A pyramid V-ABCD is cut from a cube of edge 12 in.,as shown in the figure. The
vertex V is the midpoint of an upper edge of the cube. Compute the lateral surface of the pyramid.
334.82 sq. in.
A pyramid V-ABCD is cut from A cube
oh your answer is 334.82 sq. in.
301.6 in^2
To compute the lateral surface of the pyramid, we need to find the area of its four triangular faces.
First, let's find the height of the pyramid. Since the vertex V is the midpoint of an upper edge of the cube, the height of the pyramid, VD, would be half of the cube's edge.
Height (VD) = 12 in / 2 = 6 in
Now, let's find the base of each triangular face. The base length of each triangular face of the pyramid is the same as the edge length of the cube, which is given as 12 in.
Base length (AB) = 12 in
Given that the base is an equilateral triangle in this case, we can use the formula for the area of an equilateral triangle, which is:
Area of equilateral triangle = (sqrt(3) / 4) * (side length)^2
Area of triangular face (ABCDE) = (sqrt(3) / 4) * (12 in)^2
Area of triangular face (ABCDE) = (sqrt(3) / 4) * (144 in^2)
Now, since there are four triangular faces, the total lateral surface area of the pyramid is:
Lateral surface area = 4 * Area of triangular face (ABCDE)
Lateral surface area = 4 * (sqrt(3) / 4) * (144 in^2)
Lateral surface area = 4 * sqrt(3) * (36 in^2)
Lateral surface area = 144 * sqrt(3) in^2
So, the lateral surface area of the pyramid is 144 * sqrt(3) square inches.
draw the figure. Note that it is relatively easy to find the slope height of each face, by using the slope height as the hypotenuse of a right trangle whose legs are distances to a opposite corners.
area of each face then can be determined by Area face= 1/2 b*slope height.
add the four faces areas.