1. Surface area:
We need to split trinagular prism into 5 different polygons. Then calculate the area for each of them, and finally add up the five results to find the total surface area.
We have 2 triangles and 3 rectangles.
Both triangles are equal, so, we only need to find the area for one of them:
- Triangle:
[tex]A_t\text{ = }\frac{b\cdot h}{2}[/tex]We know h (h = 5 in), but we don't know b, and here is where the Pythagorean theorem comes in:
[tex]hypotenuse^2=side^2_1+side^2_2[/tex]For our trinagle, we know the hypotenuse length (13 in) and one side length (5 in), so:
[tex]\begin{gathered} \text{missing}_{}\text{ side = }\sqrt[]{hypotenuse^2-side^2_1} \\ \text{missing}_{}\text{ side = }\sqrt[]{13^2-5^2} \\ \text{missing}_{}\text{ side = }\sqrt[]{169\text{ - 25}}\text{ = }\sqrt[]{144}\text{ = 12} \end{gathered}[/tex]And so, going back to the triangle area, we have:
[tex]A_t\text{ = }\frac{b\cdot h}{2}\text{ = }\frac{12\cdot5}{2}=30in^2[/tex]- "Front" rectangle area:
[tex]A_r\text{ = }length\cdot\text{ width = 13 }\cdot20=260in^2[/tex]- "Right" rectangel area:
[tex]A_r\text{ = }length\cdot\text{ width = 20 }\cdot5=100in^2[/tex]- "Left" rectangle area:
[tex]A_r\text{ = }length\cdot\text{ width = 20 }\cdot12=240in^2\text{ (12 comes from Pythagorean theorem result)}[/tex]Finally, the surface area of the triangle prism is:
30 + 30 + 260 + 100 + 240 = 660 in^2
2. Volume
The volume is just the area of the base triangle times the height of the prism. So:
[tex]V=A_t\cdot\text{ h = 30 }\cdot20=600in^3[/tex]