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NCERT Class 9 Mathematics Exercise Set 5.5 Solutions: I’m Up and Down, and Round and Round
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 Exercise Set 5.5 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.
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.
Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.
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Ex 5.5 Q1
Find the length of a chord 6 cm from the centre of a circle of radius 7 cm.
- Let O be the centre of the circle and AB be the chord at a distance of 6 cm from O.
- Draw OM perpendicular to AB, so OM = 6 cm and M is the midpoint of AB.
- Join OA, which is the radius of the circle, so OA = 7 cm.
- 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.
- Substitute the values: 7² = 6² + AM².
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.
- Simplify the equation: 49 = 36 + AM².
why?
Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
- Solve for AM²: AM² = 49 − 36 = 13.
- Find AM: AM = √13 cm.
- Since the perpendicular from the centre bisects the chord, the length of chord AB is 2 × AM.
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).
- Calculate AB: AB = 2√13 cm ≈ 7.21 cm.
Answer: AB = 2√13 cm ≈ 7.21 cm
Ex 5.5 Q2
Explain why a chord at perpendicular distance d from the centre of a circle of radius r has length 2√(r² − d²).
- Let O be the centre of the circle with radius r, and let AB be a chord at perpendicular distance d from O.
- Draw OM perpendicular to AB, where M is the point of intersection on the chord.
- Since the perpendicular from the centre to a chord bisects the chord, AM = MB.
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).
- Therefore, the length of the chord AB is 2 × AM.
- Join OA to form the right-angled triangle OMA.
- In △ OMA, OA is the hypotenuse with length r, OM is the perpendicular with length d, and AM is the base.
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.
- By 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.
- Substituting the values, we get r² = d² + AM².
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.
- Rearranging the equation gives AM² = r² − d².
- Taking the square root, AM = √(r² − d²).
- Thus, the length of the chord AB = 2 × AM = 2√(r² − d²).
Answer: The length of the chord is 2√(r² − d²).
Ex 5.5 Q3
Chord AB is twice as far from the centre as chord CD. Can we conclude that CD = 2AB? Give reasons.
- Let the radius be r and let chord CD be at distance d from the centre, so chord AB is at distance 2d (both less than r).
- A chord at distance x from the centre has length 2√(r² − x²), so CD = 2√(r² − d²) and AB = 2√(r² − 4d²).
- Test an example: r = 5 cm, d = 2 cm. Then CD = 2√(25 − 4) = 2√21 cm ≈ 9.17 cm.
- AB is 4 cm from the centre: AB = 2√(25 − 16) = 2 × 3 = 6 cm, so 2AB = 12 cm.
- Here CD ≈ 9.17 cm is not 12 cm, so CD = 2AB is false in this case.
- In general CD = 2AB would need r² − d² = 4(r² − 4d²), that is 15d² = 3r², or d = r/√5; it holds only for that one special distance, so it cannot be concluded.
Answer: No. Example r = 5 cm: CD at 2 cm gives CD = 2√21 cm ≈ 9.17 cm, AB at 4 cm gives AB = 6 cm, and 2AB = 12 cm ≠ CD.
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 5.6 — 3 questions
- End-of-chapter exercises — 26 questions
← I’m Up and Down, and Round and Round: notes, practice and all exercises
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