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A field is in the shape of a trapezium having parallel sides 90 meters and 30 meters. These sides meet the third side at right angles. The length of the fourth side is 100 meters. If it costs Rs. 4 to plough $1{\text{ }}{{\text{m}}^2}$ of the field, find the total cost of ploughing the field.

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Answer
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Hint: Trapezium is a quadrilateral having four vertices and four sides with two of the sides as parallel and two non-parallel sides. The area of the trapezium is calculated as half of the product of the sum of the length of the two parallel sides and the distance between the parallel sides. Mathematically, $A = \dfrac{1}{2}\left( {a + b} \right) \times h$.
In this question, the total cost of ploughing is asked where we need to multiply the total area enclosed by the field by the cost of ploughing per square meters.

Complete step by step solution:
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$ BE = BD + DE \\
  90 = 30 + DE \\
  DE = 90 - 30 \\
   = 60{\text{ meters}} \\ $
Now, to find the distance between the parallel sides AC and BE, Pythagoras theorem has been used in the triangle CDE as:
$ C{E^2} = C{D^2} + D{E^2} \\
  {100^2} = {x^2} + {60^2} \\
  {x^2} = {100^2} - {60^2} \\
  {x^2} = (100 + 60)(100 - 60) \\
  {x^2} = 160 \times 40 \\
  x = \sqrt {6400} \\
   = 80{\text{ meters}} \\ $
Substitute 80 meters for the distance between the parallel sides and 30 meters and 90 meters as the length of the parallel sides in the equation $A = \dfrac{1}{2}\left( {a + b} \right) \times h$ to determine the area of the field as:
$ A = \dfrac{1}{2}\left( {a + b} \right) \times h \\
   = \dfrac{1}{2} \times \left( {90 + 30} \right) \times 80 \\
   = 40 \times 120 \\
   = 4800{\text{ }}{{\text{m}}^2} \\ $
Now, as the area of the field is 4800 square meters and the cost of ploughing per square meter is 4 then, multiply both the quantities to determine the total cost of ploughing as:
$ C = 4800 \times 4 \\
   = 19200 \\ $

Hence, the total cost of ploughing the field which is in the shape of the trapezium is Rs. 19200.

Note: It is to be noted here that the distance between the heights is not equivalent to the numerical value of the length of the non-parallel sides. Candidates often confuse these lengths.