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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 3

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  • #10-ii
    Write the value of ∠A + ∠E
    Ans : Since AB ∥ ED
    ∴ ∠A + ∠E = 180˚
  • #10-iii
    Find angles B, C and D.
    Ans : Let ∠B = 5x ∠C = 6x ∠D = 7x
    ∴ 5x + 6x + 7x + 180˚ = 540˚ (∠A + ∠E = 180˚) Proved in (ii)
    ⇒ 18x = 540˚ - 180˚
    ⇒ 18x = 360˚
    ⇒ x = 20˚∴ ∠B = 5 × 20˚ = 100˚, ∠C = 6 × 20˚ = 120˚
    ∠D = 7 × 20 = 140˚
  • Qstn #11
    Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.
    Ans : Let number of sides = n
    Sum of interior angles = (n - 2) × 180˚
    = 180n - 360Ëš
    Sum of 2 right angles = 2 × 90˚
    = 180Ëš
    ∴ Sum of other angles = 180˚n - 360˚ - 180˚
    = 180Ëšn - 540Ëš
    No. of vertices at which these angles are formed = n - 2
    ∴ Each interior angle = (180n - 540)/(n - 2)
    ∴ (180n - 540)/(n - 2) = 120˚
    ⇒ 180n - 540 = 120n - 240
    ⇒ 180n - 120n = -240 + 540
    ⇒ 60n = 300
    ⇒ n = 300/60
    ⇒ n = 5
  • Qstn #12
    In a hexagon ABCDEF, side AB is parallel to side FE and ∠B: ∠C: ∠D: ∠E = 6 : 4: 2: 3. Find ∠B and ∠D.
    Ans :
    Given: Hexagon ABCDEF in which AB∥ EF and ∠B: ∠C: ∠D: ∠E = 6: 4: 2 : 3.
    To find: ∠B and ∠D
    Proof: No. of sides n = 6
    ∴ Sum of interior angles = (n - 2) × 180˚
    = (6 - 2) × 180˚
    = 720Ëš
    ∵ AB ∥ EF (Given)
    ∴ ∠A + ∠F = 180˚
    But ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720˚ (Proved)
    ∠B + ∠C + ∠D + ∠E + ∠180˚ = 720˚
    ∴ ∠B + ∠C + ∠D + ∠E = 720˚ - 180˚ = 540˚
    Ratio = 6 : 4 : 2 : 3
    Sum of parts = 6 + 4 + 2 + 3 = 15
    ∴ ∠B = 6/15 × 540 = 216˚
    ∠D = 2/15 × 540 = 72˚
    Hence, ∠B = 216˚; ∠D = 72˚
  • Qstn #13
    The angles of a hexagon are x + 10°, 2x + 20°, 2x - 20°, 3x - 50°, x + 40° and x + 20°. Find x.
    Ans : Angles of a hexagon are x + 10Ëš, 2x + 20Ëš, 2x - 20Ëš, 3x - 50Ëš, x + 40Ëš and x + 40Ëš and x + 20Ëš
    ∴ But sum of angles of a hexagon = (x - 2) × 180˚
    = (6 - 2) × 180˚
    = 4 × 180˚
    = 720Ëš
    But sum = x + 10 + 2x + 20Ëš + 2x - 20Ëš + 3x - 50Ëš + x + 40 + x + 20
    = 10x + 90 - 70
    = 10x + 20
    ∴ 10x + 20 = 720˚
    ⇒ 10x = 72 - 20 = 700
    ⇒ x = 700˚/10 = 70˚
    ∴ x = 70˚
  • Qstn #14
    In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3: 7. Find the biggest angle of the pentagon.
    Ans : In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3× 180°
    = 540° - 40° - 60° = 540° - 100° = 440°
    Ratio in these 3 angles =1 : 3 : 7
    Sum of ratios =1 + 3 + 7 = 11
    Biggest angle = (440 × 7)/11 = 280°
  • #
    Section : B
  • Qstn #1
    Fill in the blanks :
    In case of regular polygon, with :
    Ans :
    Explanation
    i) Each exterior angle = 360Ëš/8 = 45Ëš
    Each interior angle = 180Ëš - 45Ëš = 135Ëš
    (ii) Each exterior angle = 360Ëš/12 = 30Ëš
    Each interior angle = 180Ëš - 30Ëš = 150Ëš
    (iii) Since each exterior = 72Ëš
    ∴ Number of sides = 360˚/72˚ = 5
    Also interior angle = 180Ëš - 72Ëš = 108Ëš
    (iv) Since each exterior angle = 45Ëš
    ∴ Number of sides = 360˚/45˚ = 8
    Interior angle = 180Ëš - 45Ëš = 135Ëš
    (v) Since interior angle = 150Ëš
    ∴ Exterior angle = 180˚ - 150˚ = 30˚
    ∴ Number of sides = 360˚/30˚ = 12
    (vi) Since interior angle = 140Ëš
    ∴ Exterior angle = 180˚ - 140˚ = 40˚
    ∴ Number of sides = 360˚/40˚ = 9
  • Qstn #2
    Find the number of sides in a regular polygon, if its each interior angle is :
  • #2-i
    160°
    Ans : Let no. of sides of regular polygon be n.
    Each interior angle = 160Ëš
    ∴ (n - 2)/n × 180˚ = 360˚
    ⇒ 180˚n - 360˚ = 160n
    ⇒ 20n = 360˚
    ⇒ n = 18
  • #2-ii
    135°
    Ans : No. of sides = n
    Each interior angle = 135Ëš
    (n - 2)/n × 180˚ = 135˚
    ⇒ 180n - 360˚ = 135n
    ⇒ 180n - 135n = 360˚
    ⇒ 45n = 360˚
    ⇒ n = 8
  • #2-iii
    1 1/5 of a right-angle
    Ans : No. of sides = n
    Each interior angle = 1 1/5 right angles
    = 6/5 × 90˚
    = 108Ëš
    ∴ (n - 2)/n × 180˚ = 108˚
    ⇒ 180n - 360 ˚ = 108n
    ⇒ 180n -108n = 360˚
    ⇒ 72n = 360˚
    ⇒ n = 5
  • Qstn #3
    Find the number of sides in a regular polygon, if its each exterior angle is :
  • #3-i
    1/3 of a right angle
    Ans : Each exterior angle = 1/3 of a right angle
    = 1/3 ×90
    = 30°
    Let number of sides = n
    ∴ 360˚/n = 30˚
    ∴ n = 360˚/30˚
    ⇒ n = 12
  • #3-ii
    two-fifth of a right-angle.
    Ans : Each exterior angle = 2/5 of a right-angle
    = 2/5 × 90˚= 36˚
    Let number of sides = n
    ∴ 360˚/n = 36˚
    ⇒ n = 360˚/36˚
    ⇒ n = 10
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