+1348 I'm stuck on this
A right pyramid has a square base. The area of each triangular face is one-third the area of the square face. If the total surface area of the pyramid is $432$ square units, then what is the volume of the pyramid in cubic units?
+129895 Surface area = b^2 + 4 * ( b^2 /3)
432 = (7/3)b^2
b^2 = 432 ( 3/7) = 1296 / 7
b = sqrt (1296/ 7) = 36 /sqrt 7
To find the slant height, s, of the pyramid
area of face = 1/2 b * s
b^2 / 3 = (1/2) 36 / sqrt 7 * s
1296 / (7*3) = 18 / sqrt 7 * s
s = 1296 / 21 * sqrt 7 / 18 = 24 / sqrt 7
To find the height, h, of the pyramid by the Pythagorean Theorem
h = sqrt ( s^2 - (b/2)^2 )
h = sqrt [ (24/sqrt 7)^2 - (18/sqrt 7)^2 ]
h = 6
Volume of pyramid =
(1/3) base area * height =
(1/3) (b^2) * h
(1/3) (1296 / 7) * 6 = 2592 / 7 units^3 ≈ 370.29 units^3
