Question

good evening friends


please solve the above question .....

Answer 1

$\huge\mathfrak{Answer}$

$\sf{Solution}$

=> From the given fig, DE =3cm, BC = 9cm and also they are parallel.
Again , from the fig,
we have,

<BAC = <DAE
and..
<D=<B
<E=<C... corresponding angles..

hence by..AAA criteria

$\frac{ar(ADE)}{ar(ABC)}$ = $\frac{DE^{2}}{BC^{2}}$= $\frac{AD^{2}}{AB^{2}}$ = $\frac{AE^{2}}{AC^{2}}$

$\frac{ar(ADE)}{ar(ABC)}$ = $\frac{DE^{2}}{BC^{2}}$

$\frac{30}{ar(ABC)}$ = $\frac{3^{2}}{9^{2}}$

$\frac{30}{ar(ABC)}$ = $\frac{9}{81}$

$\frac{30}{ar(ABC)}$ = $\frac{1}{9}$

=> ar∆ABC = $30\times 9$

=> ar∆ABC = 270$cm^{2}$

ar∆ABC = 270 $cm^{2}$

from the fig,,

Ar of trapezium DECB = ar∆ABC - ar∆ADE

ar(DECB) = 270 - 30

ar(DECB) = 240$cm^{2}$...

There fore, area of DECB is 240$cm^{2}$....