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 \(area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16 \)