When extending the sides $AB,BC,CA$ of $\Delta ABC$ to $B',C',A'$ respectively, such that $AB' = 2AB$ , $CC' = 2BC$ , $AA' = 3CA$ .

Question

When extending the sides $AB,BC,CA$ of $\Delta ABC$ to $B',C',A'$ respectively, such that $AB' = 2AB$ , $CC' = 2BC$ , $AA' = 3CA$ . If $[\Delta ABC] = 1$, find $[\Delta A'B'C']$ .

What I Tried: Here is a picture :-

Unfortunately, I got no idea for this problem. The fact that only the area of the triangle of $\Delta ABC$ is given is what I find difficult to solve it. Maybe you can assume each sides $AB,BC,CA$ to be $x,y,z$ respectively and it's area from heron's formula will be :- $$\sqrt{\bigg(\frac{x+y+z}{2}\bigg)\bigg(\frac{x+y+z}{2} - x\bigg)\bigg(\frac{x+y+z}{2} - y\bigg)\bigg(\frac{x+y+z}{2} - z\bigg)} = 1$$ $$\rightarrow \bigg(\frac{x+y+z}{2}\bigg)\bigg(\frac{x+y+z}{2} - x\bigg)\bigg(\frac{x+y+z}{2} - y\bigg)\bigg(\frac{x+y+z}{2} - z\bigg) = 1$$

I guess that looks fine but will get complicated. Also I suppose to find $[\Delta A'B'C']$, I have to find the areas of each of the triangles $[C'BB'] , [B'A'A] , [C'CA']$ first. But how do I do that?

Can anyone help me? Thank You.

Answer 1

Chop it like so:

The purple triangles and pink triangles have the same area as the one in the centre. The green and brown(?) ones double, and the blue one triple.

Answer 2

See my answer to this question. We can calculate areas by sliding the vertices.Join $A'B$, $B'C$, $C'A$. $\frac{[A'AB]}{[ABC]} = \frac{A'A}{AC} = 3$, $\frac{[A'AB]}{[A'B'A]} = \frac{AB}{AB'} = \frac{1}{2}$. Therefore, $[A'B'A] = 6×[ABC]$. Similarly we can calculate $[BC'B'] = 3×[ABC]$, $[A'C'C] = 8×[ABC]$. Therefore $[A'B'C'] = [ABC] + [A'AB'] + [BC'B'] + [A'C'C] = \boxed{18}$

as $[ABC] = 1$.