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Question

# A field is in the shape of a trapezium whose parallel sides are 25cm and 10cm. If its non-parallel sides are 14cm and 13cm. 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=(15–x)2+h2 ............(ii) Substituting the value of h2 from (i) in (ii), we get, 142=(15−x)2+132−x2 196−169=225+x2−30x−x2 27=225−30x x=19830=6.6 Substituting the value of x as 6.6 in (i), we get, 132=6.62+h2 169−43.56=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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