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:

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

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

$V = 36\sqrt{2}$

Answer 2

The volume of a pyramid whose base is a square of area 36 and whose four faces are equilateral triangles is; V = 36√2

The volume of a pyramid is given by the formula;

$v = \frac{ah}{3}$

• where a = area of the base,

• h = height of the pyramid

The length of each side of the base is a square can be evaluated as follows;

$l = \sqrt{a}$

• l = √36

• l = 6

Since, each side of the base has length 6.

Consequently, each face of the pyramid is an equilateral triangle with side of length 6.

An altitude (a bisector) of the equilateral triangle can be evaluated as follows;

$c = \sqrt{ {6}^{2} - \: {3}^{2} }$

• c = √27

In essence, the vertical distance from the vertex of the pyramid to the center of the base is the height of the pyramid and is evaluated as follows;

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

• h = √(27 - 9)

• h = 3√2

Ultimately, the Volume, V of the pyramid is;

• V = ah/3

$v = \frac{(36 \times 3 \sqrt{2} )}{3}$

• V = 36√2.

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