3 Tutor System
Starting just at 265/hour

# The vertices of a $$∆ \ ABC$$ 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 $$∆ \ ADE$$ and compare it with area of $$∆ \ ABC$$ . (Recall Theorem 6.2 and Theorem 6.6) Given, the vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2) and,
$$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$
$$\Rightarrow \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4$$
$$\Rightarrow \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4$$
$$\Rightarrow 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4$$
$$\Rightarrow \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3$$

$$\therefore$$ Point D and E divides AB and AC in ratio 1 : 3

Now coordinates of D calculated by section formula are,
$$x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4}$$ and,
$$y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4}$$

Similarly, coordinates of E are,
$$x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4}$$ and,
$$y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5$$

Now area of triangle ADE is,

$$\frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ |$$
= $$\frac{1}{2} \ | \ [(12-4+7)] \ |$$
= $$\frac{15}{2}$$ sq. units

Similarly, area of triangle ABC is,

$$\frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ |$$
= $$\frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ |$$
= $$\frac{15}{32}$$ sq. units

$$\therefore$$ ratio of $$area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16$$