Square and Triangle

From the picture above, we can conclude that

Assume that the real $AB$ is $n$

$ABF$ Area $=$ $\frac {AB \times BF }{2}$

$25 = \frac {n \times 2n}{2}$

$25 = n^2$

$n = 5$

So $AB = 5$ and $BF = 10$ , now move to the next step

From the figure above, we also knew that Triangle $ABF$ is congruent with Triangle $ACE$ , so ;

$\frac{ AB}{BF} = \frac { AC }{CE}$

$\frac { 5 }{10 } = \frac { 15 } { CE }$

$CE = 30$

Now find the area of the triangle,

$\frac { 15 \times 30 }{2} = 225$

Now find the Triangular Prisms Surface Area

$AC \times h + CE \times h + AE \times h + 2 ( ACE$ Area $) =$ Surface Area

$15 \times 15 + 30 \times 15 + \sqrt { 15^2 + 30 ^2 } \times 15 + 2 ( 225 )=$ Surface Area

$1628. 115 ... =$ Surface Area

By rounding the answer to zero decimal

$\color{#20A900}{\boxed {1628}} =$ Surface Area