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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 6

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  • #19-iii
    its exterior angle exceeds its interior angle by 60°.
    Ans : Let interior angle = x
    Then exterior angle = x + 60
    ∴ x + x + 60˚ = 180˚
    ⇒ 2x = 180˚ - 60˚ = 120˚
    ⇒ x = 120˚/2 = 60˚
    ∴ Exterior angle = 60˚ + 60˚ = 120˚
    ∴ Number of sides = 360˚/120˚ = 3
  • Qstn #20
    The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.
    Ans : Sum of interior angles = 3 x Sum of exterior angles
    Let exterior angle = x
    The interior angle = 3x
    x + 3x=180°
    ⇒ 4x = 180°
    ⇒ x = 180˚/4
    ⇒ x = 45°
    Number of sides = 360/45 = 8
  • #
    Section : C
  • Qstn #1
    Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
    Ans : Let the other angle = x°
    According to given,
    89° + 113° + x° + x° = 360°
    ⇒ 2x° = 360° - 202°
    ⇒ 2x° = 158°
    ⇒ x° = 158/2
    other two angles = 79° each
  • Qstn #2
    Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5:7; find the measure of each of them.
    Ans : Two angles are 68° and 76°
    Let other two angles be 5x and 7x
    68° + 76°+ 5x + 7x = 360°
    ⇒ 12x + 144° = 360°
    ⇒ 12x = 360° - 144°
    ⇒ 12x = 216°
    ⇒ x = 18°
    angles are 5x and 7x
    i.e. 5×18° and 7×18° i.e. 90° and 126°
  • Qstn #3
    Angles of a quadrilateral are (4x)°, 5(x + 2)°, (7x - 20)° and 6(x + 3)°. Find : (i) the value of x. (ii) each angle of the quadrilateral.
    Ans : Angles of quadrilateral are,
    (4x)Ëš, 5(x + 2)Ëš, (7x - 20)Ëš and 6(x + 3)Ëš
    ∴ 4x + 5(x + 2) + (7x - 20)+ 6(x + 3) = 360˚
    ⇒ 4x + 5x + 10 + 7x - 20 + 6x + 18 = 360˚
    ⇒ 22x + 8 = 360˚
    ⇒ 22x = 360˚ - 8˚
    ⇒ 22x = 352˚
    ⇒ x = 16˚
    Hence angle are,
    (4x)˚ = (4 × 16)˚ = 64˚
    5(x + 2)Ëš = 5(16 + 2)Ëš = 90Ëš
    (7x - 20)˚ = (7 × 16 - 20)˚ = 92˚
    6(x + 3)Ëš = 6(16 + 3) = 114Ëš
  • Qstn #4
    Use the information given in the following figure to find : (i) x (ii) ∠B and ∠C
    Answer
    ∵ ∠A = 90˚ (Given)
    ∠B = (2x + 4˚)
    ∠C = (3x - 5˚)
    ∠D = (8x - 15˚)
    ∠A + ∠B + ∠C + ∠D = 360˚
    90Ëš + (2x + 4) + (3x - 5) + (8x - 15) = 360
    ⇒ 90 + 2x + 4 + 3x - 5 + 8x - 15 = 360
    ⇒ 74˚ + 13x = 360˚
    ⇒ 13x = 360˚ - 74˚
    ⇒ 13x = 286˚
    ⇒ x = 22˚
    ∵ ∠B = 2x + 4 = 2×22˚ + 4 = 48˚
    ∠C = 3x - 5 = 3 × 22˚ - 5 = 61˚
    Hence (i)22˚ (ii) ∠B = 48˚, ∠C = 61˚
  • Qstn #5
    In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5 (i) Calculate each angle of the quadrilateral. (ii) Assign a special name to quadrilateral ABCD
    Ans :
    ∵ ∠A : ∠D = 1 : 2
    Let ∠A = x and ∠B = 2x
    ∵ ∠C : ∠B = 4 : 5
    Let ∠C = 4y and ∠B = 5y
    ∵ AB ∥ DC
    ∴ ∠A + ∠D = 180˚
    x + 2x = 180Ëš
    ⇒ 3x = 180˚
    ⇒ x = 60˚
    ∴ A = 60˚
    ∠D = 2x = 2 × 60˚ = 120˚
    Again ∠B + ∠C = 180˚
    5y + 4y = 180Ëš
    ⇒ 9y = 180˚
    ⇒ y = 20˚
    ∴ ∠B = 5y = 5 × 20˚ = 100˚
    ∠C = 4y = 4 × 20˚ = 80˚
    Hence ∠A = 60˚ : ∠B = 100˚ : ∠C = 80˚ ∠D = 120˚
    (iii) Quadrilateral ABCD is a trapezium because one pair of opposite side is parallel.
  • Qstn #6
    From the following figure find ;
  • #6-i
    x
    Ans : In Quadrilateral ABCD,
    x + 4x + 3x + 4x + 48° = 360°
    ⇒ 12x = 360° - 48°
    ⇒ 12x = 312
    ⇒ x = 312/12 = 26˚
  • #6-ii
    ∠ABC
    Ans : ∠ABC = 4x
    4 × 26 = 104˚
  • #6-iii
    ∠ACD
    Ans : ∠ACD = 180˚ - 4x - 48˚
    = 180˚ - 4×26 - 48 ˚
    = 180Ëš - 104Ëš - 48Ëš
    = 180Ëš - 152Ëš
    = 28Ëš
  • Qstn #7
    Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C - 8° ; ∠A = 5(a + 2)° and ∠B = 2(2a + 7)°.
    Calculate ∠A.
    Ans : ∠C = 64° (Given)
    ∠D = ∠C - 8° = 64° - 8° = 56°
    ∠A = 5(a + 2)°
    ∠B = 2(2a + 7)°
    Now, ∠A + ∠B + ∠C + ∠D = 360°
    5(a + 2)° + 2(2a + 7)° + 64° + 56° = 360°
    ⇒ 5a + 10 + 4a + 14° + 64° + 56° = 360°
    ⇒ 9a + 144° = 360°
    ⇒ 9a = 360° - 144°
    ⇒ 9a = 216°
    ⇒ a = 24°
    ∠A = 5 (a + 2) = 5(24 + 2) = 130°
  • Qstn #8
    In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.
    Answer
    Stun of angles of quadrilateral = 360°
    70° + a + 2a + 15 + 3a + 5 = 360°
    ⇒ 6a + 90° = 360°
    ⇒ 6a = 270°
    ⇒ a = 45°
    b = 2a + 15 = 2×45 + 15 = 105°
    c = 3a + 5 = 3×45 + 5 = 140°
    Hence ∠b and ∠c are 105° and 140°
  • Qstn #9
    Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
    Ans : Let each equal angle be x°
    x + x + x + 69° = 360°

    3x = 360° - 69
    ⇒ 3x = 291
    ⇒ x = 97°
    Each, equal angle = 97°
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