[tex]\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}[/tex]
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
Construction -
[tex]draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE[/tex]
Now , we can clearly see that AECD is a parallelogram !
[tex] \therefore [/tex] AE = CD = 13 cm
Now ,
[tex]AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm[/tex]
Now , In ∆ BCE ,
[tex]semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm[/tex]
Now , by Heron's formula
[tex]area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2} [/tex]
Also ,
[tex]area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2} [/tex]
Since we've obtained the height now , we can easily find out the area of trapezium !
[tex]Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2} [/tex]
hope helpful :D
