A chocolate dessert is in the shape of a square pyramid. The pyramid is represented with a net. How many square inches of icing are needed for the faces of the dessert? (1 point) Responses 39 in.2 39 in. squared 30 in.2 30 in. squared 69 in.2 69 in. squared 60 in.2
The net of a square pyramid includes a square base and 4 triangular faces.
Let's assume the base of the square pyramid has sides of length x.
The area of the square base = x^2
Each triangular face is an isosceles triangle with base x and height equal to the slant height of the pyramid.
The slant height, l, of the pyramid can be found using the Pythagorean theorem:
l^2 = (x/2)^2 + h^2
l^2 = x^2 / 4 + h^2
The height, h, of the pyramid can be found using the Pythagorean theorem in the same way:
l^2 = (x/2)^2 + h^2
h^2 = l^2 - (x^2 / 4)
The area of each triangular face = (1/2) * x * h
Total area of the 4 triangular faces = 4 * (1/2) * x * h = 2 * x * h
Given that the net represents a square pyramid, the square base area should be equal to the total area of the 4 triangular faces:
x^2 = 2 * x * h
x = 2h
Substitute x = 2h back into the Pythagorean theorem for l:
l^2 = x^2 / 4 + h^2
l^2 = (2h)^2 / 4 + h^2
l^2 = 4h^2 / 4 + h^2
l^2 = h^2 + h^2
l = sqrt(2)h
Area of the square base = x^2 = (2h)^2 = 4h^2
Total icing needed = area of the squares + area of the 4 triangles = 4h^2 + 2 * (x * h)
Total icing needed = 4h^2 + 2 * ((2h) * h) = 4h^2 + 4h^2 = 8h^2
Therefore, the total square inches of icing needed for the faces of the dessert is 8 times the height squared. Thus, the square inches of icing needed for the faces of the dessert would be 8 times h^2. Since the specific height of the pyramid is not provided, the area of the icing can't be determined without the height.