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  • The vertices of a Delta ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD by AB equals AE by AC Calculate the area of the Delta ADE and compare it with the area of Delta ABC.

6.   The vertices of a \Delta ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4} Calculate the area of the \Delta ADE and compare it  with the area of \Delta ABC. 

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best_answer

From the figure:

triangle ratio

Given ratio:

\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}

Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of 1:3.

Section formula:

P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )

Then, coordinates of point D:

D(x_{1},y_{1})= \left (\frac{1\times1+3\times 4}{1+3} , \frac{1\times 5+3\times 6}{1+3} \right )

Coordinates of point E:

E(x_{2},y_{2})= \left (\frac{1\times7+3\times 4}{1+3} , \frac{1\times 2+3\times 6}{1+3} \right )

= \left ( \frac{19}{4}, \frac{20}{4} \right )

Then, the area of a triangle:

= \frac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]

Substituting the values in the above equation,

Area\ of\ \triangle ADE = \frac{1}{2}\left [ 4\left ( \frac{23}{4} - \frac{20}{4}\right )+\frac{13}{4}\left ( \frac{20}{4} - 6 \right )+\frac{19}{4}\left (6-\frac{23}{4} \right )\right ]= \frac{1}{2}\left [ 3-\frac{13}{4} +\frac{19}{16}\right ] = \frac{1}{2}\left [ \frac{48-52+19}{16} \right ] = \frac{15}{32}\ square\ units.

Area\ of\ \triangle ABC = \frac{1}{2}\left [ 4(5-2)+1(2-6)+7(6-5) \right ]

= \frac{1}{2}\left [ 12-4+7 \right ] = \frac{15}{2}\ Square\ units.

Hence the ratio between the areas of \triangle ADE and \triangle ABC is 1:16.

Posted by

Divya Prakash Singh

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