Question: Prove that there exists no triangle with heights 4,7, and 10 units. I am completely puzzled.

Answer: 
The area of the triangle is half the product of a side of the triangle and the height of the triangle on that side. So calling the area of the triangle $A$, the sides would have to be

$$\frac{2\cdot A}{10}, \; \frac{2\cdot A}{7},\; \frac{2\cdot A}{4}.$$

But

$$\frac{1}{4} > \frac{1}{7} + \frac{1}{10},$$

so these "sides" could not make a triangle, since the longest is longer than the sum of the two shorter.
Hint: Well let the sides of the triangle be $a, b, c$ - ordered so that twice the area of the triangle is $4a=7b=10c$ (base times height). Can you use this information to express all the sides of the triangle as multiples of $a$?