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Question

A field is in the shape of a trapezium whose parallel sides are 25m and 10m. If its non-parallel sides are 14m and 13m. Find its area ___ (in cm2).

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Solution

Let ABCD be the trapezium given to us with AB = 10cm, CD = 25cm, AD = 13cm and BC = 14cm.
If we drop perpendiculars from points A and B on the line CD it intersects the line CD on the points E and F respectively, which means that the length of the line segment EF is also 10cm. Let DE = x cm, so that,
FC = (25 - 10 - x)cm = (15 - x)cm ( as CD = 25cm = EF + DE + CF).
Let the height of the trapezium be h cm, and AE = BF = h cm.

In ADE, using Pythagoras theorem,
AD2=DE2+AE2
132=x2+h2 ......................(i)
In BFC, using Pythagoras theorem,
BC2=FC2+BF2
142=(15x)2+h2 ............(ii)
Substituting the value of h2 from (i) in (ii), we get,
142=(15x)2+132x2
196169=225+x230xx2
27=22530x
x=19830=6.6
Substituting the value of x as 6.6 in (i), we get,
132=6.62+h2
16943.56=h2
h2=125.44
h = 11.2 cm
Area of a trapezium = 12×(Sum of parallel sides of a trapezium)×(height of the trapezium)
= 12×(25+10)×(11.2)
= 35×5.6= 196cm2


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