Question

Find the area of the trapezium whose parallel sides are 11 m and 25 m and the non-parallel sides are 15 m and 13 m.

Answer 1

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 the question, the distance between the parallel sides is not given and so we need to calculate the distance by using the properties of the quadrilateral and Pythagoras theorem in the right-angle triangles.

Complete step by step solution:

Let, “h” be the height between the parallel sides of the trapezium.

As, AFED forms a square, then, $ AF = DE = 11{\text{ m}} $

Also, $ BC = BD + DE + EC $

Consider, $BD = x$ then,
$
  BC = BD + DE + EC \\
  25 = x + 11 + EC \\
  EC = 14 - x \\
 $

Now, from the right-angled triangle ABD, applying Pythagoras theorem, we get:

$
  {\left( {AB} \right)^2} = {\left( {BD} \right)^2} + {\left( {AD} \right)^2} \\
  {13^2} = {x^2} + {h^2} - - - - - (i) \\
 $

Again in right-angled triangle FEC, applying Pythagoras theorem, we get:

$
  {\left( {FC} \right)^2} = {\left( {EC} \right)^2} + {\left( {FE} \right)^2} \\
  {15^2} = {(14 - x)^2} + {h^2} \\
  225 = 196 + {x^2} - 28x + {h^2} \\
  {x^2} + {h^2} - 28x = 225 - 196 - - - - (ii) \\
 $

Substitute $ {13^2} = {x^2} + {h^2} $ from equation (i) in equation (ii) to determine the value of x:
$
  {13^2} - 28x = 29 \\
  28x = 169 - 29 \\
  28x = 140 \\
  x = \dfrac{{140}}{{28}} \\
   = 5 \\
 $

Now, to determine the value of “h”, substitute $ x = 5 $ in the equation $ {13^2} = {x^2} + {h^2} $, we get:

$
  {13^2} = {5^2} + {h^2} \\
  196 = 25 + {h^2} \\
  {h^2} = 196 - 25 \\
  h = \sqrt {171} \\
   = 13.07{\text{ m}} \\
 $

Now, to determine the area of the trapezium, substitute $ h = 13.07{\text{ m}} $ as the height between the parallel sides and $ a = 11{\text{ m;b = 25 m}} $ as the length of the parallel sides in the formula $ A = \dfrac{1}{2}\left( {a + b} \right) \times h $ we get,

$
  A = \dfrac{1}{2}\left( {a + b} \right) \times h \\
   = \dfrac{1}{2}\left( {11 + 25} \right) \times 13.07 \\
   = \dfrac{1}{2} \times 36 \times 13.07 \\
   = 18 \times 13.07 \\
   = 235.26{\text{ }}{{\text{m}}^2} \\
 $

Hence, the area of the trapezium is 235.26 square meters.

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