• question_answer 27) In fig. 8, the vertices of $\Delta \,ABC$ are $A(4,6),\,\,B(1,5)$and $C(7,2)$. A line ? segment DE is drawn to intersect the sides AB and AC at D and E respectively such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$. Calculate the area of $\Delta \,ADE$ and compare it with area of $\Delta \,ABC$.

 We have, the vertices of $\Delta \text{ }ABC$ as $A(4,6),\,\,B(1,5)$ and $C(7,2)$ And $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$ Then, coordinated of D are $\left( \frac{1(1)+2(4)}{1+2},\frac{1(5)+2(6)}{1+2} \right)$ $\left( \frac{1+8}{3},\frac{5+12}{3} \right)\,i.e.,\,\,D\left( 3,\frac{17}{3} \right)$ And coordinates of E are $\left( \frac{1(7)+2(4)}{1+2},\frac{1(2)+2(6)}{1+2} \right)$ $\left( \frac{7+8}{3},\frac{2+12}{3} \right)\,\,i.e.,\,\,E\left( 5,\frac{14}{3} \right)$ Now, Area of $\Delta \,ADE$ $=\frac{1}{2}\left[ 4\left( \frac{17}{3}-\frac{14}{3} \right)+3\left( \frac{14}{3}-6 \right)+5\left( 6-\frac{17}{3} \right) \right]$ $=\frac{1}{2}\left[ 4(1)+3\left( -\frac{4}{3} \right)+5\left( \frac{1}{3} \right) \right]$ $=\frac{5}{6}$ units And Area of $\Delta \,ABC$ $=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]$ $=\frac{1}{2}[4(3)+1(-4)+7(1)]=\frac{15}{2}$ units. $\therefore \,\,\frac{ar\,(\Delta \,ADE)}{ar(\Delta \,ABC)}=\frac{5/6}{15/2}=\frac{1}{9}$ i.e., $ar\,(\Delta \,ADE):ar\,(\Delta \,ABC)=1:9$