Question

What is the volume of a pyramid whose base is a square of area 36 and whose four faces are equilateral triangles?

Answer 1

Answer:

$V = 36\sqrt{2}$

Step-by-step explanation:

The volume of a pyramid is

$V = \dfrac{1}{3}Bh$

where B = area of the base, and h = height of the pyramid

The base is a square.

$A = s^2$

$s = \sqrt{A}$

$s = \sqrt{36}$

$s = 6$

The side of the base has length 6. Each face of the pyramid is an equilateral triangle with side of length 6.

An altitude of this triangle measures

$a = \sqrt{6^2 - 3^2}$

$a = \sqrt{27}$

The pyramid is formed by 4 equilateral triangles whose bases form a square. The tips of the faces meet at a single point on top. The vertical distance from that point to the center of the base is the height of the pyramid.

$h = \sqrt{(\sqrt{27})^2 - 3^2 }$

$h = \sqrt{27 - 9}$

$h = \sqrt{18}$

$h = 3 \sqrt{2}$

Volume of the pyramid:

$V = \dfrac{1}{3}Bh$

$V = \dfrac{1}{3}(36)(3\sqrt{2})$

$V = 36\sqrt{2}$