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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions -
Qstn# A-4 Prvs-QstnNext-Qstn
  • #4
    Is it possible to have a polygon; whose sum of interior angles is : (i) 870° (ii) 2340° (iii) 7 right-angles (iv) 4500° (i) 870° (ii) 2340° (iii) 7 right-angles (iv) 4500°
    Ans : (i) Let no. of sides = n
    Sum of angles = 870°
    (n - 2) × 180° = 870°
    ⇒ n - 2 = 870/180
    ⇒ n - 2 = 29/6
    ⇒ n = 29/6 + 2
    ⇒ n = 41/6
    Which is not a whole number.
    Hence it is not possible to have a polygon, the sum of whose interior angles is 870° (ii) Let no. of sides = n
    Sum of angles = 2340°
    (n - 2) × 180° = 2340°
    ⇒ n - 2 = 2340/180
    ⇒ n - 2 = 13
    ⇒ n = 13 + 2 = 15
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 2340°. (iii) Let no. of sides = n
    Sum of angles = 7 right angles = 7 ×90 = 630°
    (n - 2) × 180° = 630°
    ⇒ n - 2 = 630/180
    ⇒ n - 2 = 7/2
    ⇒ n = 7/2 + 2
    ⇒ n = 11/2
    Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles. (iv) Let no. of sides = n
    (n - 2)×180° = 4500°
    ⇒ n - 2 = 4500/180
    ⇒ n - 2 = 25
    ⇒ n = 25 + 2
    ⇒ n = 27
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 4500°. (i) Let no. of sides = n
    Sum of angles = 870°
    (n - 2) × 180° = 870°
    ⇒ n - 2 = 870/180
    ⇒ n - 2 = 29/6
    ⇒ n = 29/6 + 2
    ⇒ n = 41/6
    Which is not a whole number.
    Hence it is not possible to have a polygon, the sum of whose interior angles is 870° (ii) Let no. of sides = n
    Sum of angles = 2340°
    (n - 2) × 180° = 2340°
    ⇒ n - 2 = 2340/180
    ⇒ n - 2 = 13
    ⇒ n = 13 + 2 = 15
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 2340°. (iii) Let no. of sides = n
    Sum of angles = 7 right angles = 7 ×90 = 630°
    (n - 2) × 180° = 630°
    ⇒ n - 2 = 630/180
    ⇒ n - 2 = 7/2
    ⇒ n = 7/2 + 2
    ⇒ n = 11/2
    Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles. (iv) Let no. of sides = n
    (n - 2)×180° = 4500°
    ⇒ n - 2 = 4500/180
    ⇒ n - 2 = 25
    ⇒ n = 25 + 2
    ⇒ n = 27
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.
  • #4-i
    870°
    Ans : Let no. of sides = n
    Sum of angles = 870°
    (n - 2) × 180° = 870°
    ⇒ n - 2 = 870/180
    ⇒ n - 2 = 29/6
    ⇒ n = 29/6 + 2
    ⇒ n = 41/6
    Which is not a whole number.
    Hence it is not possible to have a polygon, the sum of whose interior angles is 870°
  • #4-ii
    2340°
    Ans : Let no. of sides = n
    Sum of angles = 2340°
    (n - 2) × 180° = 2340°
    ⇒ n - 2 = 2340/180
    ⇒ n - 2 = 13
    ⇒ n = 13 + 2 = 15
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.
  • #4-iii
    7 right-angles
    Ans : Let no. of sides = n
    Sum of angles = 7 right angles = 7 ×90 = 630°
    (n - 2) × 180° = 630°
    ⇒ n - 2 = 630/180
    ⇒ n - 2 = 7/2
    ⇒ n = 7/2 + 2
    ⇒ n = 11/2
    Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.
  • #4-iv
    4500°
    Ans : Let no. of sides = n
    (n - 2)×180° = 4500°
    ⇒ n - 2 = 4500/180
    ⇒ n - 2 = 25
    ⇒ n = 25 + 2
    ⇒ n = 27
    Which is a whole number.
    Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.