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If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

The vertices ...

Question

The vertices of a $Δ$ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB at D, such that AD/AB= $\frac{1}{4}$ . Calculate the area of the $Δ$ ADC.

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Solution

So D is the point which divides AB in the ratio 1: 3.

Using section formula we get the coordinates of D(3.25,5.75)

Area of a triangle = $\frac{1}{2}$ [ $x_{1}$ ( $y_{2}$ - $y_{3}$ ) + $x_{2}$ ( $y_{3}$ - $y_{1}$ ) + $x_{3}$ ( $y_{1}$ - $y_{2}$ )]

= $\frac{1}{2}$ [4(5.75 - 2) +3.25(2 - 6) + 7(6 - 5.75)]

= 1.875 sq units

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