The Vertices of ∆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 Ad/Ab=Ae/Ac=1/4 .Calculate the Area of ∆Ade and - Mathematics

Sum

The vertices of ∆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 ∆ADE and compare it with the area of ∆ABC

Solution

We have,

\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\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{3}

⇒ AD : DB = AE : EC = 1 : 3

⇒ D and E divide AB and AC respectively in the ratio 1 : 3.

So, the co-ordinates of D and E are

( \frac{1+12}{1+3},\ \frac{5+18}{1+3} )=( \frac{13}{4},\frac{23}{4})\text{ and }( \frac{7+12}{1+3},\frac{2+18}{1+3})=( 19/ {4},\ 5)

We have,

∴

=\frac{1}{2}|( \frac{92}{4}+\frac{65}{4}+\frac{114}{4})-( \frac{78}{4}+\frac{437}{16}+20 )|

=\frac{1}{2}| \frac{271}{4}-\frac{1069}{16}|

=\frac{1}{2}\times \frac{15}{16}=\frac{15}{32}

Also, we have

∴

⇒

⇒ Area of ∆ABC = \frac { 1 }{ 2 } |64 – 49| =

\therefore \frac{Area\ of\ \Delta ADE}{Area\ of\ \Delta ABC}=\frac{15/32}{15/2}=\frac{1}{16}

Hence, Area of ∆ADE : Area of ∆ABC = 1 : 16.

Concept: Area of a Triangle
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