The volume of a Pyramid
Given a pyramid of base area A and height H, the volume is calculated as:
[tex]V=\frac{A\cdot H}{3}[/tex]
The base of this pyramid is a right triangle, with a hypotenuse of c=19.3 mm and one leg of a=16.8 mm. The other leg can be calculated by using the Pythagora's Theorem:
[tex]c^2=a^2+b^2[/tex]
Solving for b:
[tex]b^{}=\sqrt[]{c^2-a^2}=\sqrt[]{19.3^2-16.8^2}=9.5\operatorname{mm}[/tex]
The area of the base is the semi-product of the legs:
[tex]A=\frac{16.8\cdot9.5}{2}=79.8\operatorname{mm}^2[/tex]
Now the volume of the pyramid:
[tex]V=\frac{79.8\operatorname{mm}\cdot12\operatorname{mm}}{3}=319.2\operatorname{mm}^3[/tex]
The volume of the figure is 319.2 cubic millimeters
