Question

A field is in the shape of a trapezium whose parallel
sides are 25 m and 10 m. The non-parallel sides
are 14 m and 13 m. Find the area of the field.​

Answer 1

AnswEr:

From C, draw CE || DA. Clearly, ADCE is a llgm having AD || CE and DC || AE such that AD = 13 m and DC = 10 m.

$\therefore$ $\tt{AE=DC=10m\:and\:CE=AD=13m}$

$\Rightarrow$ $\tt{BE=AB-AE=(25-10)m=15m}$

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Thus in BCE, we have

$\tt{BC=14m,\:CE=13m\:and\:BE=15m}$

• Let s be the semi-perimeter of ∆BCE. Then,

$\tt{2s=BC+CE+BE=14+13+15=42}$

$\tt\underline{s=21}$

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$\therefore \tt \: Area \: of \: \triangle \: BCE \\ = \tt\sqrt{21 \times (21 - 14) \times (21 - 13) \times (21 - 15)} \\ \\ \implies \tt \: Area \: of \: \triangle \: BCE = \sqrt{21 \times 7 \times 8 \times 6} \\ \\ \implies \tt \: Area \: of \: \triangle \: BCE = \sqrt{ {7}^{2} \times {3}^{2} \times {4}^{2} } \\ \\ = \sf \: 84 {m}^{2}$

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$\star$ Area of ∆BCE = $\tt\dfrac{1}{2}(BE\times\:CL)$

$\Rightarrow$ $\tt{84=}$ $\tt\dfrac{1}{2}\times\:15\times\:CL$

$\Rightarrow$ $\tt{CL=}$$\tt\dfrac{168}{15}=$$\tt\dfrac{56}{5}$

$\Rightarrow$ Height of parallelogram ADCE = $\tt{CL=}$$\tt\dfrac{56}{5}m$

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$\therefore$$\tt{Area\:of\:IIgm\:ADCE=Area\:of\:IIgm\:ADCE+Area\:of\:\triangle\:BCE}$

$\tt{(112+84)m^2=196\:m^2}$

#BAL

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