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 and AC at D and E respectively such that \(\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}\), Then ratio, \(\frac{\text{Area of}\,\Delta{ADE}}{\text{Area of}\,\Delta{ABC}}\) is

(a) 1 : 2 

(b) 1 : 4 

(c) 1 : 8 

(d) 1 : 16

Answer 1

(d) 1 : 16

Gvien \(\frac{AD}{AB}=\frac{1}{4}\) ⇒ \(\frac{AD}{DB}=\frac{1}{3}\),

i.e., D divides AB internally in the ratio 1 : 3. 

∴ Co-odinates of D are\(\bigg(\frac{1+12}{1+3},\frac{5+18}{1+3}\bigg)\)i.e.,\(\bigg(\frac{13}{4},\frac{23}{4}\bigg)\)

Also, \(\frac{AE}{AC}=\frac{1}{4}\) ⇒ \(\frac{AE}{EC}=\frac{1}{3}\), i.e., E divides AC internally in the ratio 1 : 3.

Co-ordinates of E are  \(\bigg(\frac{7+12}{1+3},\frac{2+18}{1+3}\bigg)\)i.e, \(\bigg(\frac{19}{4},5\bigg)\)

Now, area of Δ ABC = \(\frac{1}{2}\) [4(5 – 2) + 1(2 – 6)+ 7(6 – 5)]

 = \(\frac{1}{2}\) [12 – 4 + 7] = \(\frac{15}{2}\) sq. units

Area of Δ ADE, where A(4, 6), D\(\bigg(\frac{13}{4},\frac{23}{4}\bigg)\), E\(\bigg(\frac{19}{4},5\bigg)\) is 
\(\frac{1}{2}\)\(\bigg[4\bigg(\frac{23}{4}-5\bigg)+\frac{13}{4}(5-6)+\frac{19}{4}\bigg(6-\frac{23}{4}\bigg)\bigg]\)

= \(\frac{1}{2}\)\(\bigg[4\times\frac{3}{4}+\frac{13}{4}\times-1+\frac{19}{4}\times\frac{1}{4}\bigg]\)

= \(\frac{1}{2}\)\(\bigg[3-\frac{13}{4}+\frac{19}{16}\bigg]\) = \(\frac{1}{2}\) \(\bigg[\frac{84-5+19}{16}\bigg]\)= \(\frac{15}{32}\) sq. units

∴ \(\frac{\text{Area of}\,\Delta{ADE}}{\text{Area of}\,\Delta{ABC}}\) = \(\frac{\frac{15}{32}}{\frac{15}{2}}\) = 1 : 16.