Finding the radius of a triangle's incircle: confusions on scale factors 
The radius of the incircle of a triangle whose sides are $18$, $24$, and $30$ cms is :
(a) $2$
(b) $4$
(c) $6$
(d) $8 $

Note that

*

*the radius of the incircle of the triangle is given by $\text{area}/\text{semi-perimeter}$


*$P$ is the half of the perimeter or $(a+b+c)/2$, so $P = (18+24+30)/2 = 36$


*$A= \sqrt{(36(36−18)(36−24)(36−30)}= \sqrt{36×18×12×6} =216 \text{ cm.}^{2}$
Therefore, the radius of incircle of the triangle is $216/36$, or 6cm.
I simplified the lengths to $3$ $4$ $5$ instead of $18$ $24$ $30$. So after solving I am getting $1$ cm.$^2$ as answer.
I am not getting why I have to multiply it with $6$ at the end when I am finding both area and semi perimeter w.r.t. the ratios.
 A: My understanding is that you're observing that this is a $3$-$4$-$5$ triangle, scaled up by a factor of $6$, and are observing that you are concerned with a ratio problem -- so in your eyes, that scale factor should not matter.
However, it does, quite a bit - because the areas and perimeters are not scaled up in the same way.
Consider a square of side length $1$ cm. It has perimeter $4$ cm. and area $1$ cm$^2$.
What happens if you scale up its dimensions by $6$? Then it becomes a square of side length $6$ cm.: hence, it has perimeter $24$ cm. $(=6 \times 4)$, and area $36$ cm$^2$ ($= 6^2 \times 1$).
Notice what happened here: the perimeter scaled by $6$, but the area scaled by $6^2$.
This phenomenon happens in general: if you magnify a length parameter of a shape by $\alpha$, and scale the rest of the shape accordingly, then all length-based things (like perimeters) are scaled by $\alpha$; all area-based things (like areas) are scaled by $\alpha^2$; all volume-based things are scaled by $\alpha^3$; and so on.
This is why you have to multiply your final answer by $6$: to restore the original ratios of the problem, you need to multiply your final answer by
$$\frac{6^2 }{6}  = 6$$
where $6^2$ accounts for the area scaling, and $6$ the semi-perimeter.
