![16\sqrt{3}](/tpl/images/0643/0998/b59cf.png)
![cm^{3}](/tpl/images/0643/0998/61c31.png)
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
As we know, The volume of a pyramid =
base area × its height
Given:
Side lenght of the base is;
![4\sqrt{3} cm](/tpl/images/0643/0998/64d0a.png)
=> The area of the base is ![12\sqrt{3}](/tpl/images/0643/0998/d480e.png)
![cm^{2}](/tpl/images/0643/0998/82289.png)
In Δ ACB measure of angle ACB is 90° and m∠CAB is 30°
We use: ![tan(30) = \frac{BC}{AC}](/tpl/images/0643/0998/308fc.png)
<=> BC = ![4\sqrt{3}*tan(30)](/tpl/images/0643/0998/5f173.png)
= 4 cm
And BC is the height of the the pyramid
=> The volume of a pyramid = ![\frac{1}{3}](/tpl/images/0643/0998/5506e.png)
![12\sqrt{3}](/tpl/images/0643/0998/d480e.png)
* 4 cm
= ![16\sqrt{3}](/tpl/images/0643/0998/b59cf.png)
![cm^{3}](/tpl/images/0643/0998/61c31.png)
![What is the volume of the pyramid? 12StartRoot 3 EndRoot cm3 16StartRoot 3 EndRoot cm3 24StartRoot 3](/tpl/images/0643/0998/647a8.jpg)