Chapter 5 exercise answers: I’m Up and Down, and Round and Round

Open NCERT p. 98Checked by computer

1. The angles of the triangle add up to 180°, so ∠C = 180° − 70° − 60° = 50°.
2. All three angles, 70°, 60° and 50°, are less than 90°, so ΔABC is an acute triangle.
3. The circumcentre of an acute triangle always lies inside the triangle.

AnswerThe circumcentre lies inside the triangle.

Watch this explained “Where the centre lands”, 7:13 into Three points not in a line: exactly one circle (Theorem 1) · हिंदी में देखें

Open NCERT p. 98Checked by computer

1. ∠A = 100°, which is more than 90°, so ∠A is obtuse.
2. A triangle can have only one angle of 90° or more, so ∠A is the largest angle of ΔABC.
3. The circumcentre of an obtuse triangle always lies outside the triangle.

AnswerThe circumcentre lies outside the triangle.

Watch this explained “Where the centre lands”, 7:13 into Three points not in a line: exactly one circle (Theorem 1) · हिंदी में देखें

Open NCERT p. 98Checked by computer

1. O is the circumcentre, so OA = OB = OC = R, the radius of the circumcircle.
2. CA = CB = 7 cm, so C lies on the perpendicular bisector of AB, and so does O. Let M be the midpoint of AB. Then AM = 3 cm, CM ⊥ AB, and O lies on CM.
3. By the Baudhāyana–Pythagoras theorem in ΔCMA: CM² = 7² − 3² = 49 − 9 = 40, so CM = 2√10 cm.
4. OC = R, so OM = 2√10 − R. In the right triangle OMA: OA² = OM² + AM², so R² = (2√10 − R)² + 3² = 40 − 4√10R + R² + 9.
5. The R² terms cancel, leaving 4√10R = 49, so R = 49/(4√10) = 49√10/40 cm.

AnswerOA = OB = OC = 49√10/40 cm ≈ 3.87 cm, so a careful drawing measures about 3.9 cm for each.

Watch this explained “One circle, many triangles”, 9:09 into Three points not in a line: exactly one circle (Theorem 1) · हिंदी में देखें

Open NCERT p. 98Checked by computer

1. Every circle through A and B has its centre on the perpendicular bisector of AB.
2. If the centre is at height h above the midpoint of AB, the radius is √(h² + (AB/2)²) by the Baudhāyana–Pythagoras theorem.
3. This radius is smallest when h = 0, i.e., when the centre is the midpoint of AB itself.
4. At h = 0, the radius is AB/2, and AB is then a diameter of the circle.

AnswerThe least possible radius is AB/2, with AB as a diameter.

Watch this explained “And it is the smallest”, 2:16 into Two points: infinitely many circles, centres on the perpendicular bisector · हिंदी में देखें

Open NCERT p. 100One way to think about it

1. Let AB be any chord of a circle with centre O.
2. Join OA and OB; both are radii of the same circle.
3. All radii of a circle are equal in length, so OA = OB.
4. A triangle with two equal sides is isosceles, so ΔOAB is isosceles.

In shortΔOAB is isosceles because OA = OB (both are radii of the same circle).

Watch this explained “The triangle it brings with it”, 0:49 into Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) · हिंदी में देखें

Open NCERT p. 100One way to think about it

1. Let ΔOAB and ΔOCD be two such triangles from the previous question, so OA = OB = OC = OD, all being radii of the same circle.
2. Given: the bases are equal in length, so AB = CD.
3. Compare the triangles: OA = OC, OB = OD (matching radii), and AB = CD (given bases).
4. All three sides of ΔOAB match all three sides of ΔOCD, so by the SSS test, ΔOAB ≅ ΔOCD.

In shortΔOAB ≅ ΔOCD by the SSS congruence test.

Watch this explained “Same length, same angle”, 2:29 into Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) · हिंदी में देखें

Open NCERT p. 101One way to think about it

1. In Fig. 5.12, C is the centre, AB is the chord, and CM ⊥ AB, so ∠CMA = ∠CMB = 90°.
2. CA and CB are both radii of the circle, so CA = CB.
3. CM is a side common to both ΔCMA and ΔCMB.
4. In the right triangles CMA and CMB, the hypotenuse CA = hypotenuse CB, and the side CM is shared, so by the RHS congruence, ΔCMA ≅ ΔCMB.
5. Congruent triangles have matching sides, so AM = BM.

In shortAM = BM, so the perpendicular from the centre bisects the chord.

Watch this explained “The tool the converse gets”, 4:54 into The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5) · हिंदी में देखें

Open NCERT p. 101One way to think about it

1. Since AB = AC, the point A is the same distance from B and C.
2. The centre O is also the same distance from B and C, since OB = OC (both are radii).
3. So A and O both lie on the perpendicular bisector of BC, the line of all points equidistant from B and C.
4. The altitude from A is the line through A perpendicular to BC. There is only one such line, and the perpendicular bisector of BC passes through A and is perpendicular to BC, so the altitude from A is the perpendicular bisector of BC.
5. So the altitude from A passes through O. (If ∠A is obtuse, O lies beyond BC, on the altitude extended.)

