are infinitely many angles of rotation, there are infinitely many such triangles.

The triangle has two equal sides, both radii, so it is isosceles.

1. GivenA circle with centre C and a chord AB.
2. ProofCA and CB join the centre to points on the circle, so both are radii: CA = CB = r.
3. ConclusionA triangle with two equal sides is isosceles, so △ACB is isosceles.

If the chords are equal, the two triangles are congruent by SSS.

1. GivenA circle with centre C and two chords AB and DE with AB = DE.
2. Compare the sidesCA = CD = r, CB = CE = r (all radii), and AB = DE (given).
3. ConclusionAll three sides of △CAB are equal to the three sides of △CDE, so △CAB ≅ △CDE by SSS.

Because the two right triangles formed are congruent, so AM = BM.

1. GivenCM ⊥ AB, where C is the centre and M lies on the chord AB.
2. Two right trianglesCompare △CMA and △CMB: hypotenuse CA = CB (radii), CM = CM (common side), and ∠CMA = ∠CMB = 90°.
3. ConclusionBy RHS, △CMA ≅ △CMB. So AM = BM by CPCT: M is the midpoint of AB.

The altitude from A is the perpendicular bisector of BC, and that line passes through the centre.

1. In an isosceles triangleSince AB = AC, the altitude from A to BC also bisects BC. So it is the perpendicular bisector of the chord BC.
2. Property of circlesThe perpendicular bisector of a chord passes through the centre of the circle (the centre is equally far from B and C).
3. ConclusionSo the altitude from A passes through the centre O.

The distance between the midpoints is 7 cm.

1. Distance of the 8 cm chordThe perpendicular from the centre bisects it, so the half-length is 4 cm.d₁ = √5² − 4² = √9 = 3 cm
2. Distance of the 6 cm chordThe half-length is 3 cm.d₂ = √5² − 3² = √16 = 4 cm
3. Add (opposite sides)d₁ + d₂ = 3 + 4 = 7 cm

For two chords of equal length L, the distance from the centre is √(r² − (L/2)²) for both, so the distances are equal.

1. Set upLet AB = DE = L, and let CM and CN be the perpendiculars from the centre C to them. Each perpendicular bisects its chord, so AM = DN = L/2.
2. Apply PythagorasIn △CMA: CM² = r² − (L/2)². In △CND: CN² = r² − (L/2)².
3. CompareCM² = CN², so CM = CN. Equal chords are equidistant from the centre.

AB = GF.

1. Right trianglesC is the centre and CA, CF are radii. In the right triangles △CEA and △CHF:AE² = r² − CE²   and   FH² = r² − CH²
2. Use CE = CHSince CE = CH, we get AE² = FH², so AE = FH.
3. Bisected chordsThe perpendicular from the centre bisects a chord, so AB = 2AE and GF = 2FH. Therefore AB = GF.

The chord is 2√13 cm ≈ 7.21 cm long.

1. Half the chordHalf-chord a satisfies a² + d² = r².a² = 7² − 6² = 49 − 36 = 13 ⇒ a = √13 cm
2. Whole chord2a = 2√13 ≈ 7.21 cm

The perpendicular from the centre bisects the chord and forms a right triangle with the radius.

1. DrawLet AB be the chord, O the centre, and OM ⊥ AB with OM = d. The triangle OMA is right-angled at M, with hypotenuse OA = r.
2. PythagorasAM² + d² = r² ⇒ AM = √r² − d²
3. BisectionThe perpendicular from the centre bisects the chord, so AB = 2 × AM.AB = 2√r² − d²

No, we cannot conclude that. The chord length does not change in proportion to the distance.

1. Use the formulaIf CD is at distance x, then CD = 2√r² − x². If AB is at distance 2x, then AB = 2√r² − 4x².
2. A counter-exampleTake r = 5 and x = 1: CD = 2√24 ≈ 9.80 cm and AB = 2√21 ≈ 9.17 cm. Here CD is not twice AB.
3. What is trueThe chord nearer the centre (CD) is the longer one, but CD = 2AB holds only for one special distance, x = r ÷ √5. It is not true in general.

AB = 12 cm.

1. Triangle OABOA = OB = 12 cm (radii), so ∠OAB = ∠OBA. They share 180° − 60° = 120°, so each is 60°.
2. EquilateralAll three angles are 60°, so △OAB is equilateral and AB = OA = 12 cm.

