Question

Question 6
The vertices of triangle 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 triangle ADE and compare it with area of triangle ABC.

Answer 1

Point D divides AB in ratio 1 : 3
Point E divides AC in the same ratio.
Hence, $m_1$ = 1 and $m_2$ = 3
Coordinates of D can be calculated as follows:
$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$
$=\frac{3\times 4+1\times 1}{4}=\frac{13}{4}$
$y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}$
$=\frac{3\times 6+1\times 5}{4}=\frac{23}{4}$
Coordinates of E can be calculated as follows:
$x=\frac{1\times 7+3\times 4}{4}=\frac{19}{4}$
$y=\frac{1\times 2+3\times 6}{4}=5$
Area of triangle ABC can be calculated as follows:
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
$=\frac{1}{2}\left[4\right(5-2)+1(2-6)+7(6-5\left)\right]$
$=\frac{1}{2}(12-4+7)=\frac{15}{2}squnit$
Area of triangle ADE can be calculated as follows:
$=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
$=\frac{1}{2}[4(\frac{23}{4}-5)+\frac{13}{4}(5-6)+\frac{19}{4}(6-\frac{23}{4})]$
$=\frac{1}{2}\left[4\right(\frac{23-20}{4})-\frac{13}{4}+\frac{19}{4}(\frac{24-23}{4}\left)\right]$
$=\frac{1}{2}(3-\frac{13}{4}+\frac{19}{16})$
$=\frac{1}{2}\left[\frac{48-52+19}{16}\right]$
$=\frac{1}{2}\times \frac{15}{16}=\frac{15}{32}squnit$
Hence, ratio of area of triangle ADE to area of triangle ABC
= $\frac{\frac{15}{32}}{\frac{15}{2}}$ = $\frac{1}{16}$ = 1 : 16