3 Ways To Calculate The Volume Of A Pyramid

Remember, V=13lwh=13Abh{\displaystyle V={\frac {1}{3}}lwh={\frac {1}{3}}A_{b}h}, so you need to know l{\displaystyle l} and w{\displaystyle w} first. l=4cm{\displaystyle l=4,{\text{cm}}} w=3cm{\displaystyle w=3,{\text{cm}}}

Remember, V=13Abh{\displaystyle V={\frac {1}{3}}A_{b}h}, so you need to know Ab{\displaystyle A_{b}}. You can find this by plugging in l=4cm{\displaystyle l=4,{\text{cm}}} and w=3cm{\displaystyle w=3,{\text{cm}}} from the previous step. Ab=lw{\displaystyle A_{b}=lw} Ab=(4cm)(3cm)=12cm2{\displaystyle A_{b}=(4,{\text{cm}})(3,{\text{cm}})=12,{\text{cm}}^{2}}

Remember, V=13Abh{\displaystyle V={\frac {1}{3}}A_{b}h}, so you need to know Abh{\displaystyle A_{b}h}. You can find this using Ab{\displaystyle A_{b}} from the previous step. Ab=12cm2{\displaystyle A_{b}=12,{\text{cm}}^{2}} h=4cm{\displaystyle h=4,{\text{cm}}} Abh=(12cm2)(4cm)=48cm3{\displaystyle A_{b}h=(12,{\text{cm}}^{2})(4,{\text{cm}})=48,{\text{cm}}^{3}}

Remember, V=13lwh=13Abh{\displaystyle V={\frac {1}{3}}lwh={\frac {1}{3}}A_{b}h}. You can plug in Abh=48cm3{\displaystyle A_{b}h=48,{\text{cm}}^{3}} from the previous step. V=13Abh{\displaystyle V={\frac {1}{3}}A_{b}h} V=(13)(48cm3)=16cm3{\displaystyle V=({\frac {1}{3}})(48,{\text{cm}}^{3})=16,{\text{cm}}^{3}}

If the length and width are not perpendicular and you don’t know the height of the triangle, there are a few other methods you can try to calculate the area of a triangle. Remember, V=13lwh=13Abh{\displaystyle V={\frac {1}{3}}lwh={\frac {1}{3}}A_{b}h}, so you need to know l{\displaystyle l} and w{\displaystyle w} first. l=width of pyramid base=base of triangle, orb=2cm{\displaystyle l={\text{width of pyramid base}}={\text{base of triangle, or}},b=2,{\text{cm}}} w=length of pyramid base=height of triangle, orh=4cm{\displaystyle w={\text{length of pyramid base}}={\text{height of triangle, or}},h=4,{\text{cm}}}

Remember, V=13lwh=13Abh{\displaystyle V={\frac {1}{3}}lwh={\frac {1}{3}}A_{b}h}, so you need to know Ab{\displaystyle A_{b}}. You can find this using b{\displaystyle b} and h{\displaystyle h} from the previous step. Ab=12bh{\displaystyle A_{b}={\frac {1}{2}}bh} Ab=(12)(2cm)(4cm){\displaystyle A_{b}=({\frac {1}{2}})(2,{\text{cm}})(4,{\text{cm}})} Ab=(12)(8cm2){\displaystyle A_{b}=({\frac {1}{2}})(8,{\text{cm}}^{2})} Ab=4cm2{\displaystyle A_{b}=4,{\text{cm}}^{2}}

Remember, V=13Abh{\displaystyle V={\frac {1}{3}}A_{b}h}, so you need to know Abh{\displaystyle A_{b}h}. You can find this using Ab{\displaystyle A_{b}} from the previous step. Ab=area of triangular base=4cm2{\displaystyle A_{b}={\text{area of triangular base}}=4,{\text{cm}}^{2}} h=height of pyramid=5cm{\displaystyle h={\text{height of pyramid}}=5,{\text{cm}}} Abh=(4cm2)(5cm)=20cm3{\displaystyle A_{b}h=(4,{\text{cm}}^{2})(5,{\text{cm}})=20,{\text{cm}}^{3}}

Remember, V=13lwh=13Abh{\displaystyle V={\frac {1}{3}}lwh={\frac {1}{3}}A_{b}h}. You can plug in Abh=20cm3{\displaystyle A_{b}h=20,{\text{cm}}^{3}} from the previous step. V=(13)Abh{\displaystyle V=({\frac {1}{3}})A_{b}h} V=(13)(20cm3)=6. 67cm3{\displaystyle V=({\frac {1}{3}})(20,{\text{cm}}^{3})=6. 67,{\text{cm}}^{3}}