This doesnt compute. I'm finding the area of a paralleogram thats divided into a rectangle and 2 rtriangle. the base of the right triangle w/o the rectangle is 13 ft and in the left corner with the hypotenuse and the base is 17 degrees. I can't find the height or hypotenuse, it doesnt compute.
+118723 $$\\tan 17^0 = \frac{h}{13}\\\\
13\times tan17^0=h$$
$${\mathtt{13}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{17}}^\circ\right)} = {\mathtt{3.974\: \!498\: \!858\: \!967}}$$
So the height is 3.97 feet correct to 2 dec places.
you can find the slant height using the pythagorean theorem. 
Why do you have 2 right triangles? If you cut off a right triangle from the end of a parallelogram, would you not have just 1 triangle and one rectangle.
Thr height of the figure is easy enough to find, and from there the area of the triangle, but don't you need to tell us the length of the rectangle ?
+118723 $$\\tan 17^0 = \frac{h}{13}\\\\
13\times tan17^0=h$$
$${\mathtt{13}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{17}}^\circ\right)} = {\mathtt{3.974\: \!498\: \!858\: \!967}}$$
So the height is 3.97 feet correct to 2 dec places.
you can find the slant height using the pythagorean theorem.
