Answer:
[tex]16\sqrt{3}[/tex] [tex]cm^{3}[/tex]
Step-by-step explanation:
I think your question missed key information, allow me to add in and hope it will fit the orginal one. Please have a look at the attached photo,
A solid oblique pyramid has an equilateral triangle as a base with an edge length of 4StartRoot 3 EndRoot cm and an area of 12StartRoot 3 EndRoot cm2.
What is the volume of the pyramid?
My answer:
As we know, The volume of a pyramid = [tex]\frac{1}{3}[/tex]base area × its height
Given:
- Side lenght of the base is; [tex]4\sqrt{3} cm[/tex]
=> The area of the base is [tex]12\sqrt{3}[/tex] [tex]cm^{2}[/tex]
- In Δ ACB measure of angle ACB is 90° and m∠ CAB is 30°
We use: [tex]tan(30) = \frac{BC}{AC}[/tex]
<=> BC = [tex]4\sqrt{3}*tan(30)[/tex]
= 4 cm
And BC is the height of the the pyramid
=> The volume of a pyramid = [tex]\frac{1}{3}[/tex] [tex]12\sqrt{3}[/tex] [tex]cm^{2}[/tex] * 4 cm
= [tex]16\sqrt{3}[/tex] [tex]cm^{3}[/tex]
