SAT Math 2 subject test question?
A regular pyramid has a 4 inch by 4 inch square base and a height of 8 inches. What is the length, in inches, of an altitude of one of the triangular faces of the pyramid?
It would be great if someone helped. Thanks:)
- micatkieLv 72 months agoFavorite Answer
Refer to the figure below which refers to the pyramid.
AB = 8 in
BC = 4/2 in = 2 in
AC² = AB² + BC² (Pythagorean theorem)
AC = √(8² + 2²) in
Length of an altitude of one of the Δ faces, AC = 8.25 in
The answer: D) 8.25
- 2 months ago
Use pythagoras theorema a²+b²=c²
from the center to a side is 2 inch
side face altitude is 8.25 inch
- PopeLv 72 months ago
What you described would not be a regular pyramid, as only one of its five faces is a regular polygon. However, it could be a right pyramid. In that case, all lateral faces would be isosceles triangles. The distance from the pyramid vertex to the midpoint of one of the base edges would then be one altitude of a lateral face.
Let the pyramid vertex be V.
Let the center of the base be O.
Let M be the midpoint of one base edge.
Length VM is an altitude of a lateral face.
OV = 8 in
OM = 2 in
∠VOM = 90°
VM² = OV² + OM²
VM² = (8 in)² + (2 in)²
VM² = 68 in²
VM = √(68) in ≈ 8.25 in
- stanschimLv 72 months ago
A diagram should convince you that there is a right triangle whose hypotenuse is the desired altitude. The legs of this triangle are 2 and 8 inches. So, the Pythagorean Theorem gives √(2^2 + 8^2) = √68 = about 8.25 (D).