Home · Schooling · CBSE · Class 9 · Mathematics · I’m Up and Down, and Round and Round · Exercise Set 5.6

NCERT Class 9 Mathematics Exercise Set 5.6 Solutions: I’m Up and Down, and Round and Round

Step-by-step working for every question of Exercise Set 5.6 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 5.6 Q1

   In a circle of radius 12 cm with centre O, ∠AOB = 60°. Find the chord AB.

   1. In triangle AOB, OA and OB are radii of the circle, so OA = OB = 12 cm.
   2. Since OA = OB, triangle AOB is an isosceles triangle.
   3. The angles opposite to equal sides are equal, so ∠OAB = ∠OBA.
   4. The sum of angles in a triangle is 180°, so ∠OAB + ∠OBA + ∠AOB = 180°.
   5. Substituting the known values, 2∠OAB + 60° = 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 ∠OAB gives 2∠OAB = 120°, so ∠OAB = 60°.
   7. Since all three angles of triangle AOB are 60°, it is an equilateral triangle.
   8. In an equilateral triangle, all sides are equal, so AB = OA = OB.
   9. Therefore, the length of chord AB is 12 cm.

   Answer: AB = 12 cm

   Report an error in this solution

2. Ex 5.6 Q2

   A, B on a circle with centre O. (i) Can points X, Y on the circle on the same side of AB give ∠AXB ≠ ∠AYB? (ii) If ∠AXB = ∠AYB, must X and Y be on the same side of AB? (iii) If ∠AXB = ∠AYB and X, Y are not on the circle, does the circle through A, B, X also pass through Y?

   1. (i) All points of the circle on one side of AB lie on the same arc, and an arc subtends the same angle at every point of the circle outside it.
   2. (i) So for X, Y on the same side, ∠AXB = ∠AYB; no such pair with different angles exists.
   3. (ii) If X and Y are on opposite sides, AXBY is a cyclic quadrilateral, so ∠AXB + ∠AYB = 180°.
      why?

      The opposite angles of a cyclic quadrilateral (all four corners on one circle) add up to 180°.

   4. (ii) Then ∠AXB = ∠AYB forces both to be 90°, which happens exactly when AB is a diameter.
   5. (ii) So the statement is true when AB is not a diameter; if AB is a diameter, X and Y may be on opposite sides.
   6. (iii) If X and Y are on the same side of AB and ∠AXB = ∠AYB, then A, B, X, Y are concyclic (equal angles on the same side of a segment), so the circle through A, B, X passes through Y.
   7. (iii) If they are on opposite sides, the circle through A, B, X passes through Y only when the angles are 90° (then Y is on the circle with diameter AB); otherwise it does not.

   Answer: (i) No: points on the same side give equal angles; (ii) yes, unless AB is a diameter (then both angles are 90° on either side); (iii) yes when X and Y are on the same side of AB; on opposite sides only if both angles are 90°.

   Report an error in this solution

3. Ex 5.6 Q3

   Fig. 5.26: cyclic quadrilateral ABCD with ∠ADC = 100°; find x = ∠ABC.

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

   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. Here, ∠ADC and ∠ABC are opposite angles.
   3. So, ∠ADC + ∠ABC = 180°.
   4. Substitute the given value: 100° + x = 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. Solve for x: x = 180° − 100° = 80°.

   Answer: x = 80°

   Report an error in this solution

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