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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

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    Section : A
  • Qstn #1
    Find the area of a triangle, whose sides are :
  • #1-i
    10 cm, 24 cm and 26 cm
    Ans : Sides of ∆ are
    a = 10 cm
    b = 24 cm
    c = 26 cm
    S = (a + b + c)/2 = (10 + 24 + 26)/2
    = 60/2
    = 30
  • #1-ii
    18 mm, 24 mm and 30 mm
    Ans : Sides of ∆ are
    a = 18 mm
    b = 24 mm
    c = 30 mm
    S = (a + b + c)/2
    = (18 + 24 + 30)/2
    = 72/2
    = 36
  • #1-iii
    21 m, 28 m and 35 m
    Ans : Sides of ∆ are
    a = 21 m
    b = 28 m
    c = 35 m
    S = (a + b + c)/2
    = (21 + 28 + 35)/2
    = 84/2
    = 42
  • Qstn #2
    Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm; find : (i) area of the triangle (ii) height of the triangle corresponding to 8 cm side.
    Ans :
    BC = 6 cm
    Height AD = 4 cm
    Area of ∆ = ½ × base × height
    = 1//2 × BC × AD
    = ½ × 6 × 4
    = 12 cm2
    Again area of ∆ = ½ × AC × BE
    12 = ½ × 8 × BE
    ∴ BE = (12 × 2)/8
    BE = 3 cm
    ∴ (i) 12 cm2
    (ii) 3 cm
  • Qstn #3
    The sides of a triangle are 16 cm, 12 cm and 20 cm. Find :
    Ans : Sides of ∆ are
    a = 20 cm
    b = 12 cm
    c = 16 cm
    S = (a + b + c)/2
    = (20 + 12 + 16)/2
    = 48/2
    = 24

    AD is height of ∆ corresponding to largest side.
    ∴ ½ × BC × AD = 96
    ⇒ ½ × 20 × AD = 96⇒ AD = (96 × 2)/20
    ⇒ AD = 9.6 cm
    BE is height ∆ corresponding to smallest side.
    ∴ ½ × AC × BE = 96
    ⇒ ½ × 12 × BE = 96
    ⇒ BE = (96 × 2)/12
    ⇒ BE = 16 cm
  • #3-i
    area of the triangle ; (ii) height of the triangle, corresponding to the largest side ; (iii) height of the triangle, corresponding to the smallest side.
    Ans : 96cm2(ii) 9.6 cm(iii) 16 cm
  • Qstn #4
    Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m; find : (i) area of the triangle ; (ii) height of the triangle corresponding to 6.4 m side.
    Ans :
    ABC is the ∆ in which BC = 4.8 m
    AC = 6.4 m and AD = 6 m
    ∴ area of ∆ABC = ½ × BC × AD
    = ½ × 4.8 × 6
    = 14.4 m2
    BE is height of ∆ corresponding to 6.4 m
    ∴ ½ × AC × BE = 14.4
    ⇒ ½ × 6.4 × B.E = 14. 4
    ⇒ BE = (14.4 × 2)/6.4
    ⇒ BE = 14.4/3.2
    = 9/2
    = 4.5 m
    Hence, (i)14.4 m2(ii)4.5 m
  • Qstn #5
    The base and the height of a triangle are in the ratio 4 : 5. If the area of the triangle is 40 m2; find its base and height.
    Ans : Let base of ∆ = 4x m
    and height of ∆ = 5x m
    area of ∆ = 40 m2
    ∵ ½ × base × height = area of ∆
    ⇒ ½ × 4x × 5x = 40
    ⇒ 10x2 = 40
    ⇒ x2 = 4
    ⇒ x = √4
    ⇒ x = 2
    ∴ base = 4x = 4 × 2 = 8 m
    Height = 5x = 5 × 2 = 10m
    ∴ 8 m; 10 m
  • Qstn #6
    The base and the height of a triangle are in the ratio 5 : 3. If the area of the triangle is 67.5 m2; find its base and height.
    Ans : Let base = 5x m
    height = 3x m
    Area of ∆ = ½ × base × height
    ∴ ½ × 5x × 3x = 6.75
    ⇒ x2 = (67.5 × 2)/15
    ⇒ x2 = 4.5 × 2
    ⇒ x2 = 9.0
    ⇒ x = √9
    ⇒ x = 3
    base = 5x = 5× 3 = 15 m
    height = 3x = 3× 3 = 9 m
  • Qstn #7
    The area of an equilateral triangle is 144√3 cm2; find its perimeter.
    Ans : Let each side of an equilateral triangle = x cm
    ∴ Its area = √3/4 (side)2
    = √3/4 x2
    = 144√3 (given)
    ⇒ x2 = 144√3 × 4/√3
    ⇒ x2 = 144 × 4
    ⇒ x2 = 576
    ⇒ x = √576
    = 24 cm
    Each side = 24 cm
    Hence perimeter = 3(24) = 72 cm
  • Qstn #8
    The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.
    Ans : Let each side of the equilateral traingle = x
    ∴ Its area = √3/4 x2
    Area perimeter = 3x
    By the given condition = √3/4 x2 = 3x
    x2 = 3x × 4/√3
    ⇒ x2 = (3x × 4 × √3)/( √3 × √3)
    = (3x × 4 × √3)/3
    = 4x√3
    ⇒ x2 = √3 (4x)
    ⇒ x = 4√3 [∵ x ≠ 0]
    ∴ Perimeter = 12√3 units
    = 12(1.732)
    = 20.784
    = 20.78 units
  • Qstn #9
    A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.
    Ans : Since ∠A = 90°
    By Pythagoras Theorem,
    In ∆ABD,


    Now, area of ∆ABD = ½(18)(24)
    = (18)(12)
    = 216 m2
    Again in ∆BCD;
    ⇒ By Pythagoras Theorem √CBD = 90˚
    [∴ DC2 = BD2 + BC2, Since (50)2 = (30)2 + (40)2]
    ∴ Area of ∆BCD = 1/2(40)(30)
    = 600 m2
    Hence, area of quadrilateral ABCD = Area of ∆ABD + area of ∆BCD
    = 216 + 600
    = 816 m2
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