A field is in the shape of trapezium whose parallell side are 10cm and 25cm respectively and non parellel sides are 13cm and 14cm, find its area?

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Solution

DC = DE + EF + FC
25 = DE + 10 + FC
DE + FC = 15
Let FC = x
$∴$ DE = 15 - x
In $Δ A E D$
$h^{2} = 13^{2} - (15 - x)^{2}$ ...(1)
In $Δ B F C$
$h^{2} = 14^{2} - x^{2}$ ...(2)
Equating the values of $h^{2}$ from (1) and (2), we get
$13^{2} - (15 - x)^{2} = 14^{2} - x^{2}$
$169 - 225 + 30 x - x^{2} = 196 - x^{2}$
$- 56 + 30 x = 196 30 x = 252 x = 8.4 h^{2} + x^{2} = 14^{2} h^{2} = 14^{2} - {8.4}^{2} = 196 - 70.56 = 125.44 h = \sqrt{125.44} = 11.2$
$A r e a = \frac{1}{2} \times (A B + C D) \times h = \frac{1}{2} \times (35) (11.2) = 196 m^{2}$

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