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Home · Schooling · CBSE · Class 9 · Mathematics · I’m Up and Down, and Round and Round · End-of-chapter exercises

NCERT Class 9 Mathematics I’m Up and Down, and Round and Round: Exercise Solutions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “I’m Up and Down, and Round and Round”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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Exercises Q1
A chord is 5 cm from the centre of a circle of radius 13 cm. Find its length.
1. Let O be the centre of the circle and AB be the chord of length 2x cm.
2. Draw OM perpendicular to AB, so OM = 5 cm and AM = MB = x cm.
3. In right-angled triangle OMA, OA is the radius, so OA = 13 cm.
4. Apply the Pythagorean theorem: OA² = OM² + AM².
5. Substitute the values: 13² = 5² + x².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Calculate the squares: 169 = 25 + x².
7. Solve for x²: x² = 169 − 25 = 144.
8. Find x: x = √144 = 12 cm.
9. The length of the chord AB is 2x = 2 × 12 = 24 cm.
Answer: 24 cm
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Exercises Q2
An arc subtends 70° at the centre. What angle does it subtend at a point on the circle?
1. The angle subtended by an arc at the centre is given as 70°.
2. The angle subtended by the same arc at any point on the remaining part of the circle is half the angle at the centre.
3. Calculate the angle at the point: 70° ÷ 2 = 35°.
Answer: 35°
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Exercises Q3
A circle of diameter 26 cm has a chord of 24 cm. Find the distance of the chord from the centre.
1. Let O be the centre of the circle and AB be the chord of length 24 cm.
2. Draw a perpendicular from O to the chord AB, meeting it at point M.
3. Since the perpendicular from the centre bisects the chord, AM = 24 ÷ 2 = 12 cm.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
4. The radius of the circle is half the diameter, so OA = 26 ÷ 2 = 13 cm.
5. In the right-angled triangle OMA, apply the Pythagorean theorem: OM² + AM² = OA².
6. Substitute the known values: OM² + 12² = 13².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
7. Calculate the squares: OM² + 144 = 169.
8. Solve for OM²: OM² = 169 − 144 = 25.
9. Take the square root to find OM: OM = √25 = 5 cm.
10. The distance of the chord from the centre is the length of OM.
Answer: 5 cm
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Exercises Q4
A chord is 9 cm from the centre of a circle of radius 15 cm. Find its length.
1. Let O be the centre of the circle and AB be the chord of length 2x cm.
2. Draw the perpendicular from O to the chord AB, meeting it at M.
3. The perpendicular from the centre to a chord bisects the chord, so AM = MB = x cm.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
4. The distance from the centre to the chord is given as OM = 9 cm.
5. The radius of the circle is OA = 15 cm.
6. In the right-angled triangle OMA, apply the Pythagoras theorem: OA² = OM² + AM².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
7. Substitute the known values: 15² = 9² + x².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
8. Calculate the squares: 225 = 81 + x².
9. Rearrange to find x²: x² = 225 − 81 = 144.
10. Take the square root to find x: x = √144 = 12 cm.
11. The total length of the chord AB is 2x = 2 × 12 = 24 cm.
Answer: 24 cm
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Exercises Q5
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
1. Let AB be a chord of a circle with centre O.
2. Let M be the midpoint of chord AB, so AM = MB.
3. Draw the perpendicular bisector of AB passing through M.
4. Join OA and OB to form triangles OMA and OMB.
5. In triangles OMA and OMB, OA = OB because both are radii of the same circle.
6. AM = MB because M is the midpoint of chord AB.
7. OM is common to both triangles OMA and OMB.
8. Therefore, triangle OMA is congruent to triangle OMB by the SSS congruence criterion.
9. By corresponding parts of congruent triangles, angle OMA is equal to angle OMB.
10. Since angle OMA and angle OMB form a linear pair, their sum is 180 degrees.
11. Thus, angle OMA = angle OMB = 90 degrees.
12. This proves that the line OM is perpendicular to the chord AB.
13. There is only one line through M perpendicular to AB, so the perpendicular bisector of AB is the line OM.
