Resoluçao:
- 20171003_220458-1.jpg (164.12 KiB) Exibido 6251 vezes
Teorema de Pitágoras([tex3]\Delta [/tex3]
destacado)
[tex3]h^{2}+2^{2}=(3\sqrt{2})^{2}[/tex3]
[tex3]h^{2}+4=18\rightarrow h=\sqrt{14}[/tex3]
Area da base:
[tex3]A_{b}=\frac{(B+b)h}{2}=\frac{(8+6)\sqrt{14}}{2}=\frac{14.\sqrt{14}}{2}=7\sqrt{14}[/tex3]
Volume da piramide:
[tex3]V=\frac{1}{3}A_{b}.h=\frac{1}{3}.7\sqrt{14}.9=21\sqrt{14}[/tex3]
[tex3]\therefore V=21\sqrt{14}[/tex3]