In shortThe altitude from A to BC passes through the centre O, because the altitude and the perpendicular bisector of BC are the same line, and O lies on that line.

Watch this explained “The height that finds the centre”, 8:50 into The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5) · हिंदी में देखें

Open NCERT p. 101Checked by computer

1. Let O be the centre, AB the 8 cm chord with midpoint M, and CD the 6 cm chord with midpoint N.
2. The line joining the centre to the midpoint of a chord is perpendicular to the chord, so OM ⊥ AB and ON ⊥ CD.
3. In the right triangle OMA: OM² = OA² − AM² = 5² − 4² = 9, so OM = 3 cm.
4. In the right triangle ONC: ON² = OC² − CN² = 5² − 3² = 16, so ON = 4 cm.
5. The chords are parallel, so the line through O perpendicular to AB is also perpendicular to CD. So M, O and N all lie on one straight line.
6. The chords are on opposite sides of the centre, so O lies between M and N. So MN = OM + ON = 3 + 4 = 7 cm.

AnswerThe distance between the midpoints of the chords is 7 cm.

Watch this explained “One word changes the answer”, 7:57 into The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5) · हिंदी में देखें

Open NCERT p. 104One way to think about it

1. Let AB and CD be two equal chords of a circle with centre O, so AB = CD.
2. Draw OM ⊥ AB and ON ⊥ CD; M and N are then the midpoints of AB and CD (the perpendicular from the centre bisects a chord).
3. So AM = AB/2 and CN = CD/2, and since AB = CD, AM = CN.
4. In right ΔOMA, by the Baudhāyana–Pythagoras theorem: OA² = OM² + AM². In right ΔONC: OC² = ON² + CN².
5. OA and OC are both radii, so OA = OC, giving OM² + AM² = ON² + CN².
6. Since AM = CN, their squares are equal, so OM² = ON², which gives OM = ON.

In shortOM = ON, so equal chords are equidistant from the centre — this is Theorem 6.

Watch this explained “One equation behind all three”, 6:04 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 104One way to think about it

1. In Fig. 5.15, E is the foot of the perpendicular from centre C to chord AB, and H is the foot of the perpendicular from C to chord GF (H lies on GF, so 'GH' means the part of GF from G to H).
2. Since CE ⊥ AB, E is the midpoint of AB, so AE = AB/2. Since CH ⊥ GF, H is the midpoint of GF, so GH = GF/2.
3. CA and CG are both radii of the circle, so CA = CG.
4. In right triangles CEA and CHG: hypotenuse CA = hypotenuse CG, and CE = CH (given), so by the RHS congruence, ΔCEA ≅ ΔCHG.
5. Congruent triangles have matching sides, so AE = GH.
6. So AB/2 = GF/2, which gives AB = GF.

In shortAB = GF, proved using the RHS congruence of ΔCEA and ΔCHG.

Watch this explained “Now turn it round”, 3:31 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 104One way to think about it

1. E and H are the midpoints of AB and GF (the perpendicular from the centre bisects a chord), so AB = 2·AE and GF = 2·GH.
2. In right ΔCEA, by the Baudhāyana–Pythagoras theorem: CA² = CE² + AE². In right ΔCHG: CG² = CH² + GH².
3. CA and CG are both radii, so CA = CG, which gives CE² + AE² = CH² + GH².
4. Since CE = CH (given), their squares are equal, so AE² = GH², which gives AE = GH (lengths are positive).
5. Multiplying both sides by 2: AB = GF.

In shortAB = GF, found directly from the Pythagoras theorem applied to the two half-chord right triangles.

Watch this explained “One equation behind all three”, 6:04 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 105Checked by computer

1. Let half the chord be x. The radius, the perpendicular distance and half the chord form a right triangle.
2. By the Baudhāyana–Pythagoras theorem: 7² = 6² + x², so x² = 49 − 36 = 13.
3. So x = √13 cm, and the full chord is 2x = 2√13 cm.

AnswerThe length of the chord is 2√13 cm ≈ 7.21 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 105One way to think about it

1. Let AB be a chord of a circle with centre O and radius r, and let M be the foot of the perpendicular from O to AB, so OM = d.
2. The perpendicular from the centre to a chord bisects the chord, so M is the midpoint of AB, and AM = AB/2.
3. ΔOMA is right-angled at M, with OA = r (a radius) and OM = d.
4. By the Baudhāyana–Pythagoras theorem: OA² = OM² + AM², so r² = d² + AM².
5. So AM² = r² − d², which gives AM = √(r² − d²).
6. Since AB = 2·AM, the chord length is AB = 2√(r² − d²).

In shortThe chord length is 2√(r² − d²), from applying Pythagoras to the half-chord right triangle.

