How do you find the inner radius of a cone? Suppose you have a cone with it's bottom cut, so the top radius is $r_1$ and the bottom radius is $r_2$. The height of the cone is $h$. What is the radius of a cross sectional area $y$ units above $r_2$?
I've seen an answer that looks like $r = r_2+\frac{y}{h}\cdot(r_1-r_2)$, but I have no idea how it was obtained.
 A: Hints:
$1.$ Use the similitude of triangles $\mathop {ABC}\limits^\vartriangle$ and $\mathop {AB'C'}\limits^\vartriangle$ to find $x$.
$2.$ Next, use the similitude of triangles $\mathop {ABC}\limits^\vartriangle$ and $\mathop {AB''C''}\limits^\vartriangle$ to find the relation between $r$ and $y$.
$\qquad \qquad \qquad \qquad \qquad \quad$ 
A: Hint: Look at the the picture in H.R.'s answer. Fix a Cartesian coordinate system with $y$-axis along $h$, $x$-axis parallel to $r_1$, and origin at $A$.
What is the slope of the line joining $C$ and $C'$?
A: Notice, let $r_y$ be the radius of the cross section at a height $y$ from the bottom then
consider two similar right triangles having corresponding bases as $\color{blue}{r_y-r_1}$ & $\color{blue}{r_2-r_1}$, and the corresponding heights $\color{blue}{h-y}$ & $\color{blue}{h}$ then we have $$\frac{r_y-r_1}{r_2-r_1}=\frac{h-y}{h}$$   $$r_y-r_1=(r_2-r_1)\left(1-\frac{y}{h}\right)$$
$$r_y=r_2-r_1-(r_2-r_1)\frac{y}{h}+r_1$$
$$r_y=r_2-(r_2-r_1)\frac{y}{h}$$
$$\color{red}{r_y=r_2+(r_1-r_2)\frac{y}{h}}$$
hence, your answer is correct
