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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 6

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  • Qstn #13
    The area of a rhombus is equal to the area of a triangle. If base of ∆ is 24 cm, its corresponding altitude is 16 cm and one of the diagonals of the rhombus is 19.2 cm. Find its other diagonal.
    Ans : Area of a rhombus = Area of a triangle Base of triangle = 24 cm
    and altitude = 16 cm
    ∴ Area = ½ × base × altitude
    = ½ × 24 × 16
    = 192 cm2
    ∴ Area of rhombus = 192 cm2
    One diagonal = 19.2 cm
    ∴ Second diagonal = (Area × 2)/(One diagonal)
    = (192 × 2)/(19.2)
    = (192 × 10 × 2)/192
    = 20 cm
  • Qstn #14
    Find the area of the trapezium ABCD in which AB//DC, AB = 18 cm, ∠B = ∠C = 90°, CD = 12 cm and AD = 10 cm.
    Ans : In trapezium ABCD,
    AB ∥ DC, AB = 18 cm
    ∠B = ∠C = 90˚, CD = 12 cm and AD = 10 cm
    Area of trapezium ABCD
    Draw DL ⊥ AB
    ∴ AL = 18 - 12 = 6 cm
    AL = BC

    Now area of trapezium = ½ (AB + CD) × AL
    = 1/2(18 + 12) × 8 cm2
    = 1/2 × 30 × 8
    = 120 cm2
  • #
    Section : D
  • Qstn #1
    Find the radius and area of a circle, whose circumference is :
  • #1-i
    132 cm
    Ans : Circumference of circle = 132 cm
    2Ï€r = 132
    ⇒ 2× 22/7 x r = 132
    ⇒ r = (132 × 7)/(2 × 22)
    ⇒ r = 21 cm∴ Area of circle = πr2
    = 22/7 × 21 × 21
    = 1386 cm2
  • #1-ii
    22 m
    Ans : Circumference of circle = 22m
    ∴ 2πr = 22
    ⇒ 2 × 22/7 × r = 22
    ⇒ r = (22 × 7)/(2 × 22)
    ⇒ r = 7/2
    ⇒ r = 3.5 m
    Area of circle = πr2
    = 22/7 × 3.5 × 3.5
    = 38.5 m2
  • Qstn #2
    Find the radius and circumference of a circle, whose area is :
  • #2-i
    154 cm2
    Ans : Area of circle = 154 cm2
    πr2 = 154
    ⇒ r2 = 154/π
    ⇒ r2 = 154/22 × 7
    ⇒ r2 = 7 × 7
    ⇒ r = 7 cm
    ∴ Circumference = 2πr
    = 2 × 22/7 × 7
    = 44cm
    Hence, 7 cm; 44 cm
  • #2-ii
    6.16 m2
    Ans : Area of circle = 6.16 m2
    πr2 = 6.16
    ⇒ 22/7 r2 = 616/100
    ⇒ r2 = 616/100 × 7/22
    ⇒ r2 = 196/100
    ⇒ r2 = 1.96
    ⇒ r = √1.96
    ⇒ r = 1.4 m
    Circumference = 2Ï€r
    = 2 × 22/7 × 1.4
    = 8.8 m
    ∴ 1.4 m; 8.8 m
  • Qstn #3
    The circumference of a circular table is 88 m. Find its area.
    Ans : Circumference of circle = 88 m
    2Ï€r = 88 m
    ⇒ 2 × 22/7 × r = 88
    ⇒ r = (88 × 7)/(2 × 22)
    ⇒ r = 14 m
    Area of circle = πr2
    = 22/7 × 14 × 14
    = 616 m2
  • Qstn #4
    The area of a circle is 1386 sq. cm; find its circumference.
    Ans : Area of circle = 1386 cm2
    πr2 = 1386
    ⇒ 22/7 r2 = 1386
    ⇒ r2 = 1386 × 7/22
    ⇒ r2 = 441
    ⇒ r = √441
    ⇒ r = 21 cm
    Circumference = 2Ï€r
    = 2 × 22/7 × 21
    = 132 m
  • Qstn #5
    Find the area of a flat circular ring formed by two concentric circles (circles with same centre) whose radii are 9 cm and 5 cm.
    Ans : External radius, r1 = 9 cm
    Internal radius, r2 = 5 cm
    Area of ring = πr12 - πr22
    = π(r12 - r22)
    = π(92 - 52)
    = 22/7 ×(81 - 25)= 22/7 × 56
    = 176 cm2
  • Qstn #6
    Find the area of the shaded portion in each of the following diagrams :
    Ans : (i)

    Radius of circle, r = 7 cm
    ∴ Side of square = 7 + 7 = 14 cm
    Area of circle = πr2
    = 22/7 × 7 × 7
    = 154 cm2
    Area of square = 14 × 14
    = 196 cm2
    ∴ Area of shaded portion = 196 - 154
    = 42 cm2
    (ii) Radii of concentric circles are
    r1 = 4.5 m
    r2 = 2.5 m

    ∴ Area of shaded portion = πr12 - πr22
    = π(r12 - r22)
    = 22/7 [(4.5)2 - (2.5)2]
    = 22/7 × (4.5 + 2.5)(4.5 - 2.5)
    = 22/7 × 14 = 44 cm2
  • Qstn #7
    The radii of the inner and outer circumferences of a circular running track are 63 m and 70 m respectively. Find : (i) the area of the track ;
    (it) the difference between the lengths of the two circumferences of the track.
    Ans :
    Outer radius, r1 = 70 m
    Inner radius , r2 = 63 m
    ∴ Area of track = πr12 - πr22
    = 22/7 [(70)2 - (63)2]
    = 22/7(70 + 63)(70 - 63)]
    = 22/7 × 133 × 7
    = 2926 m2
    Length of outer edge i.e., circumference = 2Ï€r1
    = 2 × 22/7 × 70
    = 440 m
    Length of inner edge = 2Ï€r2
    = 2× 22/7 ×63 = 396 m
    Difference between lengths of two circumferences = 440 - 396 = 44 m
    Hence, (i) 2926 m2(ii) 44 m
  • Qstn #8
    A circular field cf radius 105 m has a circular path of uniform width of 5 m along and inside its boundary. Find the area of the path.
    Ans :
    Radius of circular field, r1 = 105 m
    Width of path = 50 m
    ∴ Radius of inner circle, r2 = 105 - 5 = 100m
    ∴ Area of path = πr12 - πr22
    = 22/7 (105)2 - (100)2
    = 22/7 (105 + 100)(105 - 100)
    = 22/7 × 205 × 5
    = 22550/7 m2
    = 3221 3/7 m2
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