14. Hence the perpendicular bisector passes through the centre O (if AB is a diameter, M is O itself and the result is immediate).
Answer: The perpendicular bisector of a chord passes through the centre of the circle.
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Exercises Q6
AB is a diameter and C is a point on the circle. Find ∠ACB and explain.
1. AB is the diameter of the circle and C is a point on the circumference.
2. The angle subtended by a diameter at any point on the circle is a right angle.
3. Therefore, the angle ACB is 90°.
Answer: 90°
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Exercises Q7
ABCD is a cyclic quadrilateral with ∠A = 75° and ∠B = 110°. Find ∠C and ∠D.
1. In a cyclic quadrilateral, the sum of opposite angles is 180°.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
2. Given ∠A = 75° and ∠B = 110°.
3. Since ∠A and ∠C are opposite, ∠A + ∠C = 180°.
4. Substitute the value of ∠A: 75° + ∠C = 180°.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Solving for ∠C: ∠C = 180° − 75° = 105°.
6. Since ∠B and ∠D are opposite, ∠B + ∠D = 180°.
7. Substitute the value of ∠B: 110° + ∠D = 180°.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
8. Solving for ∠D: ∠D = 180° − 110° = 70°.
Answer: ∠C = 105°, ∠D = 70°
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Exercises Q8
PQRS is cyclic with ∠P = (2x + 10)° and ∠R = (3x − 20)°. Find x, ∠P and ∠R.
1. In a cyclic quadrilateral, the sum of opposite angles is 180°.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
2. Therefore, ∠P + ∠R = 180°.
3. Substitute the given expressions: (2x + 10) + (3x − 20) = 180.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
4. Simplify the equation: 5x − 10 = 180.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. Add 10 to both sides: 5x = 190.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
6. Divide by 5: x = 38.
7. Calculate ∠P: 2(38) + 10 = 76 + 10 = 86°.
8. Calculate ∠R: 3(38) − 20 = 114 − 20 = 94°.
Answer: x = 38, ∠P = 86°, ∠R = 94°
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Exercises Q9
A chord of 16 cm is 6 cm from the centre. Find the radius.
1. Let O be the centre of the circle and AB be the chord of length 16 cm.
2. Draw OM perpendicular to AB from the centre O, where M is the midpoint of AB.
3. Since the perpendicular from the centre bisects the chord, AM = 16 ÷ 2 = 8 cm.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
4. The distance from the centre to the chord is given as OM = 6 cm.
5. In the right-angled triangle OMA, apply the Pythagoras theorem: OA² = OM² + AM².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
6. Substitute the values: OA² = 6² + 8².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
7. Calculate the squares: OA² = 36 + 64.
8. Add the numbers: OA² = 100.
9. Take the square root to find the radius: OA = √100 = 10 cm.
Answer: 10 cm
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Exercises Q10
A cyclic quadrilateral has sides 5, 5, 12 and 12 units. Find its area.
1. Let the sides of the cyclic quadrilateral be a = 5, b = 5, c = 12, and d = 12.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
2. Calculate the semi-perimeter s using the formula s = (a + b + c + d) ÷ 2.
3. Substitute the values: s = (5 + 5 + 12 + 12) ÷ 2 = 34 ÷ 2 = 17.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
4. Use Brahmagupta's formula for the area of a cyclic quadrilateral: Area = √((s − a)(s − b)(s − c)(s − d)).
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
5. Substitute the values into the formula: Area = √((17 − 5)(17 − 5)(17 − 12)(17 − 12)).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Simplify the terms inside the square root: Area = √(12 × 12 × 5 × 5).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Calculate the product: 12 × 12 = 144 and 5 × 5 = 25, so Area = √(144 × 25).
8. Take the square root: √144 = 12 and √25 = 5, so Area = 12 × 5 = 60.
Answer: 60 square units
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Exercises Q11
For a cyclic quadrilateral, how can we tell, without drawing the circumcircle, whether its centre lies inside or outside the quadrilateral?
1. Let ABCD be cyclic with circumcentre O. Look at one side, say AB, and the angle it subtends at a vertex not on it, ∠ACB.
