Question 1200556
.
Find, in simplest radical form, the shortest possible
distance in centimeters that could be traced from A to B
on the surface of this 12 cm x 16 cm x 36 cm block.
https://ibb.co/SPd5gsB
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
Unfold the parallelepiped/(the block) on a plane and draw a straight line 
connecting the points A and B on the unfold.


    +------------------------------------------------------------+
    |   The segment AB on the plane is the shortest distance.    |
    +------------------------------------------------------------+


On the unfold, you have right-angled triangle with the legs  (36+12) = 48 cm and 16 cm
and the hypotenuse AB.


Now it is easy to calculate the distance 

    d = {{{sqrt((36+12)^2+ 16^2)}}} = {{{sqrt(48^2 + 16^2)}}} = {{{sqrt(16^2*(3^2+1))}}} = {{{16*sqrt(10)}}}.


It is about  50.6 cm.


Compare it with the solution by the other tutor  {{{4*sqrt(97)+12}}} = 51.4 cm (approximately).
</pre>

Solved.