Truncated pyramid

How many cubic meters is volume of a regular four-side truncated pyramid with edges one meter and 60 cm and high 250 mm?

Correct result:

V =  0.163 m3

Solution:

a1=1 m a2=60/100=35=0.6 m h=250/1000=14=0.25 m  S1=a12=12=1 m2 S2=a22=0.62=925=0.36 m2  V=h/3⋅ (S1+S1⋅ S2+S2)=0.25/3⋅ (1+1⋅ 0.36+0.36)=49300=0.163 m3a_{1}=1 \ \text{m} \ \\ a_{2}=60/100=\dfrac{ 3 }{ 5 }=0.6 \ \text{m} \ \\ h=250/1000=\dfrac{ 1 }{ 4 }=0.25 \ \text{m} \ \\ \ \\ S_{1}=a_{1}^{ 2 }=1^{ 2 }=1 \ \text{m}^2 \ \\ S_{2}=a_{2}^{ 2 }=0.6^{ 2 }=\dfrac{ 9 }{ 25 }=0.36 \ \text{m}^2 \ \\ \ \\ V=h/3 \cdot \ (S_{1}+\sqrt{ S_{1} \cdot \ S_{2} }+S_{2})=0.25/3 \cdot \ (1+\sqrt{ 1 \cdot \ 0.36 }+0.36)=\dfrac{ 49 }{ 300 }=0.163 \ \text{m}^3

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