The net surface area of the is equal to the sum of the area of the all the figure.The net surface area of the given right triangular prism is 209 squared cm.
Given information-
The width of the rectangle is 7 feet.
The length of the 3rd rectangle is 10 feet.
The base of the right angle triangle is 6 feet.
The height of the right angle triangle is 6 feet.
Area of the triangle
Area of the triangle half of the product of the height and the base. Thus the area of the triangle is,
[tex]A=\dfrac{1}{2} \times6\times8[/tex]
[tex]A=24[/tex]
Thus the area of the triangle is 24 squared feet.
As the length of the first rectangle is equal to the third one for the triangular prism which is equal to the 10 cm.
Area of the rectangle
The area of the rectangle is equal to the product of the length and width. Thus,
[tex]A=10\times7\\A=70[/tex]
The length of the middle rectangle is equal to the base of the triangle which is equal to the 3 cm. Thus the area of the middle rectangle,
[tex]A=3\times7\\
A=21[/tex]
Now the net surface area is equal to the area of all the 2 similar triangle, 2 similar rectangle and one middle rectangle. Thus the net surface area is,
[tex]A_n=2\times24+2\times70+21\\
A_n=48+140+21\\
A_n=209[/tex]
Thus the net surface area of the given right triangular prism is 209 squared cm.
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