Tetrahedral prism - rhomboid base

Calculate the area and volume tetrahedral prism that has base rhomboid shape and its dimensions are:
a = 12 cm, b = 70 mm, v_a = 6 cm, v_h = 1 dm.

Result

S =  5.24 dm2
V =  0.72 dm3

Solution:

a=12/10=65=1.2 b=70/100=710=0.7 v=6/10=35=0.6 h=1 S1=a⋅ v=1.2⋅ 0.6=1825=0.72 S2=2⋅ (a+b)⋅ h=2⋅ (1.2+0.7)⋅ 1=195=3.8 S=2⋅ S1+S2=2⋅ 0.72+3.8=13125=5.24 dm2a=12/10=\dfrac{ 6 }{ 5 }=1.2 \ \\ b=70/100=\dfrac{ 7 }{ 10 }=0.7 \ \\ v=6/10=\dfrac{ 3 }{ 5 }=0.6 \ \\ h=1 \ \\ S_{1}=a \cdot \ v=1.2 \cdot \ 0.6=\dfrac{ 18 }{ 25 }=0.72 \ \\ S_{2}=2 \cdot \ (a+b) \cdot \ h=2 \cdot \ (1.2+0.7) \cdot \ 1=\dfrac{ 19 }{ 5 }=3.8 \ \\ S=2 \cdot \ S_{1}+S_{2}=2 \cdot \ 0.72+3.8=\dfrac{ 131 }{ 25 }=5.24 \ \text{dm}^2
V=S1⋅ h=0.72⋅ 1=1825=0.72 dm3V=S_{1} \cdot \ h=0.72 \cdot \ 1=\dfrac{ 18 }{ 25 }=0.72 \ \text{dm}^3

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