Problem:
On a 4×4×3 rectangular parallelepiped, vertices A,B, and C are adjacent to vertex D. The perpendicular distance from D to the plane containing A,B, and C is closest to
Answer Choices:
A. 1.6
B. 1.9
C. 2.1
D. 2.7
E. 2.9
Solution:
Let h be the required distance. Find the volume of pyramid ABCD as a third of the area of a triangular base times the altitude to that base in two different ways, and equate these volumes. Use the altitude AD to △BCD to find that the volume is 8 . Next, note that h is the length of the altitude of the pyramid from D to △ABC. Since the sides of △ABC are 5,5 , and 42​, by the Pythagorean Theorem the altitude to the side of length 42​ is a=17​. Thus, the area of △ABC is 234​, and the volume of the pyramid is 234​h/3. Equating the volumes yields
234​h/3=8, and thus h=12/34​≈2.1.
OR
Imagine the parallelepiped embedded in a coordinate system as shown in the diagram. The equation for the plane (in intercept form) is 4x​+4y​+3z​=1. Thus, it can be expressed as 3x+3y+4z−12=0. The formula for the distance d from a point (a,b,c) to the plane Rx+Sy+Tz+U=0 is given by
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