Watch this explained “One equation behind all three”, 6:04 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 106Checked by computer

1. Let the radius be r and let CD be at distance d from the centre. Then AB is at distance 2d (given), with 2d < r so that AB is a chord.
2. The perpendicular from the centre bisects a chord, so by the Baudhāyana–Pythagoras theorem: CD = 2√(r² − d²) and AB = 2√(r² − 4d²).
3. Try r = 5 cm and d = 2 cm (so AB is 4 cm from the centre): CD = 2√21 ≈ 9.17 cm and AB = 2√9 = 6 cm. Then 2 AB = 12 cm, which is not equal to CD.
4. So CD = 2 AB is not true in general. It holds only in one special position: CD = 2 AB needs r² − d² = 4(r² − 4d²), that is, d² = r²/5.

AnswerNo. In general CD ≠ 2 AB, because a chord's length depends on r² − d², not on its distance in proportion. (CD = 2 AB happens only in the special case d² = r²/5.)

Watch this explained “The doubling trap”, 8:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 110Checked by computer

1. OA and OB are both radii, so OA = OB = 12 cm.
2. ΔOAB is isosceles with OA = OB and the angle between them, ∠AOB, equal to 60°.
3. An isosceles triangle with a 60° angle between the two equal sides has its other two angles also equal to (180° − 60°)/2 = 60°, so all three angles are 60° and the triangle is equilateral.
4. In an equilateral triangle all sides are equal, so AB = OA = OB = 12 cm.

AnswerThe length of chord AB is 12 cm.

Watch this explained “Double the angle, and see”, 8:13 into Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) · हिंदी में देखें

Open NCERT p. 111Checked by computer

(i) such that ∠AXB is different from ∠AYB?

1. X and Y are on the circle and on the same side of AB, so both lie on the same arc cut off by AB, and neither lies on the other arc AB.
2. By Theorem 9, the angle that other arc AB subtends at a point of the circle outside it is half the angle it subtends at the centre O. So ∠AXB and ∠AYB are both half of the same angle.
3. So ∠AXB = ∠AYB for any two such points: angles in the same segment are equal.

AnswerNo. Points of the circle on the same side of AB always see AB at the same angle.

(ii) X and Y lie on the same side of the circle?

1. Here X and Y are points on the circle, as in (i), and 'same side' means the same side of AB.
2. If X and Y are on opposite sides of AB, each angle is half the angle at O of the arc it does not lie on. These two arcs make up the whole circle, so their angles at O add to 360°, and ∠AXB + ∠AYB = 180°.
3. So for points on opposite sides, ∠AXB = ∠AYB only when both angles are 90°, which happens exactly when AB is a diameter (the angle subtended by a diameter at any point on the circle is 90°).
4. Take AB a diameter, X on one side and Y on the other: ∠AXB = ∠AYB = 90°, but X and Y are on opposite sides. So the statement fails in this case.

AnswerNo, not always. If AB is not a diameter, equal angles do put X and Y on the same side of AB; but if AB is a diameter, every point of the circle sees AB at 90°, so X and Y can be on opposite sides.

(iii) does the circle through A, B and X also pass through Y?

1. The question does not say that X and Y are on the same side of AB, so test a case where they are not.
2. Let AB = 4 cm with midpoint M, and let the given circle be the one with AB as diameter (radius 2 cm). Take X 3 cm directly above M and Y 3 cm directly below M. Neither is on the circle, since MX = MY = 3 cm, not 2 cm.
3. Y is the mirror image of X in the line AB, so ∠AXB = ∠AYB.
4. The circle through A, B and X has its centre P on XM, the perpendicular bisector of AB. With PM = p: PA² = p² + 2² and PX = 3 − p, so p² + 4 = (3 − p)², giving p = 5/6 cm and radius 3 − 5/6 = 13/6 cm.
5. But PY = 3 + 5/6 = 23/6 cm, not 13/6 cm, so this circle does not pass through Y.
6. If X and Y are on the same side of AB, the answer is yes: that is Theorem 10, proved in the next section.

AnswerNo, not necessarily. When X and Y are on opposite sides of AB, the circle through A, B and X can miss Y; it does pass through Y when X and Y are on the same side of AB (Theorem 10).

Watch this explained “Running it backwards, with a piece missing”, 7:25 into Equal angles in the same segment: the arc looks the same from every point beyond it · हिंदी में देखें

Open NCERT p. 111Checked by computer

1. Fig. 5.26 shows a cyclic quadrilateral ABCD with ∠D = 100° and ∠B = x.
2. ∠B and ∠D are opposite angles of the cyclic quadrilateral, so they add up to 180°.
3. x = 180° − 100° = 80°.

Answerx = 80°.

Watch this explained “Putting numbers in”, 5:08 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. Let half the chord be x. The radius, the distance from the centre, and half the chord form a right triangle.
2. By the Baudhāyana–Pythagoras theorem: 13² = 5² + x², so x² = 169 − 25 = 144.
3. So x = 12 cm, and the chord is 2x = 24 cm.

AnswerThe length of the chord is 24 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. The angle an arc subtends at the centre is double the angle it subtends at any point on the circle (on the other part of the circle).
2. So the angle at a point on the circle = 70° ÷ 2 = 35°.

