Question

$A(3,-4),B(5,-2),C(-1,8)$ are the vertices of $△ABC.D,E,F$ are the mid-points of sides $\overline{BC},\overline{CA}$ and $\overline{AB}$ respectively. Find area of $△ABC$. Using coordinates of $D,E,F$, find area of $△DEF$. Hence show that the $ABC=4\left(DEF\right)$

Answer 1

$D=\left(\frac{5-1}{2}\right),\left(\frac{-2+8}{2}\right)$

$\Rightarrow (2,3)$

$E=(\frac{-1+3}{2},\frac{8-4}{2})$

$\Rightarrow (1,2)$

$F=(\frac{3+5}{2},\frac{-4-2}{2})$

$\Rightarrow (4,-3)$

In $△ABC\Rightarrow a=\sqrt{(-1-5{)}^2+(P-(-2){)}^2}$

$=\sqrt{6^2+{10}^2}$

$=\sqrt{136}=2\sqrt{34}$

$b=\sqrt{(-1-3{)}^2+(8-(-4){)}^2}$

$=\sqrt{4^2+{12}^2}$

$=\sqrt{160}=4\sqrt{10}$

$c=\sqrt{(5-3{)}^2+(-2+4{)}^2}$

$=\sqrt{2^2+2^2}$

$\sqrt{8}=2\sqrt{2}$

$s=\frac{a+b+c}{2},A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=(\sqrt{34}-2\sqrt{10}+\sqrt{2})(\sqrt{34}+2\sqrt{10}-\sqrt{2})$

$\Rightarrow 16sq=ar\left(△ABC\right)$

In $△DEF,r=\sqrt{(4-1{)}^2+(-3-2{)}^2}$

$\sqrt{34}=\sqrt{34}$

$c=\sqrt{(4-2{)}^2+(-3-3{)}^2}$

$=\sqrt{40}=2\sqrt{10}$

$f=\sqrt{(1-2{)}^2+(2-3{)}^2}$

$=\sqrt{1^2+1^2}$

$=\sqrt{2}$

$=\frac{d+e+f}{2},A=\sqrt{s(s-d)(s-e)(s-f)}$

$=\sqrt{\left(\frac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}\right)(\frac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-\sqrt{34})}\times (\frac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-2\sqrt{10})\times (\frac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-\sqrt{2})$

$A^{\prime }=4squnit=ar\left(△DEF\right)$

$\Rightarrow ar\left(△ABC\right)=16squnit$

$=4\times 4$

$=4\times ar\left(△DEF\right)$

Hence proved