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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 4

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  • Qstn #4
    Is it possible to have a regular polygon whose each interior angle is :
  • #4-i
    170°
    Ans : No. of sides = n
    each interior angle = 170°
    ∴ (n - 2)/n × 180˚ = 170˚
    ⇒ 180n - 360˚ = 170n
    ⇒ 180n - 170n = 360˚
    ⇒ 10n = 360˚
    ⇒ n = 360˚/10˚
    ⇒ n = 36
    which is a whole number.
    Hence it is possible to have a regular polygon whose interior angle is 170Ëš.
  • #4-ii
    138°
    Ans : Let no. of sides = n
    Each interior angle = 138Ëš
    ∴ (n - 2)/n × 180˚ = 138˚
    ⇒ 180n - 360˚ = 138n
    ⇒ 180n - 138n = 360˚
    ⇒ 42n = 360˚
    ⇒ n = 360˚/42
    ⇒ n = 60˚/7
    Which is not a whole number.
    Hence it is not possible to have a regular polygon having interior angle of 138°.
  • Qstn #5
    Is it possible to have a regular polygon whose each exterior angle is :
  • #5-i
    80°
    Ans : Let no. of sides = n each exterior angle = 80°
    360Ëš/n = 80Ëš
    ⇒ n = 360˚/80˚
    ⇒ n = 9/2
    Which is not a whole number.
    Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.
  • #5-ii
    40% of a right angle.
    Ans : Let number of sides = n
    Each exterior angle = 40% of a right angle
    = 40/100 × 90
    = 36Ëš
    n = 360Ëš/36Ëš
    ⇒ n = 10
    Which is a whole number.
    Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.
  • Qstn #6
    Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
    Ans : Let each exterior angle or interior angle be = x°

    ∴ x + x = 180˚
    ⇒ 2x = 180˚
    ⇒ x = 90˚
    Now, let no. of sides = n
    ∵ Each exterior angle = 360˚/n
    ∴ 90˚ = 360˚/n
    ⇒ n = 360˚/90˚
    ⇒ n = 4
  • Qstn #7
    The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
    Ans : Let interior angle = x°
    Exterior angle = 1/3 x°

    ∴ x + 1/3x = 180˚
    ⇒ 3x + x = 540
    ⇒ 4x = 540
    ⇒ x = 540/4
    ⇒ x = 135˚
    ∴ Exterior angle = 1/3 × 135˚ = 45˚
    Let no. of sides = n
    ∵ Each exterior angle = 360˚/n
    ∴ 45˚ = 360˚/n
    ∴ n = 360˚/45˚
    ⇒ n = 8
  • Qstn #8
    The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find : (i) measure of each interior angle ; (ii) measure of each exterior angle and (iii) number of sides in the polygon.
    Ans : Let exterior angle = x°
    Interior angle = 5x°
    x + 5x = 180°
    ⇒ 6x = 180°
    ⇒ x = 30°
    Each exterior angle = 30°
    Each interior angle = 5 x 30° = 150°
    Let no. of sides = n
    ∵ each exterior angle = 360˚/n
    30Ëš = 360Ëš/n
    ⇒ n = 360˚/30˚
    ⇒ n = 12
    Hence, (i) 150Ëš (ii) 30Ëš (iii) 12
  • Qstn #9
    The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
    Ans : Interior angle : exterior angle = 2 : 1
    Let interior angle = 2x° & exterior angle = x°

    ∴ 2x˚ + x˚ = 180˚
    ⇒ 3x = 180˚
  • #9-i
    each exterior angle of the polygon ;
    Ans : x = 60Ëš
    ∴ Each exterior angle = 60˚
    Let angle of sides = n
    ∴ 360˚/n = 60˚
    ⇒ n = 360˚/60˚
    dlt29025704036958965620">⇒ n = 360˚/60˚
  • #9-ii
    number of sides in the polygon
    Ans : n = 6
    ∴ (i) 60˚ (ii) 6
  • Qstn #10
    The ratio between the exterior angle and the interior angle of a regular polygon is 1: 4. Find the number of sides in the polygon.
    Ans : Let exterior angle be x° & interior angle be 4x°

    ∴ 4x + x = 180˚
    ⇒ 5x = 180˚
    ⇒ x = 36˚
    ∴ Each exterior angle = 36˚
    Let no. sides = n
    ∴ 360˚/n = 36˚
    ⇒ n = 360˚/36
    ⇒ n = 10
  • Qstn #11
    The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
    Ans : Let number of sides = n
    Sum of exterior angles = 360°
    Sum of interior angles = 360° × 2 = 720°
    Sum of interior angles = (n - 2) × 180°
    720° = (n - 2)× 180°
    ⇒ n - 2 = 720/180
    ⇒ n - 2 = 4
    ⇒ n = 4 + 2
    ⇒ n = 6
  • Qstn #12
    AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
    Ans :
    ∵ Polygon is regular (Given)
    ∴ AB = BC
    ⇒ ∠BAC = ∠BAC [∠s opp. to equal sides]
    But ∠BAC = 20˚
    ∴ ∠BCA = 20˚
    i.e., In ΔABC,
    ∠B + ∠BAC + ∠BCA = 180˚
    ⇒ ∠B + 20˚ + 20˚ = 180˚
    ⇒ ∠B = 180˚ - 40˚
    ⇒ ∠B = 140˚
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