AnswerThe angle subtended at a point on the circle is 35°.

Watch this explained “The numbers it gives you”, 7:48 into Major and minor arcs, and why an arc's central angle is double what it subtends on the circle (Theorem 9) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. The radius is half the diameter: r = 26/2 = 13 cm.
2. Half the chord is 24/2 = 12 cm.
3. The radius, the distance from the centre, and half the chord form a right triangle, so by Pythagoras: 13² = d² + 12², giving d² = 169 − 144 = 25.
4. So d = 5 cm.

AnswerThe distance from the centre to the chord is 5 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. Let half the chord be x. The radius, the distance from the centre, and half the chord form a right triangle.
2. By Pythagoras: 15² = 9² + x², so x² = 225 − 81 = 144.
3. So x = 12 cm, and the chord is 2x = 24 cm.

AnswerThe length of the chord is 24 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 114One way to think about it

1. Let AB be a chord of a circle with centre O. OA = OB, because both are radii.
2. If AB is a diameter, O is the midpoint of AB, so O lies on the perpendicular bisector of AB and there is nothing more to prove. From here on AB is not a diameter, so O is not on AB.
3. Let M be the midpoint of AB and join OM. Compare triangles OMA and OMB: OA = OB (radii), MA = MB (M is the midpoint) and OM is common. So △OMA ≅ △OMB (SSS).
4. Hence ∠OMA = ∠OMB. These two angles lie along the straight line AB, so they add up to 180°. Being equal, each is 90°, so OM ⊥ AB.
5. So the line OM passes through the midpoint of AB and is perpendicular to AB. Only one line does both, so OM is the perpendicular bisector of AB, and it passes through the centre O.
6. The same fact in one line: every point that is equally far from A and B lies on the perpendicular bisector of AB, and the centre is such a point because OA = OB.

In shortThe centre O is equally far from both ends of the chord (OA = OB, radii). Joining O to the midpoint M of AB gives two congruent triangles (SSS), so OM ⊥ AB. The line OM is therefore the perpendicular bisector of AB, and it passes through the centre.

Watch this explained “Finding a centre never marked”, 5:57 into The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. Let O be the centre. AB is a diameter, so O is the midpoint of AB and the radii OA and OB point in opposite directions.
2. Take the arc from A to B that does not contain C. Turning the radius OA along this arc until it lies along OB sweeps a straight angle, so this arc subtends 180° at the centre.
3. C is on the circle outside this arc, so by Theorem 9 the angle the arc subtends at C is half the angle at the centre: ∠ACB = ½ × 180° = 90°.
4. This does not depend on where C is, as long as C is not A or B: the angle in a semicircle is always 90°.

Answer∠ACB = 90°. The arc from A to B that does not contain C is a semicircle, which subtends 180° at the centre, and the angle at C is half of that.

Watch this explained “The whole derivation”, 2:55 into The Corollary: a diameter stands on a right angle wherever you take the point · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. In any cyclic quadrilateral, opposite angles add up to 180°.
2. ∠A and ∠C are opposite, so ∠A + ∠C = 180°, giving ∠C = 180° − 75° = 105°.
3. ∠B and ∠D are opposite, so ∠B + ∠D = 180°, giving ∠D = 180° − 110° = 70°.

Answer∠C = 105° and ∠D = 70°.

Watch this explained “Opposite, not next door”, 0:39 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. In quadrilateral PQRS, ∠P and ∠R are opposite angles (P,Q,R,S in order around the circle), so ∠P + ∠R = 180°.
2. (2x + 10) + (3x − 20) = 180
3. 5x − 10 = 180, so 5x = 190, so x = 38.
4. ∠P = 2(38) + 10 = 86°.
5. ∠R = 3(38) − 20 = 94°.
6. Check: 86° + 94° = 180° ✓

Answerx = 38, ∠P = 86°, ∠R = 94°.

Watch this explained “Putting numbers in”, 5:08 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. The perpendicular from the centre to a chord bisects the chord, so half the chord is 16 ÷ 2 = 8 cm.
2. The radius, the half-chord and the distance from the centre form a right triangle (right angle at the midpoint of the chord), with the radius as the hypotenuse.
3. radius² = 8² + 6² = 64 + 36 = 100
4. radius = √100 = 10 cm.

AnswerThe radius is 10 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 114Checked by computer

1. The question gives the four sides but not their order. Either the two 5s are next to each other (5, 5, 12, 12) or they alternate (5, 12, 5, 12). Both give the same area.
2. Order 5, 5, 12, 12: let AB = BC = 5 and CD = DA = 12, and join BD. Triangles ABD and CBD have AB = CB, AD = CD and BD common, so they are congruent (SSS) and ∠A = ∠C.
3. ABCD is cyclic, so ∠A + ∠C = 180°. Since ∠A = ∠C, each is 90°.
4. So triangles ABD and CBD are right-angled at A and C, each with legs 5 and 12. Area = ½ × 5 × 12 + ½ × 5 × 12 = 30 + 30 = 60 square units. (The diagonal BD is √(5² + 12²) = 13.)
5. Order 5, 12, 5, 12: both pairs of opposite sides are equal, so ABCD is a parallelogram and its opposite angles are equal. Being cyclic, opposite angles also add to 180°, so every angle is 90°. It is a 5 by 12 rectangle, with area 5 × 12 = 60 square units.