2. ∠ACB is half the angle AB subtends at O on the other side. If ∠ACB < 90°, O lies on the same side of AB as C and D; if ∠ACB = 90°, AB is a diameter and O is on AB; if ∠ACB > 90°, O lies on the far side of AB.
3. Do this for all four sides: ∠ACB for AB, ∠BDC for BC, ∠CAD for CD and ∠DBA for DA.
4. If all four angles are acute, O is on the inner side of every side, so O lies inside the quadrilateral.
5. If one of them is obtuse, O is outside, beyond that side (at most one can be obtuse, as the four arcs add up to 360°).
Answer: Measure the angle each side subtends at an opposite vertex (∠ACB, ∠BDC, ∠CAD, ∠DBA): all acute → centre inside; one equal to 90° → centre on that side; one obtuse → centre outside, beyond that side.
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Exercises Q12
Two equal chords of a circle intersect. Show that the two pieces of one chord equal the two pieces of the other.
1. Let AB and CD be equal chords of a circle with centre O, meeting at P. Let M and N be the feet of the perpendiculars from O to AB and CD.
2. The perpendicular from the centre bisects a chord, so M and N are the midpoints; since AB = CD, AM = MB = CN = ND; call this half-length h.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
3. Equal chords are equally far from the centre, so OM = ON.
4. Triangles OMP and ONP have right angles at M and N, OM = ON and the common hypotenuse OP, so they are congruent (RHS) and MP = NP; call this length t.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
5. On AB, P is t away from the midpoint M, so the two pieces of AB are h + t and h − t (P is inside the circle, so t < h; if P = M, then t = 0).
6. In the same way the two pieces of CD are h + t and h − t.
7. So the pieces of one chord are equal to the pieces of the other: the longer piece of AB equals the longer piece of CD and the shorter equals the shorter.
Answer: Both chords are cut into pieces of lengths h + t and h − t (h = half the chord, t = MP = NP), so the pieces of one chord equal the corresponding pieces of the other.
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Exercises Q13
Draw a circle in which a 6 cm chord is 3 cm from the centre (hint: think of it as the circumcircle of a suitable triangle).
1. Let AB be the 6 cm chord, M its midpoint and O the centre; then AM = 3 cm and OM ⟂ AB with OM = 3 cm.
2. By the Baudhāyana–Pythagoras theorem, OA² = 3² + 3² = 9 + 9 = 18, so the radius is OA = √18 = 3√2 cm ≈ 4.24 cm.
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
3. Construction: draw AB = 6 cm and its perpendicular bisector; on it mark O at 3 cm from M.
4. With centre O and radius OA (≈ 4.24 cm) draw the circle; it passes through A and B, and the chord AB is 3 cm from O.
5. Link with the hint: OM = AM = MB = 3 cm makes ∠AOB = 45° + 45° = 90°, so this circle is the circumcircle of any triangle ABX with X on the major arc, for which ∠AXB = 45°.
Answer: Radius = 3√2 cm ≈ 4.24 cm; centre on the perpendicular bisector of the chord, 3 cm from its midpoint.
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Exercises Q14
Show that the rectangle is the only parallelogram that can be inscribed in a circle.
1. Let ABCD be a parallelogram inscribed in a circle.
2. In a parallelogram, opposite angles are equal, so ∠A = ∠C and ∠B = ∠D.
3. Since ABCD is a cyclic quadrilateral, the sum of opposite angles is 180°.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
4. Therefore, ∠A + ∠C = 180°.
5. Substituting ∠A = ∠C, we get ∠A + ∠A = 180°, which means 2∠A = 180°.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
6. Solving for ∠A gives ∠A = 90°.
7. Adjacent angles of a parallelogram add up to 180°, so ∠B = 180° − 90° = 90°; and ∠C = ∠A = 90°, ∠D = ∠B = 90°.
8. A parallelogram with all four angles equal to 90° is a rectangle.
9. Thus, the only parallelogram that can be inscribed in a circle is a rectangle.
Answer: The rectangle is the only parallelogram that can be inscribed in a circle.
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Exercises Q15
Show that the diagonals of a rectangle inscribed in a circle meet at the centre of the circle.
1. Let ABCD be a rectangle inscribed in a circle with centre O, and let its diagonals AC and BD meet at P.
2. The diagonals of a rectangle are equal and bisect each other, so PA = PC = AC/2 and PB = PD = BD/2.
