# Find triangel-area with Cavalieri's principle

A triangle is given by $A=(0,0), B=(5,1)$ and $C=(2,4)$. I already know $\lambda^2(\Delta ABC)=9$. Now I want to compute the area by using Cavalieri's principle.

I know how to start when I have to use the principle for volumes, but I don't know how to start here. I thought about intercept theorem, but then I don't know how to find the ratios.

Any hints are appreciated.

If we are allowed to use "The Area of a Triangle as Half a Rectangle", one side of the rectangle has the length $\sqrt{(5-0)^2+(1-0)^2}=\sqrt26$ unit

The equation of the line containing the side is $$\frac{y-0}{x-0}=\frac{1-0}{5=0}\implies x-5y=0$$

The other side contains $(2,4),$ so the length of the perpendicular side will be $$\left|\frac{2-5\cdot4}{\sqrt{5^2+1^2}}\right|=\frac{18}{\sqrt{26}}$$

So, the are of the rectangle will be $\sqrt26\cdot \frac{18}{\sqrt{26}}=18$ Sq Unit.

Perhaps you could use shear mapping twice to get a right triangle. Shearing preserves areas, which relates this to Cavalieri's principle.
$A=(0,0)$, $B=(5,1)$, $C=(2,4)$
$A'=(0,0)$, $B'=(5,0)$, $C'=(2,\frac{18}5)$
$A''=(0,0)$, $B''=(5,0)$, $C''=(0,\frac{18}5)$
Right triangle is a half of the rectangle
$A''=(0,0)$, $B''=(5,0)$, $C''=(0,\frac{18}5)$, $D''=(5,\frac{18}5)$
so the area will be half of the area of rectangle: $$\frac{5\cdot\frac{18}5}2=9$$