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NCERT Class 9 Mathematics Exercise Set 5.4 Solutions: I’m Up and Down, and Round and Round

Step-by-step working for every question of Exercise Set 5.4 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 5.4 Q1

   Use the Baudhāyana–Pythagoras theorem to show that equal chords of a circle are equally far from the centre.

   1. Let AB and CD be two equal chords of a circle with centre O.
   2. Draw perpendiculars OM and ON from the centre O to the chords AB and CD respectively.
   3. Since the perpendicular from the centre to a chord bisects the chord, M and N are the midpoints of AB and 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).

   4. Therefore, AM = AB/2 and CN = CD/2.
   5. Given that AB = CD, it follows that AM = CN.
   6. Consider the right-angled triangles OMA and ONC.
   7. In these triangles, OA = OC because both are radii of the same circle.
   8. Also, AM = CN as proved above.
   9. By the Baudhāyana–Pythagoras theorem, in △OMA, 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.

   10. In △ONC, OC² = ON² + CN².
   11. Since OA = OC and AM = CN, we have OM² + AM² = ON² + CN².
   12. Substituting AM = CN, we get OM² = ON².
       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.

   13. Taking the square root of both sides, we get OM = ON.
       why?

       If two non-negative quantities are equal, their square roots are equal. A square has two roots, + and −; a length or a count cannot be negative, so the negative root is dropped there.

   14. Thus, the perpendicular distances from the centre to the equal chords are equal.

   Answer: Equal chords of a circle are equally far from the centre.

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2. Ex 5.4 Q2

   Fig. 5.15: circle with centre C, chords AB and GF, CE ⟂ AB at E, CH ⟂ GF at H, and CE = CH. Show that AB = GF.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Given: Circle with centre C, chords AB and GF, CE ⟂ AB at E, CH ⟂ GF at H, and CE = CH.
   2. To prove: AB = GF.
   3. In △CEB and △CHG, CE = CH (given).
   4. ∠CEB = ∠CHG = 90° (since CE ⟂ AB and CH ⟂ GF).
   5. CB = CG (radii of the same circle).
   6. Therefore, △CEB ≅ △CHG by RHS congruence criterion.
   7. Hence, BE = HG (corresponding parts of congruent triangles).
   8. Since CE ⟂ AB, E is the midpoint of AB, so AB = 2 × BE.
   9. Since CH ⟂ GF, H is the midpoint of GF, so GF = 2 × HG.
   10. Since BE = HG, we have 2 × BE = 2 × HG, i.e., AB = GF.

   Answer: AB = GF

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3. Ex 5.4 Q3

   Prove the same result (CE = CH gives AB = GF, Fig. 5.15) using the Baudhāyana–Pythagoras theorem.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. In Fig. 5.15, CE ⟂ AB and CH ⟂ GF, so E is the midpoint of AB and H is the midpoint of GF (the perpendicular from the centre bisects a 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).

   2. Join CB and CG; both are radii, so CB = CG = r.
   3. In right triangle CEB, by the Baudhāyana–Pythagoras theorem, EB² = CB² − CE² = r² − CE².
      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. In right triangle CHG, likewise HG² = CG² − CH² = r² − CH².
   5. Since CE = CH, the right sides are equal, so EB² = HG² and EB = HG (lengths are positive).
   6. Then AB = 2 × EB and GF = 2 × HG, so AB = GF.

   Answer: EB² = r² − CE² = r² − CH² = HG², so EB = HG and AB = 2EB = 2HG = GF.

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