3. Since AC = BD, PA = PB = PC = PD.
4. So P is the centre of a circle through A, B and C.
5. A, B, C are not collinear, and there is only one circle through three non-collinear points; that circle is the given one, with centre O.
6. Hence P = O: the diagonals meet at the centre of the circle.
Answer: The diagonals of a rectangle inscribed in a circle meet at the centre of the circle.
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Exercises Q16
All chords of one fixed length in a circle: what shape do their midpoints form?
1. Consider a circle with centre O and radius R.
2. Let AB be a chord of fixed length L.
3. Let M be the midpoint of chord AB.
4. Join O to M.
5. Since the perpendicular from the centre to a chord bisects the chord, OM is perpendicular to AB.
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
6. In the right-angled triangle OMA, OA is the radius R and AM is half the chord length, so AM = L/2.
7. By the Pythagoras theorem, OM² + AM² = OA².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
8. Substituting the values, OM² + (L/2)² = R².
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
9. Rearranging gives OM² = R² − (L/2)².
10. Since R and L are fixed, the value of OM is constant.
11. This means the distance from the centre O to the midpoint M is always the same for any chord of length L.
12. The locus of points at a fixed distance from a centre is a circle.
13. Therefore the midpoints lie on the circle with centre O and radius √(R² − (L/2)²); and every point of that circle is the midpoint of such a chord (draw the chord through it perpendicular to the radius).
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
Answer: A circle concentric with the given one, of radius √(R² − (L/2)²) (just the centre if the chords are diameters).
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Exercises Q17
In a circle with centre O, chords AB and AC are equal. Explain why the centre lies on the bisector of ∠BAC.
1. Given that chords AB and AC are equal in a circle with centre O.
2. Consider triangles OAB and OAC.
3. OA is a common side to both triangles.
4. OB and OC are radii of the same circle, so OB = OC.
5. Since AB = AC (given), OB = OC, and OA = OA, triangle OAB is congruent to triangle OAC by SSS congruence criterion.
6. By CPCT (Corresponding Parts of Congruent Triangles), angle OAB is equal to angle OAC.
7. This means that AO bisects angle BAC.
8. Therefore, the centre O lies on the bisector of angle BAC.
Answer: The centre O lies on the bisector of ∠BAC because triangles OAB and OAC are congruent, making ∠OAB = ∠OAC.
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Exercises Q18
Parallel chords of 10 cm and 24 cm lie on the same side of the centre, 7 cm apart. Find the radius.
1. Let O be the centre and r the radius. On the same side of the centre, the longer chord is nearer to the centre.
2. Let the 24 cm chord be x cm from O; then the 10 cm chord is x + 7 cm from O (both perpendiculars lie on one line, as the chords are parallel).
3. Half-chords are 12 cm and 5 cm, so by the Baudhāyana–Pythagoras theorem r² = x² + 12² and r² = (x + 7)² + 5².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
4. Equate: x² + 144 = x² + 14x + 49 + 25, so 14x = 70 and x = 5.
5. Then r² = 5² + 12² = 25 + 144 = 169, so r = 13.
6. Check: the 10 cm chord is 12 cm from O and 12² + 5² = 169 as well.
Answer: Radius = 13 cm
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Exercises Q19
A regular hexagon is inscribed in a circle of radius r. Find its side and the distance of each side from the centre.
1. A regular hexagon has 6 equal sides, so the angle subtended by each side at the centre is 360° ÷ 6 = 60°.
2. Consider the triangle formed by the centre and two adjacent vertices of the hexagon.
3. The two sides of this triangle are radii of the circle, so both have length r.
4. Since the angle between these two radii is 60° and the two sides are equal, the triangle is isosceles with a vertex angle of 60°.
5. The base angles are therefore (180° − 60°) ÷ 2 = 60° each.
6. Since all three angles are 60°, the triangle is equilateral.
7. Thus, the side of the hexagon is equal to the radius, i.e., side = r.
8. To find the distance of each side from the centre, draw a perpendicular from the centre to the midpoint of a side.
