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NCERT Class 9 Mathematics Exercise Set 5.3 Solutions: I’m Up and Down, and Round and Round

Step-by-step working for every question of Exercise Set 5.3 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.

1. Ex 5.3 Q1

   Show that the perpendicular from the centre of a circle to a chord bisects the chord.

   1. Let O be the centre of the circle and AB be a chord.
   2. Draw OM perpendicular to AB, meeting AB at M.
   3. Join OA and OB to form triangles OMA and OMB.
   4. OA and OB are radii of the same circle, so OA = OB.
   5. OM is common to both triangles OMA and OMB.
   6. Since OM is perpendicular to AB, angle OMA = angle OMB = 90°.
   7. By RHS congruence criterion, triangle OMA is congruent to triangle OMB.
   8. Therefore, AM = BM by CPCT.
   9. Thus, the perpendicular from the centre to the chord bisects the chord.
      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).

   Answer: Proved that the perpendicular from the centre of a circle to a chord bisects the chord.

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2. Ex 5.3 Q2

   An isosceles △ABC with AB = AC is inscribed in a circle. Show that the altitude from A to BC passes through the centre.

   1. Let O be the centre of the circle in which the isosceles triangle ABC is inscribed.
   2. Join OA, OB, and OC to form radii of the circle.
   3. Since AB = AC, the base angles of the triangle are equal, so ∠ABC = ∠ACB.
   4. In triangles OAB and OAC, OA is common, OB = OC (radii), and AB = AC (given).
   5. By SSS congruence criterion, △OAB ≅ △OAC.
   6. Therefore, corresponding angles are equal, so ∠BAO = ∠CAO.
   7. This means AO is the angle bisector of ∠BAC.
   8. In an isosceles triangle, the angle bisector of the vertex angle is also the altitude to the base.
   9. Thus, AO is perpendicular to BC.
   10. Since O lies on AO, the altitude from A to BC passes through the centre O.

   Answer: The altitude from A to BC passes through the centre O.

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3. Ex 5.3 Q3

   Parallel chords of 6 cm and 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Find the distance between their midpoints.

   1. Let O be the centre of the circle with radius 5 cm.
   2. Let AB be the chord of length 6 cm and CD be the chord of length 8 cm.
   3. Let M and N be the midpoints of AB and CD respectively.
   4. Since the perpendicular from the centre to a chord bisects it, OM is perpendicular to AB and ON is perpendicular to CD.
      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).

   5. In right-angled triangle OMA, OA = 5 cm and AM = 6/2 = 3 cm.
   6. Using Pythagoras theorem, OM = √(OA² − AM²) = √(25 − 9) = √16 = 4 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.

   7. In right-angled triangle OND, OD = 5 cm and DN = 8/2 = 4 cm.
   8. Using Pythagoras theorem, ON = √(OD² − DN²) = √(25 − 16) = √9 = 3 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.

   9. OM ⟂ AB and ON ⟂ CD, and AB ∥ CD, so OM and ON lie on one line through O perpendicular to both chords; as the chords are on opposite sides of the centre, O lies between M and N.
   10. The distance between the midpoints is MN = OM + ON = 4 + 3 = 7 cm.

   Answer: The midpoints are 7 cm apart.

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