AnswerThe area is 60 square units, whichever order the sides come in.

Watch this explained “What it earns”, 8:50 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 114One way to think about it

1. There is more than one way to decide this; here is one that needs only a ruler and a protractor.
2. Let the cyclic quadrilateral be ABCD. Draw one diagonal, AC. It splits ABCD into triangles ABC and ACD.
3. A, B and C lie on the quadrilateral's circle, and only one circle passes through three points that are not in a line (Theorem 1). So that circle is the circumcircle of triangle ABC, and in the same way of triangle ACD. Its centre O is the circumcentre of both triangles.
4. Recall where a triangle's circumcentre lies: inside the triangle if the triangle is acute-angled, at the midpoint of the longest side if it is right-angled, and outside if it is obtuse-angled.
5. If either triangle ABC or ACD is acute-angled, O is inside that triangle, and so inside the quadrilateral.
6. ∠B + ∠D = 180° (opposite angles), so at most one of them is more than 90°. If ∠B = ∠D = 90°, both triangles are right-angled with longest side AC, so O is the midpoint of AC: inside the quadrilateral, on the diagonal.
7. Otherwise one of them, say ∠D, is more than 90°. Then triangle ACD is obtuse and O is outside it, so O can only be inside the quadrilateral by being inside triangle ABC. If ABC is right-angled (at A or C), O is the midpoint of a side (BC or AB) of the quadrilateral, so it is on the boundary. If ABC is obtuse too, O is outside the quadrilateral.
8. So the best way is to draw one diagonal and measure the angles of the two triangles it makes, with no circle needed. If one of the two triangles is acute-angled, the centre is inside.

In shortDraw one diagonal, say AC. The quadrilateral's circumcircle is also the circumcircle of triangles ABC and ACD, so its centre is the circumcentre of both. The centre is inside the quadrilateral if one of these two triangles is acute-angled (or if ∠B = ∠D = 90°, when it is the midpoint of AC). Otherwise it is outside, or on a side when a triangle has its right angle at A or C. Measuring the angles of the two triangles settles it without drawing the circle.

Watch this explained “Where the centre lands”, 7:13 into Three points not in a line: exactly one circle (Theorem 1) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let the equal chords AB and CD of a circle with centre O cross at P.
2. Drop perpendiculars OM on AB and ON on CD. The perpendicular from the centre bisects a chord (Theorem 5), so M and N are the midpoints. Call half the common length h: AM = MB = CN = ND = h.
3. Equal chords are equally far from the centre (Theorem 6), so OM = ON.
4. Triangles OMP and ONP are right-angled at M and N. By the Baudhāyana–Pythagoras theorem, MP² = OP² − OM² and NP² = OP² − ON². So MP = NP. Call this length x.
5. P lies on AB at distance x from its midpoint M, so it cuts AB into pieces of lengths h + x and h − x. In the same way it cuts CD into pieces of lengths h + x and h − x.
6. So the longer piece of one chord equals the longer piece of the other, and the shorter piece equals the shorter piece. (If x = 0, P is the midpoint of both chords and all four pieces equal h.)

In shortEach chord is cut at P into pieces h + x and h − x, where h is half the common length and x is the distance from P to the chord's midpoint. That distance is the same for both chords, because OM = ON (equal chords are equidistant from the centre) and so MP = NP. So the two chords are cut into matching equal pieces.

Watch this explained “First argument: the whole triangles”, 1:53 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. First find the radius. Let O be the centre and AB the 6 cm chord, and let M be the foot of the perpendicular from O to AB. Then OM = 3 cm, and M is the midpoint of AB (Theorem 5), so AM = 3 cm.
2. By the Baudhāyana–Pythagoras theorem, OA² = OM² + AM² = 3² + 3² = 18, so the radius is √18 = 3√2 cm, about 4.24 cm.
3. A suitable triangle (the hint): draw AB = 6 cm. At B draw a line perpendicular to AB and mark C on it with BC = 6 cm. Join AC. Triangle ABC is right-angled at B.
4. The circumcentre of a right-angled triangle is the midpoint of its hypotenuse. So find the midpoint O of AC (draw its perpendicular bisector) and draw the circle with centre O and radius OA. It passes through A, B and C.
5. Why this is the circle we want: AC = √(6² + 6²) = 6√2 cm, so the radius OA = 3√2 cm. The perpendicular OM from the centre bisects AB, so AM = 3 cm and OM² = OA² − AM² = 18 − 9 = 9, giving OM = 3 cm. The 6 cm chord AB is 3 cm from the centre.
6. A direct construction gives the same circle: draw AB = 6 cm and its perpendicular bisector through the midpoint M, mark O on it with OM = 3 cm, and draw the circle with centre O through A.

