the vertex of triangle ABC are …

Area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by
Area = {tex}\frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}\therefore{/tex} Area of {tex}\triangle{/tex} ABC is given by
Area = {tex}\frac 12{/tex}[4(5 - 2)+ 1 (2 - 6)+ 7(6 - 5)]
Area = {tex}\frac 12{/tex}[12 + (-4) + 7]
Area = {tex}\frac{1}{2}[12-4+7]{/tex}
Area = {tex}\frac{15}{2}{/tex} sq units.
Area of {tex}\triangle{/tex}ABC = {tex}\frac{15}{2} sq\ units{/tex}
In {tex}\triangle{/tex}ADE and {tex}\triangle{/tex}ABC,
{tex}\frac { A D } { A B } = \frac { A E } { E C } = \frac { 1 } { 3 }{/tex}
and {tex}\angle D A E = \angle B A C{/tex} (Common)
{tex}\therefore{/tex} {tex}\triangle A D E \sim \triangle A B C{/tex} (By SAS)
{tex}\therefore{/tex} {tex}\frac { \text { Area } \Delta A D E } { \text { Area } \Delta A B C } = \left( \frac { A D } { A B } \right) ^ { 2 } = \left( \frac { 1 } { 3 } \right) ^ { 2 } = \frac { 1 } { 9 }{/tex}
{tex}\therefore{/tex} {tex}\frac { \text { Ar } \Delta A D E } { \left( \frac { 15 } { 2 } \right) } = \frac { 1 } { 9 }{/tex}
{tex}\therefore{/tex}  Area {tex}\triangle A D E = \frac { 15 } { 2 \times 9 } = \frac { 5 } { 6 }{/tex}Sq units
Area of {tex}\triangle{/tex}ADE : Area of {tex}\triangle{/tex}ABC = {tex}\frac { 5 } { 6 } : \frac { 15 } { 2 } {/tex}= 1 : 9