9. This perpendicular bisects the side, so it meets the side at a point r/2 from each end.
10. By the Baudhāyana–Pythagoras theorem, distance² = r² − (r/2)² = (3/4)r².
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
11. So the distance of each side from the centre is (√3/2)r.
Answer: side = r; distance of each side from the centre = (√3/2)r
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Exercises Q20
MNOP is inscribed in a circle and MN is a diameter. What can you say about ∠MOP and ∠MNP? Explain.
1. The vertices are in the order M, N, O, P around the circle, so N and O lie on the same arc cut off by the chord MP.
2. Angles subtended by the same arc at points of the circle on the same side are equal, so ∠MOP = ∠MNP.
3. Since MN is a diameter, the angle it subtends at P is a right angle: ∠MPN = 90° (and also ∠MON = 90°).
4. In △MNP the angles add to 180°, so ∠MNP = 90° − ∠NMP; in particular both angles are acute.
Answer: ∠MOP = ∠MNP (both subtend chord MP from the same arc); each equals 90° − ∠NMP, because ∠MPN = 90° in the semicircle on MN.
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Exercises Q21
ABCD is a cyclic quadrilateral. Explain why an exterior angle equals the interior opposite angle (e.g. at D, the angle between DA and side CD extended beyond D equals ∠ABC).
1. Let ABCD be a cyclic quadrilateral inscribed in a circle.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
2. Extend side CD to a point E to form the exterior angle ∠ADE at vertex D.
3. The interior opposite angle to ∠ADE is ∠ABC.
4. Since ABCD is cyclic, the sum of opposite angles is 180°, so ∠ABC + ∠ADC = 180°.
5. Angles ∠ADE and ∠ADC form a linear pair on the straight line CDE, so ∠ADE + ∠ADC = 180°.
6. Comparing the two equations, ∠ABC + ∠ADC = ∠ADE + ∠ADC.
7. Subtracting ∠ADC from both sides gives ∠ABC = ∠ADE.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
8. Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
  why?
  The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.
Answer: The exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
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Exercises Q22
Justify: no chord of a circle is longer than its diameter.
1. Let AB be any chord of a circle with centre O.
2. Let CD be the diameter of the same circle.
3. If AB passes through O, it is itself a diameter, of length 2r. Otherwise O, A, B form a triangle.
4. Consider the triangle OAB formed by joining the centre to the endpoints of the chord.
5. Since OA and OB are radii of the circle, OA = OB.
6. Let the length of the radius be r, so OA = OB = r.
7. By the triangle inequality, the sum of any two sides of a triangle is greater than the third side.
8. Therefore, OA + OB > AB.
9. Substituting the values, we get r + r > AB.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. This simplifies to 2r > AB.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
11. Since the diameter CD is equal to 2r, we have CD > AB.
12. Thus every chord other than a diameter is shorter than the diameter, and a diameter is equal to it: no chord is longer than the diameter.
Answer: No chord of a circle is longer than its diameter.
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Exercises Q23
A is a point inside a circle with centre O. Show that the shortest chord through A is the one perpendicular to OA.
1. Take A ≠ O (if A = O, every chord through A is a diameter and all have the same length). Let r be the radius and OA = a, so a < r.
2. Let PQ be the chord through A perpendicular to OA. The perpendicular from O to PQ is OA itself, so A is its midpoint and PQ = 2√(r² − a²).
3. Let XY be any other chord through A and M its midpoint; OM ⟂ XY, so XY = 2√(r² − OM²).
4. In △OMA the angle at M is 90°, so OA is the hypotenuse and OM < OA = a (M ≠ A because XY is not perpendicular to OA).
  why?
  In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.
5. Then r² − OM² > r² − a², so XY > PQ.
6. Hence the chord through A perpendicular to OA is the shortest.
Answer: Every other chord through A is nearer to O than OA, so it is longer; the chord perpendicular to OA (at distance OA from O) is the shortest.
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Exercises Q24
