A rectangular pyramid is sliced so the cross section is perpendicular to its base but does not pass through its vertex.What is the shape of the cross section?squarerectangletrapezoidtriangleA sphere is sliced so that the cross section does not intersect the center of the sphere.What is the shape of the cross section?circlesemicirclesquarerectangleWhat is the area of a cross section that is parallel to face CDHG ?Enter your answer in the box. ( picture)What is the area of the two-dimensional cross section that is parallel to face ABC ?Enter your answer in the box. ( picture)This rectangular prism is intersected by a plane that contains points D, E, K, and L.What is the perimeter of the cross section?Enter your answer in the box. Round only your final answer to the nearest tenth. (picture)
Accepted Solution
A:
Problem 1)
A trapezoid forms. See "figure1" in the attachments to get a visual idea of what I mean.
=================================================================== Problem 2)
We get a circle that forms but this circle is has a smaller radius compared to the radius of the sphere. See figure 2 (attached)
=================================================================== Problem 3)
Area of bottom face CDHG = length*width = 36*12 = 432
Any face parallel to this will have the same area. The same applies to any parallel cross section.
Answer: 432 square cm
=================================================================== Problem 4)
EF = 12 ft which runs parallel to BC, so BC = 12 ft as well
area of triangle ABC = (0.5)*(base)*(height) area of triangle ABC = (0.5)*(BC)*(AB) area of triangle ABC = (0.5)*(12)*(5) area of triangle ABC = 30
Any cross section parallel to face ABC will also have the same area.
Answer: 30 square feet
=================================================================== Problem 5)
Using the pythagorean theorem we find that EK is roughly 6.40312 meters long
a^2 + b^2 = c^2 4^2 + 5^2 = c^2 c^2 = 41 c = sqrt(41) c = 6.40312
LD = EK as they are parallel DE = LK for the same reason
DE = 12 EK = 6.40312 LK = 12 LD = 6.40312
The perimeter of plane DEKL is roughly P = (DE) + (EK) + (KL) + (LD) P = (12) + (6.40312) + (12) + (6.40312) P = 36.80624 P = 36.8