## anonymous one year ago I need help!! Find the volume for the regular pyramid. V=

1. anonymous

2. Michele_Laino

are all sides 4 units long? right?

3. anonymous

yes

4. Michele_Laino

we have to understand where is the intersection point P|dw:1438359138429:dw| between the height of your pyramid and its base. Have you any idea about that?

5. Michele_Laino

I think that P is the common point of the three bisectors of the triangle

6. anonymous

i have no clue

7. Michele_Laino

|dw:1438359407990:dw| it is the base triangle

8. Michele_Laino

|dw:1438359464896:dw|

9. anonymous

oh.....

10. anonymous

Still clueless lol

11. Michele_Laino

we have to find the measure of the height of your pyramid

12. anonymous

oh ok

13. anonymous

would it be 8

14. Michele_Laino

I don't know, in order to find the height of yopur pyramid I have to find this length: |dw:1438359730025:dw| namely the distance between each vertex of the triangle and the point P

15. anonymous

oh ummmm is it 4

16. anonymous

because it looks like it may be 4

17. Michele_Laino

we have to compute it first

18. anonymous

ok

19. Michele_Laino

I'm pondering...

20. anonymous

lol okay me too

21. Michele_Laino

There exists a theorem which states that the distance x is given by the subsequent formula: $\Large x = \frac{{4\sqrt 3 }}{3}$

22. anonymous

thats okay i finished the lesson

23. Michele_Laino

so, in order to compute the height of your pyramid we can apply the theorem of Pitagora: |dw:1438360380159:dw|

24. Michele_Laino

there fore we can write this: $\Large h = \sqrt {{4^2} - {x^2}} = \sqrt {16 - \frac{{16 \times 3}}{9}} = ...?$

25. Michele_Laino

oops..therefore*

26. Michele_Laino

hint: $\Large h = \sqrt {{4^2} - {x^2}} = \sqrt {16 - \frac{{16 \times 3}}{9}} = 4\sqrt {\frac{2}{3}}$ so being the base area A equal to: $\Large A = \frac{{4 \times 2\sqrt 3 }}{2} = 4\sqrt 3$ the requested volume V is: $\Large V = \frac{{A \times h}}{3} = \frac{1}{3} \times 4\sqrt 3 \times 4\sqrt {\frac{2}{3}} = ...?$