In shortThe circle must have radius 3√2 cm (about 4.24 cm). One way to draw it: make a right-angled triangle ABC with AB = BC = 6 cm and the right angle at B. Its circumcircle, centred at the midpoint of AC, has the 6 cm chord AB exactly 3 cm from the centre.

Watch this explained “The right-angled case, sharpened”, 8:13 into Three points not in a line: exactly one circle (Theorem 1) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let ABCD be a parallelogram inscribed in a circle.
2. Since ABCD is inscribed in a circle, opposite angles are supplementary: ∠A + ∠C = 180° and ∠B + ∠D = 180°.
3. Since ABCD is a parallelogram, opposite angles are equal: ∠A = ∠C and ∠B = ∠D.
4. From ∠A + ∠C = 180° and ∠A = ∠C, we get 2∠A = 180°, so ∠A = 90°.
5. Similarly, ∠B = ∠D = 90°.
6. A parallelogram with all angles 90° is a rectangle.
7. Conversely, every rectangle has all angles 90°, so opposite angles sum to 180°, satisfying the cyclic quadrilateral condition.

In shortA parallelogram is inscribed in a circle if and only if all its angles are 90°, which means it must be a rectangle.

Watch this explained “What it earns”, 8:50 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let rectangle ABCD be inscribed in a circle with centre O, and let its diagonals AC and BD meet at X.
2. ∠ABC = 90°. B is on the circle outside the arc ADC, so by Theorem 9 that arc subtends 2 × 90° = 180° at the centre. A straight angle at O means A, O and C lie on one straight line, so AC is a diameter and O lies on AC. (Equivalently: the circumcentre of the right-angled triangle ABC is the midpoint of its hypotenuse AC.)
3. In the same way ∠BAD = 90°, so the arc BCD subtends 180° at O. BD is a diameter, so O lies on BD.
4. O lies on both diagonals. Two different lines meet in only one point, so O is the point X where the diagonals cross.

In shortEach right angle of the rectangle stands on a diagonal, and a 90° angle on the circle means its arc makes 180° at the centre. So both diagonals are diameters, both pass through the centre, and the only point they share, where they cross, is the centre.

Watch this explained “Two diameters wearing a hat”, 8:43 into The Corollary: a diameter stands on a right angle wherever you take the point · हिंदी में देखें

Open NCERT p. 115Checked by computer

1. For a chord of a fixed length L in a circle of radius r, the perpendicular distance from the centre O to the chord is also fixed: d = √(r² − (L/2)²) (from the right triangle formed by the radius, the half-chord and this distance).
2. The midpoint of any such chord is exactly the foot of this perpendicular from O — so its distance from O is always this same fixed value d, no matter which direction the chord points in.
3. A set of points all at one fixed distance from a fixed centre O is, by definition, a circle.
4. So the midpoints of all chords of length L form a circle of radius d = √(r² − (L/2)²), centred at the same centre O as the original circle.

AnswerThe midpoints form a circle, centred at the same centre as the original circle, with radius √(r² − (L/2)²) (where r is the original radius and L is the fixed chord length).

Watch this explained “One equation behind all three”, 6:04 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. AB and AC are chords from the common point A, with AB = AC (given).
2. OA, OB and OC are all radii of the circle, so OB = OC.
3. Compare triangles OAB and OAC: OA is common to both, AB = AC (given), and OB = OC (radii).
4. So triangle OAB ≅ triangle OAC (SSS congruence).
5. Congruent triangles have matching angles equal, so ∠OAB = ∠OAC.
6. ∠OAB and ∠OAC are exactly the two halves that line AO splits ∠BAC into, so AO bisects ∠BAC — meaning O lies on the angle bisector of ∠BAC.

In shortSince OA is common, AB = AC (given) and OB = OC (both radii), triangles OAB and OAC are congruent (SSS), so ∠OAB = ∠OAC — which says exactly that line AO bisects ∠BAC, so O lies on that bisector.

Watch this explained “Same length, same angle”, 2:29 into Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) · हिंदी में देखें

Open NCERT p. 115Checked by computer

1. Let the chord of length 10 cm be at distance d1 from the centre, and the chord of length 24 cm be at distance d2 from the centre, with both on the same side of the centre.
2. Half-chords: 10/2 = 5 cm and 24/2 = 12 cm. From the right triangles formed with the radius r: r² = d1² + 5² and r² = d2² + 12².
3. The longer chord (24 cm) is nearer the centre, so d1 − d2 = 7 cm (the given gap between the chords), i.e. d1 = d2 + 7.
4. Substituting: (d2+7)² + 25 = d2² + 144, so 14·d2 + 49 + 25 = 144, so 14·d2 = 70, so d2 = 5 cm. Then d1 = 12 cm.
5. r² = 12² + 5² = 144 + 25 = 169, so r = 13 cm.
6. Check with the other chord: r² = 5² + 12² = 169 too. ✓

AnswerThe radius is 13 cm.