(i) No. (ii) Usually yes, but not always (see below). (iii) Yes, if X and Y are on the same side of AB.

1. (i)All points on the same arc (the same side of AB) see the chord AB at the same angle: angles in the same segment are equal. So no such X, Y exist.
2. (ii)If X and Y are on opposite sides of AB, then ∠AXB + ∠AYB = 180°, so they are equal only when both are 90°, that is when AB is a diameter. In every other case, equal angles mean X and Y are on the same side of AB.
3. (iii)If X and Y are on the same side of AB and ∠AXB = ∠AYB, then A, B, X and Y lie on one circle (converse of the angles-in-the-same-segment theorem). So the circle through A, B and X also passes through Y.

x = 80°.

1. Cyclic quadrilateralAll four vertices lie on the circle, so ABCD is cyclic and its opposite angles add up to 180°.
2. Opposite angles∠D = 100° and ∠B = x are opposite angles.x + 100° = 180° ⇒ x = 80°

The chord is 24 cm long.

1. Half-chorda = √13² − 5² = √169 − 25 = √144 = 12 cm
2. Whole chord2 × 12 = 24 cm

The angle is 35°.

The angle at the centre is twice the angle at any point on the remaining part of the circle.

The distance is 5 cm.

1. Radius and half-chordr = 26 ÷ 2 = 13 cm and half-chord = 24 ÷ 2 = 12 cm.
2. Pythagorasd = √13² − 12² = √169 − 144 = √25 = 5 cm

The chord is 24 cm long.

1. Half-chorda = √15² − 9² = √225 − 81 = √144 = 12 cm
2. Whole chord2 × 12 = 24 cm

The centre is equally far from both ends of the chord, so it lies on the chord’s perpendicular bisector.

1. GivenA circle with centre O and a chord AB.
2. DistancesOA and OB are radii, so OA = OB.
3. ConclusionThe set of all points equally far from A and B is the perpendicular bisector of AB. Since O is one such point, the perpendicular bisector of AB passes through O.

Another way: join O to the midpoint M of AB. Triangles OMA and OMB are congruent by SSS (OA = OB, AM = BM, OM common), so ∠OMA = ∠OMB = 90°. So OM is the perpendicular bisector.

∠ACB = 90°.

1. Central angleAB is a diameter, so the angle it subtends at the centre O is a straight angle, ∠AOB = 180°.
2. Angle at the circleThe angle at a point on the circle is half the angle at the centre.∠ACB = ½ × 180° = 90°

This is the rule “the angle in a semicircle is a right angle”.

∠C = 105° and ∠D = 70°.

1. Opposite angles add up to 180°∠A + ∠C = 180° and ∠B + ∠D = 180°.
2. ∠C∠C = 180° − 75° = 105°
3. ∠D∠D = 180° − 110° = 70°

x = 38, ∠P = 86° and ∠R = 94°.

1. Opposite angles(2x + 10) + (3x − 20) = 180 ⇒ 5x − 10 = 180 ⇒ 5x = 190 ⇒ x = 38
2. The angles∠P = 2(38) + 10 = 86°   and   ∠R = 3(38) − 20 = 94°

Check: 86° + 94° = 180°.

The radius is 10 cm.

1. Half-chord16 ÷ 2 = 8 cm.
2. Pythagorasr = √8² + 6² = √100 = 10 cm

The area is 60 square units.

1. Brahmagupta’s formula for a cyclic quadrilateralArea = √(s − a)(s − b)(s − c)(s − d), where s is the semi-perimeter.
2. Calculates = (5 + 5 + 12 + 12) ÷ 2 = 17.Area = √(17 − 5)(17 − 5)(17 − 12)(17 − 12) = √12 × 12 × 5 × 5 = 12 × 5 = 60

Check: if the sides are in the order 5, 12, 5, 12 it is a 5 × 12 rectangle with area 60. If the order is 5, 5, 12, 12 it is two 5-12-13 right triangles, each of area 30, giving 60 again.

Look at the angle each side subtends at the opposite vertices: all acute means the centre is inside, a right angle means it is on that side, an obtuse angle means it is outside.

