HumanCalc

We see that the rectangle with side lengths 5 and 10 have an area of 50 and another rectangle with the same sides on the opposite side. 

The two bottom rectangles have side lengths 3.6 and 10 have a combined area of 3.6*2*10 = 72.

The two front triangles have bases of 6 and heights 2 and 4 which have a combined area of \(\dfrac{6\cdot2}{2}\cdot \dfrac{6\cdot4}{2} = 6\cdot 12= 72\)

There are also the two triangles on the back which have the same area. 

Thus the total surface area is 72+72+72=216.

We can see that \(DG = 12\) and \(DF = 5\).

By Pythagorean theorem we get:

\(a^2+b^2= c^2\)  \(\rightarrow\) \(12^2+5^2=13^2\)\(\).

Thus FG  = 13.

I found a similar thread.

https://web2.0calc.com/questions/cos-3-4-where-3-2-tan-3-4-where-0-2-what-is-the-exact_2

Since the diameter of the smaller circle is 2 units the radius is 1 unit.
We know that the radius of the larger circle is 4 times the radius of the smaller circle the radius of the larger circle is 4 units.
Notice that the area of the gray region is equal to the larger circle minus the smaller one.
Since the area of a circle is  \(\pi \cdot r^2\) we can find the area of the gray area by substituting.
\(\pi \cdot 4^2 - \pi \cdot 1^2 = 16\pi - \pi = 15\pi\).

The bottom block is 12*9*2 = 216 \(cm^3\).

The top bock is 10*9*8 = 720 \(cm^3\).

Thus the total volume is 720+216 = 936 \(cm^3\)

Thanks for the answer, but I really want to know how to solve it.