The lateral area of a right prism equals the perimeter of the base multiplied by the height of the prism. Let h be the height of both prisms.
L_(P_1) = P_(□)* h
and
L_(P_2) = P_(△)* h
The perimeter of the square equals 4 s and the perimeter of the triangle is 3 x.
L_(P_1) = 4 s h
and
L_(P_2) = 3 x h
Let's substitute x = 2sqrt(3) s3 into the lateral area above.
L_(P_1) = 4 s h
and
L_(P_2) = 3* 2sqrt(3) s3 * h
By simplifying the right-hand side of the equation, we will obtain both lateral areas in terms of the same variables.
L_(P_1) = 4 s h
and
L_(P_2) = 2sqrt(3) s h
Since 4> 2sqrt(3), we conclude that the lateral area of the square based prism is greater than the lateral area of the triangular prism.
L_(P_1) > L_(P_2)