Watch this explained “Reading it off”, 9:01 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 115Checked by computer

1. Join the centre O to all six vertices. The six sides are equal chords, so they subtend equal angles at the centre (Theorem 2), and the six angles make a full turn, so each is 360° ÷ 6 = 60°.
2. Take one side AB. In triangle OAB, OA = OB = r, so its base angles are equal, each (180° − 60°) ÷ 2 = 60°. All three angles are 60°, so the triangle is equilateral and AB = r.
3. Drop the perpendicular OM from O to AB. It bisects AB (Theorem 5), so AM = r/2.
4. In right triangle OMA, by the Baudhāyana–Pythagoras theorem, OM² = OA² − AM² = r² − r²/4 = 3r²/4, so OM = (√3/2)r.

AnswerEach side of the hexagon is r, and each side is (√3/2)r from the centre.

Watch this explained “Double the angle, and see”, 8:13 into Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) · हिंदी में देखें

Open NCERT p. 115Checked by computer

1. Here O is a vertex of the quadrilateral, not the centre. The vertices come in the order M, N, O, P round the circle, and the side MN is a diameter.
2. ∠MOP is the angle at O made by the chord MP, and ∠MNP is the angle at N made by the same chord MP.
3. The chord MP cuts the circle into two arcs. N and O both lie on the arc that runs from M through N and O to P, so both are outside the other arc MP: they are on the same side of MP.
4. By Theorem 9, ∠MNP and ∠MOP are each half the angle that this other arc MP subtends at the centre. So ∠MOP = ∠MNP: angles in the same segment are equal.
5. What the diameter adds: the angles standing on MN are right angles, ∠MON = ∠MPN = 90°. In triangle MNP, right-angled at P, ∠MNP = 90° − ∠NMP, so ∠MNP (and therefore ∠MOP) is always less than 90°.
6. The actual size depends on where O and P are, so no single number can be given, only that the two angles are equal (and acute).

Answer∠MOP = ∠MNP. Both stand on the same chord MP from the same side (N and O are on the same arc), so they are angles in the same segment. Because MN is a diameter, ∠MPN = 90°, so both angles are acute (each equals 90° − ∠NMP). Their exact size depends on where O and P are.

Watch this explained “A diameter in the question is not 90 in the answer”, 7:49 into The Corollary: a diameter stands on a right angle wherever you take the point · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let ABCD be a cyclic quadrilateral. Produce side CD beyond D to a point E, so that C, D and E are on one straight line. The exterior angle at D is the angle between side DA and the extension DE, that is ∠ADE.
2. Angles on a straight line: ∠ADC + ∠ADE = 180°.
3. Opposite angles of a cyclic quadrilateral (Theorem 11): ∠ABC + ∠ADC = 180°.
4. Both totals are 180° and both contain ∠ADC, so ∠ADE = 180° − ∠ADC = ∠ABC.
5. A note on the book's example: with E on CD produced, C, D and E are in a line, so the exterior angle is ∠ADE. The book's ∠CDE is the same exterior angle if E is taken on AD produced beyond D instead; the argument is identical (∠ADC + ∠CDE = 180°), and it also equals ∠ABC.
6. The same works at every vertex: the exterior angle there is 180° minus the interior angle, and so is the interior opposite angle, so the two are equal.

In shortAt any vertex, the exterior angle and the interior angle add to 180° (a straight line), and the interior angle and the opposite interior angle also add to 180° (cyclic quadrilateral). So the exterior angle equals the interior opposite angle. For example, with CD produced to E, ∠ADE = ∠ABC.

Watch this explained “What it earns”, 8:50 into Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let AB be any chord of a circle with centre O and radius r.
2. In triangle OAB, OA = r and OB = r are two sides, and AB is the third side.
3. By the triangle inequality, any side of a triangle is at most the sum of the other two sides: AB ≤ OA + OB = r + r = 2r.
4. 2r is exactly the length of the diameter.
5. Equality (AB = 2r) happens exactly when O, A, B are collinear — that is, exactly when AB itself passes through the centre, i.e. AB is a diameter.
6. So every chord has length at most the diameter, and only the diameter itself reaches that maximum — no chord can be longer.

In shortFor any chord AB, triangle OAB gives AB ≤ OA + OB = 2r (the diameter), by the triangle inequality — with equality only when the chord passes through the centre, i.e. is itself a diameter. So no chord can exceed the diameter.

Watch this explained “The longest chord”, 6:23 into Total rotational symmetry, and why every diameter is an axis of reflection · हिंदी में देखें

Open NCERT p. 115One way to think about it

1. Let PQ be any chord through the fixed interior point A, and let M be the midpoint of PQ, so OM ⊥ PQ (perpendicular from the centre bisects the chord).
2. Since A also lies on line PQ, triangle OMA is right-angled at M, with OA as its hypotenuse. So OM ≤ OA always, with equality exactly when M and A are the same point.
3. The chord length is PQ = 2·√(r² − OM²) (from the right triangle formed by the radius r, the distance OM, and half the chord).
4. Since OM ≤ OA for every chord through A, we get r² − OM² ≥ r² − OA², so PQ is smallest exactly when OM is as large as possible, i.e. when OM = OA.
5. OM = OA happens exactly when M = A — that is, when A itself is the midpoint of the chord, which happens exactly when the chord is perpendicular to OA at A.
6. So among all chords through A, the one perpendicular to OA has the largest possible distance from the centre, and therefore the shortest length.

