#### Question: In the figure below, the vertices of ∆ ABC are A (4, 6), B(1, 5) and C (7, 2). A line -segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \\frac{AD}{AB}) = \\frac{AE}{AC}) = \\frac{1}{3}) calculate the area of ∆ ADE and compare it with area of ∆ABC.

#### Video Explanation

#### Explanatory Answer

Let us find out coordinate of points D and E.

AD : DB = 1 : 2

and AE : EC = 1 : 2

Coordinates of D = \ ( \frac {2×4+1×1} {2+1}),\\frac {2×6+1×5} {2+1}) )

= \(\frac{9}{3}), \\frac{17}{3})) = (3, \\frac{17}{3}))

Coordinates of E = \(\frac{2 × 4 + 1 ×7}{3}), \\frac{2 × 6 + 1 ×2}{3}))

= (5, \\frac{14}{3}))

Area of a triangle whose coordinates are (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3})

= \\frac{1}{2}) {x_{1} (y_{2} - y_{3} ) - x_{2} (y_{1} - y_{3}) + x_{3} (y_{1} - y_{2})}

Area of ADE whose coordinates are (4, 6) , (3, \\frac{17}{3})) and (5, \\frac{14}{3}))

= \\frac{1}{2}) {4 \(\frac{17}{3}) - \\frac{14}{3})) - 3 (6 - \\frac{14}{3})) + 5 (6 - \\frac{17}{3}))}

= \\frac{1}{2}) {4 × 1 - 3 × \\frac{4}{3}) + 5 × \\frac{1}{3})} = \\frac{5}{6}) square units

Area of triangle ABC whose coordinates are (4, 6), (1, 5), and (7, 2)

= \\frac{1}{2}) {4 (5 - 2) - 1 (6 - 2 ) + 7 (6 - 5)}

= \\frac{1}{2}) {4 × 3 - 4 + 7} = \\frac{15}{2})

\\frac{Area of ∆ ABC}{Area of ∆ ABE}) = \\frac{15/2}{5/6})

= \\frac{15}{2}) × \\frac{6}{5}) = 9.

Sides of ADE = \\frac{1}{3}) Sides of ABC

Area of ADE = \\frac{1}{9}) Area of ABC