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ICSE-VIII-Mathematics

20: Area of Trapezium and a Polygon Class 8 Maths

with Solutions - page 2

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  • Qstn #10
    The lengths of the sides of a triangle are in the ratio 4: 5: 3 and its perimeter is 96 cm. Find its area.
    Ans : Let the sides of the triangle ABC be 4x, 5x and 3x
    Let AB = 4x, AC = 5x and BC = 3x
    Perimeter = 4x + 5x + 3x = 96
    ⇒ 12x = 96
    ⇒ x = 96/12

    ∴ x = 8
    ∴ Sides are
    BC = 3(8) = 24 cm,
    AB = 4(8) = 32 cm,
    AC = 5(8) = 40 cm
    Since, (AC)2 = (AB)2 + (BC)2
    [∵ (5x)2 = (3x)2 + (4x)2]
    ∴ By Pythagoras Theorem , ∠B = 90˚
    ∴ Area of ΔABC = ½ (BC)(AB)
    = ½ (24)(32)
    = 12 × 32
    = 384 cm2
  • Qstn #11
    One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.
    Ans : In isosceles ∆ABC
    AB = AC = 13 cm But perimeter = 50 cm

    ∴ BC = 50 - (13 + 13) cm
    = 50 - 26
    = 24 cm
    AD ⊥ BC
    ∴ AC = DC = 24/2 = 12 cm
    In right ∆ABD,
    AC2 = AD2 + BD2 (Pythagoras Theorem)
    ⇒ (13)2 = AD2 + (12)2
    ⇒ 169 = AD2 + 144
    ⇒ AD2 = 169 - 144
    = 25
    = (5)2
    ∴ AD = 5 cm
    Now area of ∆ABC = ½ × base × Altitude
    = ½ × BC × AD
    = ½ × 24 × 5
    = 60 cm2
  • Qstn #12
    The altitude and the base of a triangular field are in the ratio 6 : 5. If its cost is ₹ 49,57,200 at the rate of ₹ 36,720 per hectare and 1 hectare = 10,000 sq. m, find (in metre) dimensions of the field,
    Ans : Total cost = ₹ 49,57,200
    Rate = ₹ 36,720 per hectare
    Total area of the triangular field = 4957200/36720 × 10000 m2
    = 1350000 m2
    Ratio in altitude and base of the field = 6 : 5
    Let altitude = 6x
    and Base = 5x
    ∴ Area = ½ × Base × Altitude
    ⇒ 1350000 = ½ × 5x × 6x
    ⇒ 15x2 = 1350000
    ⇒ x2 = 1350000/15
    ⇒ x2 = 90000
    = (300)2
    ∴ x = 300
    ∴ Base = 5x = 5 × 300 = 1500 m
    And Altitude = 6x = 6 × 300 = 1800 m
  • Qstn #13
    Find the area of the right-angled triangle with hypotenuse 40 cm and one of the other two sides 24 cm.
    Ans : In right angled triangle ABC Hypotenuse AC = 40 cm
    One side AB = 24 cm


    ∴ Area = ½ × AB × BC
    = ½ × 24 × 32 cm2
    = 384 cm2
  • Qstn #14
    Use the information given in the adjoining figure to find :
    Ans : AB = 24 cm, BC = 7 cm
  • #14-i
    the length of AC.
    Ans :
  • #14-ii
    the area of a ∆ABC
    Ans : Area of ∆ABC = ½ × AB × BC
    = ½ × 24 × 7
    = 84 cm2
  • #14-iii
    the length of BD, correct to one decimal place.
    Ans : BD ⊥ AC
    Area ∆ABC = ½ × AC × BD
    84 = ½ × 25 × BD
    ⇒ BD = (84 × 2)/25
    = 168/25
    = 6.72 cm
    = 6.7 cm
  • #
    Section : B
  • Qstn #1
    Find the length and perimeter of a rectangle, whose area = 120 cm2 and breadth = 8 cm
    Ans : area of rectangle = 120 cm2
    breadth, b = 8 cm
    Area = l × b⇒ l × 8 = 120
    ⇒ l = 120/8 = 15 cm
    Perimeter = 2 (l + b) = 2(15 + 8) = 2× 23 = 46 cm
    Length = 15 cm; Perimeter = 46 cm
  • Qstn #2
    The perimeter of a rectangle is 46 m and its length is 15 m. Find its :
  • #2-i
    breadth
    Ans : Perimeter of rectangle = 46 m
    length, l = 15 m
    2 (l + b) = 46
    ⇒ 2(15 + b) = 46
    ⇒ 15 + b = 46/2 = 23
    ⇒ b = 23 - 15
    ⇒ b = 8 m
  • #2-ii
    area
    Ans : area = l × b = 15 × 8 = 120 m2
  • #2-iii
    diagonal.
    Ans : Hence, (i) 8 m(ii)120 m2(iii)17 m
  • Qstn #3
    The diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its :
    Ans : AC2 = AB2 + BC2 (By Pythagoras theorem)
    ⇒ (34)2 = l2 + (16)2
    ⇒ 1156 = l2 + 256
    ⇒ l2 = 1156 - 256
    ⇒ l2 = 900
    ⇒ l = √900 = 30 cm
    area = l × b = 30× 16 = 480 cm2
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