In shortThe chord through A perpendicular to OA has M = A, giving it the greatest possible distance OM (= OA) from the centre among all chords through A — and the greater the distance from the centre, the shorter the chord — so it is the shortest chord through A.

Watch this explained “The longer chord is the nearer one”, 4:55 into Length and distance from the centre are the same fact twice (Theorems 6–8) · हिंदी में देखें

Open NCERT p. 116One way to think about it

1. In Fig. 5.30, O is the centre, lying on the diameter (the two marked points at the ends of the base), and A is a point on the circle, with a dashed line drawn from A to O. The two base angles are marked a and b.
2. The dashed segment AO is a radius, equal in length to the two half-diameters (also radii), so each of the two smaller triangles it creates is isosceles.
3. In the left small triangle, the two equal sides make its base angle equal to the angle it cuts off at A — so that part of ∠A also equals a. Likewise, in the right small triangle, the part of ∠A there equals b.
4. So the full angle at A (of the big triangle) is a + b, since AO splits it exactly into those two parts.
5. The angle sum of the big triangle (with vertices at the two diameter ends and A) is 180°: a + b + (a + b) = 180°, i.e. 2(a+b) = 180°, so a + b = 90°.
6. Since the angle at A equals a + b, the angle at A — the angle in the semicircle — is 90°.

In shortSplitting the big triangle with the radius AO gives two isosceles triangles, so the angle at A equals a + b; the triangle's angle sum then forces a + b = 90°, i.e. the angle in the semicircle is 90°.

Watch this explained “The same answer out of two triangles”, 5:20 into The Corollary: a diameter stands on a right angle wherever you take the point · हिंदी में देखें

Open NCERT p. 116One way to think about it

1. A circle has reflection symmetry across any diameter, so reflecting the whole figure across diameter AB maps the circle onto itself.
2. Since chord CC' is perpendicular to AB, and AB passes through the centre, AB is exactly the perpendicular bisector of CC' — so reflecting across AB swaps C and C' (sends C to C', and C' to C).
3. In the same way, reflecting across AB swaps D and D' (since DD' ⊥ AB too).
4. So this reflection sends segment CD to segment C'D' (C↦C', D↦D'), and therefore sends the midpoint M of CD to the midpoint M' of C'D'.
5. Whenever a reflection across a line sends one point to another (and the two points are different), the segment joining the point to its own image is perpendicular to the mirror line.
6. So MM' (joining M to its mirror image M') is perpendicular to AB.

In shortReflection across the diameter AB swaps C with C' and D with D' (since both chords are perpendicular to AB), so it swaps M with M' too — and a point and its mirror image are always joined by a segment perpendicular to the mirror line, so MM' ⊥ AB.

Watch this explained “Every diameter, and nothing else”, 4:39 into Total rotational symmetry, and why every diameter is an axis of reflection · हिंदी में देखें

Open NCERT p. 116One way to think about it

1. In Fig. 5.31, O (the centre) is joined to all four vertices A, B, C, D of the cyclic quadrilateral, splitting it into four triangles: OAB, OBC, OCD, ODA — each isosceles, since every side to O is a radius.
2. Each isosceles triangle has two equal base angles; call the common base angle of triangle OAB as p, of OBC as q, of OCD as u, and of ODA as v (as marked in the figure).
3. The full angle at each vertex is made up of two of these base angles: ∠A = p + v (from triangles OAB and ODA), ∠B = p + q (from OAB and OBC), ∠C = q + u (from OBC and OCD), ∠D = u + v (from OCD and ODA).
4. Adding the opposite pair: ∠A + ∠C = (p+v) + (q+u) = p+q+u+v. Adding the other opposite pair: ∠B + ∠D = (p+q) + (u+v) = p+q+u+v — the same total.
5. So ∠A + ∠C = ∠B + ∠D.
6. The angles of any quadrilateral add to 360°, so (∠A+∠C) + (∠B+∠D) = 360°; since the two halves are equal, each one is 180°.
7. So ∠A + ∠C = 180° and ∠B + ∠D = 180° — the opposite angles of the cyclic quadrilateral are supplementary.

In shortJoining the centre to all four vertices splits the quadrilateral into four isosceles triangles; each vertex angle is a sum of two of their base angles (p,q,u,v), and both pairs of opposite angles add up to the same total p+q+u+v — which, since all four angles sum to 360°, must be 180° each.

Watch the lesson Cyclic quadrilaterals: opposite angles sum to 180°, and the converse (Theorems 11–12) · हिंदी में देखें