Question

I need help on a question

Answer 1

we have that

The volume of a regular pyramid is equal to

V=(1/3)Bh

where

B is the area of the base

h is the height of the pyramid

step 1

Find out the area of the triangular face

Applying the law of sines

B=(1/2)(a)(b)sin(60)

we have

a=4

b=4

B=(1/2)(4)(4)sin(60)

B=8sin(60)

Remember that

`\sin (60^o)=\frac{\sqrt[]{3}}{2}`

substitute

`\begin{gathered} B=8(\frac{\sqrt[]{3}}{2}) \\ \\ B=4\sqrt[]{3\text{ }} \end{gathered}`

step 2

Find the height of piramid

see the attached figure to better understand the problem

Find h1

Applying Pythagorean Theorem

4^2=2^2+h1^2

h1^2=4^2-2^2

h1^2=12

h1=2√3

the center of triangle is located at (1/3)h1

c=(1/3)2√3

c=(2/3)√3

Find out the height of piramid

Apply Pythagorean Theorem

h1^2=h^2+c^2

substitute

(2√3)^2=h^2+((2/3)√3)^2

12=h^2+(4/3)

h^2=12-4/3

h^2=32/3

h=√(32/3)

Find the volume

V=(1/3)Bh

V=(1/3)(4√3)(√(32/3)

V=(4/3)(4√2)

V=(16/3)√2 unit3