Question 586404
Making a bertical line down the middle would split that front wall into two congruent (symmetrical, flipped) right triangles with a horizontal leg of length 5 m, and a hypotenuse of leg 13 m.
The height of the front wall is the length of the other leg. Let's call that height in meters, h.
{{{h^2+5^2=13^2}}} --> {{{h^2+25=169}}} --> {{{h^2=169-25}}} --> {{{h^2=144}}} --> {{{h=12}}} 
So the surface area of the front wall, in square meters, is {{{A=10*12/2=60}}}
At $120 per square meter, it will cost ${{{120*60}}}=${{{highlight(7200)}}}