The verticles of a ∆ABC are …

 
Since,
{tex}\frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{1}{4}{/tex}
{tex}\therefore DE||BC{/tex} [By Thales theorem]
{tex}\therefore \triangle ADE \sim \triangle ABC{/tex}
{tex}\therefore \frac{{Area(\Delta ADE)}}{{Area(\Delta ABC)}} = \frac{{A{D^2}}}{{A{B^2}}}{/tex}
{tex} = {\left( {\frac{{AD}}{{AB}}} \right)^2} = {\left( {\frac{1}{4}} \right)^2} = \frac{1}{{16}}{/tex} .......... (i)
Now, Area ({tex}\triangle{/tex}ABC) = {tex}\frac{1}{2}\left[ {4(5 - 2) + 1(2 - 6) + 7(6 - 5)} \right]{/tex}
{tex} = \frac{1}{2}\left[ {12 - 4 + 7} \right] = \frac{{15}}{2}{/tex} sq. units .......... (ii)
From eq. (i) and (ii),
Area ({tex}\triangle{/tex}ADE) = {tex}\frac{1}{{16}} \times {/tex}Area ({tex}\triangle{/tex}ABC) = {tex}\frac{1}{{16}} \times \frac{{15}}{2} = \frac{{15}}{{32}}{/tex} sq. units
Area ({tex}\triangle{/tex}ADE) : Area ({tex}\triangle{/tex}ABC) = 1 : 16