Fig. 5.30: triangle on a diameter with centre O, apex A on the circle, OA drawn, base angles marked a and b. Use it to show the angle in a semicircle is 90°.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. In Fig. 5.30 the base is a diameter with centre O; call its ends P (angle a) and Q (angle b), with A the third vertex on the circle.
2. OP = OA (radii), so △OPA is isosceles and ∠OAP = ∠OPA = a.
3. OQ = OA (radii), so △OQA is isosceles and ∠OAQ = ∠OQA = b.
4. So the angle at A is ∠PAQ = a + b.
5. Angle sum in △PAQ: a + b + (a + b) = 180°, so 2(a + b) = 180° and a + b = 90°.
6. Hence ∠PAQ = 90°: the angle in a semicircle is a right angle.
  why?
  An angle in a semicircle (made at the circle by the two ends of a diameter) is a right angle.
Answer: ∠PAQ = a + b and a + b + (a + b) = 180°, so the angle in the semicircle is 90°.
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Exercises Q25
Chords CC′ and DD′ are perpendicular to a diameter AB. Prove that the segment MM′ joining the midpoints of CD and C′D′ is perpendicular to AB.
1. Place the circle with centre O at the origin and the diameter AB along the x-axis.
2. CC′ ⟂ AB, so CC′ is a vertical chord; the perpendicular from the centre bisects it, so its midpoint is on AB. Write C = (p, c), C′ = (p, −c).
  why?
  The perpendicular from the centre of a circle to a chord bisects the chord (the two right triangles it makes are congruent by RHS).
3. In the same way D = (q, d), D′ = (q, −d).
4. Midpoint of CD: M = ((p + q)/2, (c + d)/2). Midpoint of C′D′: M′ = ((p + q)/2, (−c − d)/2).
  why?
  The midpoint of the join of (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2): each coordinate is the average of the two ends.
5. M and M′ have the same x-coordinate, so MM′ is a vertical segment (if c + d = 0 the two points coincide on AB).
6. A vertical segment is perpendicular to the x-axis, that is, to AB.
Answer: M and M′ have the same x-coordinate (p + q)/2 when AB is the x-axis, so MM′ ⟂ AB.
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Exercises Q26
Fig. 5.31: cyclic ABCD with centre O inside, OA, OB, OC, OD drawn; marked angles p = ∠OAB, q = ∠OBC, u = ∠OCD, v = ∠ODA. Use it to show opposite angles add to 180°.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Since OA and OB are radii of the circle, OA = OB, so △OAB is isosceles.
2. In △OAB, the angles opposite equal sides are equal, so ∠OAB = ∠OBA = p.
3. Similarly, in △OBC, OB = OC, so ∠OBC = ∠OCB = q.
4. In △OCD, OC = OD, so ∠OCD = ∠ODC = u.
5. In △ODA, OD = OA, so ∠ODA = ∠OAD = v.
6. The sum of angles in △OAB is 180°, so ∠AOB = 180° − 2p.
7. The sum of angles in △OBC is 180°, so ∠BOC = 180° − 2q.
8. The sum of angles in △OCD is 180°, so ∠COD = 180° − 2u.
9. The sum of angles in △ODA is 180°, so ∠DOA = 180° − 2v.
10. The angles around centre O add to 360°, so (180° − 2p) + (180° − 2q) + (180° − 2u) + (180° − 2v) = 360°.
11. Simplifying gives 720° − 2(p + q + u + v) = 360°, so p + q + u + v = 180°.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
12. Angle A of the quadrilateral is ∠DAB = ∠OAD + ∠OAB = v + p.
13. Angle B of the quadrilateral is ∠ABC = ∠OBA + ∠OBC = p + q.
14. Angle C of the quadrilateral is ∠BCD = ∠OCB + ∠OCD = q + u.
15. Angle D of the quadrilateral is ∠CDA = ∠ODC + ∠ODA = u + v.
16. Adding opposite angles A and C: (v + p) + (q + u) = p + q + u + v = 180°.
17. Adding opposite angles B and D: (p + q) + (u + v) = p + q + u + v = 180°.
Answer: Opposite angles of cyclic quadrilateral ABCD add to 180°.
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Other exercises in this chapter

Exercise Set 5.1 — 4 questions
Exercise Set 5.2 — 2 questions
Exercise Set 5.3 — 3 questions
Exercise Set 5.4 — 3 questions
Exercise Set 5.5 — 3 questions
Exercise Set 5.6 — 3 questions

← I’m Up and Down, and Round and Round: notes, practice and all exercises

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NCERT Class 9 Mathematics I’m Up and Down, and Round and Round: Exercise Solutions

Exercises Q1

Exercises Q2

Exercises Q3

Exercises Q4

Exercises Q5

Exercises Q6

Exercises Q7

Exercises Q8

Exercises Q9

Exercises Q10

Exercises Q11

Exercises Q12

Exercises Q13

Exercises Q14

Exercises Q15

Exercises Q16

Exercises Q17

Exercises Q18

Exercises Q19

Exercises Q20

Exercises Q21

Exercises Q22

Exercises Q23

Exercises Q24

Exercises Q25

Exercises Q26