Answer :
The area of a triangular pyramid is given by [tex] V =\frac{1}{3} Ah[/tex]
Where A = area of the base, and h = height.
Knowing this, the area of your figure can be calculated by:
[tex] \frac{1}{3} * ( \frac{17*10}{2} ) *20[/tex]
=
[tex] \frac{1}{3} * 85 *20[/tex]
=
[tex] \frac{1}{3} * 1700[/tex]
=
[tex]566.(6)[/tex] reoccuring.
Or 567in^3
Where A = area of the base, and h = height.
Knowing this, the area of your figure can be calculated by:
[tex] \frac{1}{3} * ( \frac{17*10}{2} ) *20[/tex]
=
[tex] \frac{1}{3} * 85 *20[/tex]
=
[tex] \frac{1}{3} * 1700[/tex]
=
[tex]566.(6)[/tex] reoccuring.
Or 567in^3
V = 1/3AH...where A is the the area of triangle base and H is height
lets do area of base....area of triangle is (HB)/2...so the area of the base is
(10* 17) / 2 = 170/2 = 85....so in the main formula, A = 85
V = 1/3AH......A = 85 and H = 20
V = 1/3(85 * 20)
V = 1/3(1700)
V = 1700/3
V = 566.6 rounds to 567 in^3 <==
lets do area of base....area of triangle is (HB)/2...so the area of the base is
(10* 17) / 2 = 170/2 = 85....so in the main formula, A = 85
V = 1/3AH......A = 85 and H = 20
V = 1/3(85 * 20)
V = 1/3(1700)
V = 1700/3
V = 566.6 rounds to 567 in^3 <==
