Joseph will need at least d. 516.3 square cm of wrapping paper to wrap the triangular prism.
Step-by-step explanation:
Step 1:
To determine the amount of paper required to wrap the triangular prism, we need to determine the surface area of the entire prism.
The prism consists of 2 similar rectangles and 3 rectangles.
According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Assume the unknown side is x cm long.
[tex]15^{2} = 10^{2} +x^{2} , x^{2} = 225-100=125.[/tex]
[tex]x=11.08[/tex] cm.
Step 2:
The area of a triangle [tex]= \frac{1}{2} (b)(h).[/tex]
Here the triangles have a base length of 10 cm and a height of 11.08 cm.
The area of the triangle [tex]= \frac{1}{2} (10)(11.08)= 55.4[/tex] square cm.
The area of 2 such triangles [tex]= 2(55.4) = 110.8[/tex] square cm.
Step 3:
There are three rectangles with a common length i.e. 12 cm.
The area of a rectangle [tex]= (l)(w).[/tex]
The area of the rectangle with a width of 15 cm [tex]= (12)(15) = 180[/tex].
The area of the rectangle with a width of 10 cm [tex]= (12)(10) = 120.[/tex]
The area of the rectangle with a width of 11.08 cm [tex]= (12)(11.08) = 132.96.[/tex]
Step 4:
The surface area of the entire prism is the sum of all 5 side areas.
The surface area of the prism [tex]= 110.8+180+120+132.96 = 543.76[/tex].
This is closest to option d which is 516.3 square cm.