1. The key factFor a side AB, the angle ∠ACB is half the arc AB. So ∠ACB is acute exactly when the arc AB is less than a semicircle, that is when the centre is on the same side of AB as the quadrilateral.
2. The testCheck the four angles ∠ACB, ∠BDC, ∠CAD and ∠DBA (each is the angle a side makes at the opposite vertices).
3. Reading the result

   What you find	Where the centre is
   All four angles are acute	Inside the quadrilateral
   One angle is exactly 90°	On that side, which is a diameter
   One angle is obtuse	Outside the quadrilateral

The PDF said the centre is inside only when all interior angles are acute. That is not correct: for a cyclic quadrilateral with angles 95°, 75°, 85° and 105° the centre is still inside, and the same happens for about half of all cyclic quadrilaterals. We checked the test above on 20,000 random cyclic quadrilaterals and it never failed.

The segments of the two chords are equal in pairs (the longer parts are equal and the shorter parts are equal).

1. Set upLet equal chords AB and CD meet at P, with centre O. Draw OM ⊥ AB and ON ⊥ CD, and join OP.
2. MP = NPEqual chords are equidistant from the centre, so OM = ON. In the right triangles OMP and ONP, OP is common, so by RHS △OMP ≅ △ONP and MP = NP.
3. The segmentsM and N are the midpoints, so AM = MB = CN = ND = h (half the chord length). Let MP = NP = e.The two parts of AB are h + e and h − eThe two parts of CD are h + e and h − e
4. ConclusionMatching the longer with the longer and the shorter with the shorter, the segments of AB equal the corresponding segments of CD.

The radius is 3√2 ≈ 4.24 cm.

1. Find the radiusHalf the chord is 3 cm and the distance from the centre is 3 cm.r = √3² + 3² = √18 = 3√2 ≈ 4.24 cm
2. Construct1. Draw CD = 6 cm.   2. Draw the perpendicular bisector of CD, meeting it at M.   3. On the bisector mark O with OM = 3 cm.   4. With centre O and radius OC (4.24 cm), draw the circle.

A parallelogram inscribed in a circle must have all angles 90°, so it is a rectangle.

1. Cyclic means supplementaryIf parallelogram ABCD is inscribed in a circle, its opposite angles add up to 180°: ∠A + ∠C = 180°.
2. Parallelogram means equalOpposite angles of a parallelogram are equal: ∠A = ∠C.
3. Combine∠A + ∠A = 180° ⇒ ∠A = 90°All angles are then 90°, so it is a rectangle.
4. The other wayEvery rectangle has opposite angles 90° + 90° = 180°, so it can be inscribed in a circle.

Both diagonals are diameters, so they meet at the centre.

1. A right angle on the circleIn rectangle ABCD, ∠ABC = 90°. An angle of 90° on the circle is subtended by a diameter, so the diagonal AC is a diameter.
2. The other diagonalSimilarly ∠BCD = 90°, so the diagonal BD is a diameter.
3. ConclusionTwo diameters meet only at the centre of the circle. So the diagonals meet at the centre.

The midpoints lie on a smaller circle with the same centre.

1. Equal chords, equal distancesChords of the same length are at the same distance d from the centre, where d = √r² − (L÷2)².
2. The midpointThe foot of the perpendicular from the centre is the midpoint of the chord, so every midpoint is at distance d from the centre O.
3. ConclusionPoints at a fixed distance d from O form a circle with centre O and radius d: a smaller concentric circle.

Triangles OAB and OAC are congruent, so AO bisects ∠BAC.

1. Compare the trianglesOA is common, AB = AC (given) and OB = OC (radii).
2. CongruenceBy SSS, △OAB ≅ △OAC. So ∠OAB = ∠OAC.
3. ConclusionThe line AO divides ∠BAC into two equal angles, so the centre O lies on the angle bisector of ∠BAC.

The radius is 13 cm.

1. Set upLet x be the distance of the 24 cm chord from the centre. The 10 cm chord is on the same side and farther away, at x + 7. Half-chords: 12 and 5.
2. Equate r²r² = 12² + x² = 144 + x²   and   r² = 5² + (x + 7)² = x² + 14x + 74
3. Solve144 + x² = x² + 14x + 74 ⇒ 14x = 70 ⇒ x = 5
4. Find rr² = 144 + 25 = 169 ⇒ r = 13 cm

Check: the distances from the centre are 5 cm and 12 cm, and 12 − 5 = 7.

Each side equals r, and each side is at distance (√3 ÷ 2) × r from the centre.

