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NCERT Class 9 Mathematics Exercise Set 5.1 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 5.1 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.

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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  1. Ex 5.1 Q1

    Draw △ABC with AB = 5 cm, ∠A = 70°, ∠B = 60° and its circumcircle. Is the circumcentre inside or outside the triangle?

    1. Draw a line segment AB of length 5 cm.
    2. At point A, construct an angle of 70° with AB and draw a ray.
    3. At point B, construct an angle of 60° with BA and draw a ray.
    4. Let the two rays intersect at point C to complete triangle ABC.
    5. Calculate angle C using the angle sum property: 180° − (70° + 60°) = 50°.
      why?

      The three angles of any triangle add up to 180°.

    6. Since all angles (70°, 60°, 50°) are less than 90°, the triangle is acute-angled.
    7. Draw the perpendicular bisectors of sides AB and BC.
    8. Let the perpendicular bisectors intersect at point O, which is the circumcentre.
    9. Draw a circle with centre O passing through A, B, and C.
    10. Observe that the circumcentre O lies inside the triangle.

    Answer: ∠C = 50°; inside

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  2. Ex 5.1 Q2

    Draw △ABC with AB = 5 cm, ∠A = 100°, AC = 4 cm and its circumcircle. Is the circumcentre inside or outside the triangle?

    1. Draw a line segment AB of length 5 cm.
    2. At point A, construct an angle of 100° using a protractor.
    3. On the ray of the angle, mark point C such that AC = 4 cm.
    4. Join B and C to complete the triangle ABC.
    5. Draw the perpendicular bisector of side AB.
    6. Draw the perpendicular bisector of side AC.
    7. Let the two perpendicular bisectors intersect at point O.
    8. Point O is the circumcentre of the triangle.
    9. With O as the centre and OA as the radius, draw the circumcircle passing through A, B, and C.
    10. Since angle A is 100°, which is greater than 90°, the triangle is obtuse-angled.
    11. In an obtuse-angled triangle, the circumcentre lies outside the triangle.

    Answer: The circumcentre is outside the triangle.

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  3. Ex 5.1 Q3

    Draw △ABC with AB = 6 cm, BC = 7 cm, CA = 7 cm and its circumcircle with centre O. Measure OA, OB and OC.

    1. Draw a line segment AB of length 6 cm.
    2. With A as centre and radius 7 cm, draw an arc above AB.
    3. With B as centre and radius 7 cm, draw another arc intersecting the first arc at C.
    4. Join AC and BC to form the triangle ABC.
    5. Draw the perpendicular bisector of side AB.
    6. Draw the perpendicular bisector of side BC.
    7. Let the point where the two perpendicular bisectors intersect be O.
    8. With O as centre and OA as radius, draw a circle passing through A, B, and C.
    9. This circle is the circumcircle of triangle ABC.
    10. Measure OA, OB and OC with a ruler: all three are equal, because O is on the perpendicular bisectors of the sides.
    11. To know what to expect: CA = CB = 7 cm, so the perpendicular bisector of AB passes through C and meets AB at its midpoint M, with AM = 3 cm and CM = √(7² − 3²) = √40 cm.
    12. O lies on CM; call its radius R = OA = OC. In right triangle OMA, R² = 3² + (√40 − R)² = 9 + 40 − 2√40 R + R², so R = 49/(2√40) ≈ 3.87 cm.
    13. So the measurement should give OA = OB = OC ≈ 3.9 cm.

    Answer: OA = OB = OC ≈ 3.9 cm (the exact value is 49/(2√40) ≈ 3.87 cm)

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  4. Ex 5.1 Q4

    What is the smallest possible radius of a circle passing through two given points A and B?

    1. The distance between the two given points A and B is fixed.
    2. A circle passing through A and B must have a radius at least half the length of the segment AB.
    3. If the radius were smaller than half of AB, the two points would be too far apart to lie on the circle.
    4. The smallest radius is achieved when AB is the diameter of the circle.
    5. In this case, the center of the circle is the midpoint of AB.
    6. Therefore, the minimum radius is half the distance between A and B.

    Answer: The smallest possible radius is half the distance between A and B.

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Other exercises in this chapter

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