Common tangent, height of intersection point

Question

Two spheres of diameter $6 cm$ and $4 cm$ touch each other at $A$ as they rest on a horizontal table.

How high is $A$ above the table?

Answer 1

Hint: It is a trapezoid of this shape where $AD$ is the table; and $B$ and $C$ are centers of the large and small spheres, respectively: !https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTaBHxbv_N4Eyl7JHdQNr-tADerSqK6LRNjIC2MIRaXfeQDCGWsgQ

Since both spheres are on the table, suppose they touch the table at points $A$ and $D$ (assume spheres touch each other at point $A'$). If centers of spheres are $B$ (with diameter $6$) and $C$ (with diameter $4$), $BA\bot AD$ and $CD\bot AD$ (radius-point of tangent relation). Thus, $DC$ and $AB$ are parallel to each other. Also, $AB$ is $3$ $cm$ while $DC$ is $2$ $cm$ long (radii of spheres). When $C$ & $B$ are connected, $CB$ should go through $A'$, the touching point of spheres (again, radius-point of tangent relation). If we draw two parallel lines from $C$ and $A'$ perpendicular to $AB$ in such a way that they meet $AB$ at $E$ & $F$, respectively and $CE\bot AB$ and $A'F\bot AB$ ($AF$ is equal to the perpendicllar distance from table top to $A'$).

Now, consider $\Delta$ $CEB$ and $\Delta$ $A'FB$: $BA'=3$, $BC=3+2=5$, and $BE=BA-EA=BA-CD=3-2=1$. They are also two right-triangles with a common vertex at $B$. Therefore, they are similar, hence their corresponding sides are proportional: $$\frac{BF}{BE}=\frac{BA'}{BC}$$ Thus, $$BF=BE\cdot \frac{BA'}{BC}=1\times\frac{3}{5}=\frac{3}{5}$$ Therefore, $AF=AB-BF=3-\frac{3}{5}=2\frac{2}{5}$ $cm$.