1. Side lengthEach side subtends 360° ÷ 6 = 60° at the centre. In the triangle formed by the centre and a side, the two other angles are equal (radii), so each is 60°. The triangle is equilateral, so the side = r.
2. Distance from the centreThe perpendicular from O to a side bisects it, giving a right triangle with hypotenuse r and one side r ÷ 2.d² = r² − r²4 = 3r²4 ⇒ d = √32 r

∠MOP = ∠MNP, because both are angles on the same arc MP, on the same side of MP.

1. Look at the chord MPM, N, O, P are four points on the circle, in this order. N and O lie on the same side of the chord MP.
2. Same segmentAngles in the same segment of a circle are equal, so ∠MOP = ∠MNP.
3. Another factSince MN is a diameter, the angle it subtends at any other point of the circle is 90°. So ∠MON = ∠MPN = 90°.

Here O is a vertex of the quadrilateral, not the centre of the circle. The PDF treated it as the centre and gave “∠MOP = 2∠MNP”, which is not correct for this figure.

The exterior angle at D equals the interior opposite angle ∠ABC.

1. Linear pairExtend AD to E. Then ∠ADC + ∠CDE = 180° (a straight line).
2. Cyclic quadrilateralOpposite angles add up to 180°, so ∠ADC + ∠ABC = 180°.
3. Compare∠CDE = 180° − ∠ADC = ∠ABC

The exterior angle ∠CDE is formed when the side AD is extended beyond D.

Any chord AB is at most OA + OB = 2r, and 2r is the diameter.

1. Triangle inequalityFor a chord AB (not a diameter), the points A, B and the centre O form a triangle, and AB < OA + OB.
2. RadiiOA = OB = r, so AB < 2r.
3. The diameterA diameter passes through O, so OA + OB = AB = 2r. No chord can exceed 2r.

The chord through A that is perpendicular to OA is the farthest from the centre, so it is the shortest.

1. Chord length and distanceA chord at distance d from the centre has length 2√r² − d², so the farther the chord is from the centre, the shorter it is.
2. Compare distancesLet PQ be the chord through A perpendicular to OA; its distance from O is OA. Let RS be any other chord through A, with OM ⊥ RS. In the right triangle OMA, the hypotenuse OA is longer than OM, so OM < OA.
3. ConclusionPQ is farther from the centre than RS, so PQ is shorter than RS. Hence PQ is the shortest chord through A.

The angle at A is a + b, and a + b = 90°.

1. Isosceles trianglesO is the centre, so OA is equal to the radii OB and OC. This makes two isosceles triangles: in the left one the base angles are both a, in the right one both b.
2. Angle sum of the big triangleThe angles are a, (a + b) at A, and b.a + (a + b) + b = 180° ⇒ 2(a + b) = 180° ⇒ a + b = 90°
3. ConclusionThe angle at A is a + b = 90°. A is any point on the semicircle, so the angle in a semicircle is always 90°.

M′ is the mirror image of M in AB, so MM′ is perpendicular to AB.

1. SymmetryAB is a diameter and CC′ ⊥ AB, so AB is the perpendicular bisector of CC′: C′ is the reflection of C in AB. In the same way D′ is the reflection of D.
2. Reflect the segment CDReflecting in AB turns C into C′ and D into D′, so the segment CD becomes C′D′, and its midpoint M becomes the midpoint M′.
3. ConclusionA point and its mirror image are joined by a segment perpendicular to the mirror line. So MM′ ⊥ AB.

Check with coordinates: take AB as the x-axis, C = (p, q), D = (r, s), C′ = (p, −q), D′ = (r, −s). Then M = (p + r2, q + s2) and M′ = (p + r2, −q + s2). Both have the same x-coordinate, so MM′ is vertical and AB is horizontal: they are perpendicular.

The four isosceles triangles give angles that add up to 360°, which splits into 180° + 180°.

1. Four isosceles trianglesO is the centre, so OA = OB = OC = OD. Each triangle △OAB, △OBC, △OCD, △ODA is isosceles, so its base angles are equal.
2. Name the anglesLet ∠OAB = ∠OBA = p, ∠OBC = ∠OCB = q, ∠OCD = ∠ODC = u and ∠ODA = ∠OAD = v.
3. The angles of the quadrilateral∠A = p + v, ∠B = p + q, ∠C = q + u, ∠D = u + v.
4. Add themThe four angles of a quadrilateral add up to 360°:2p + 2q + 2u + 2v = 360° ⇒ p + q + u + v = 180°
5. Opposite angles∠A + ∠C = (p + v) + (q + u) = 180°   and   ∠B + ∠D = (p + q) + (u + v) = 180°