The parallel sides of a trapezium are 25 cm and 13 cm,its non- parallel sides are equal being 10 cm. Find the area of the trapezium.

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Solution

Draw perpendiculars from A and D on BC as AE and DF.
Since, non parallel sides are equal, this is an isosceles trapezium and hence BE = FC.
$B E = \frac{B C - E F}{2} = \frac{25 - 13}{2} = 6$ .
From right angled triangle ABE, $A B^{2} = B E^{2} + A E^{2}$
or $A E^{2} = A B^{2} - B E^{2} = 10^{2} - 6^{2} = 64$
$\to$ AE = 8. Note that AE is height of trapezium.
Area of trapezium = $\frac{1}{2} \times$ (sum of parallel sides) $\times$ height
$= \frac{1}{2} \times (25 + 13) \times 8 = 19 \times 8 = 152 s q . c m .$

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