The quadrilateral pyramid

The quadrilateral pyramid has a rectangular base of 24 cm x 3.2dm and a body height of 0.4m. Calculate its volume and surface area.

Result

V =  10.24 dm3
S =  31.383 dm2

Solution:

a=24 cm=24/10 dm=2.4 dm b=3.2 dm h=0.4 m=0.4 10 dm=4 dm  S1=a b=2.4 3.2=19225=7.68 dm2  V=13 S1 h=13 7.68 4=25625=10.24 dm3a=24 \ cm=24 / 10 \ dm=2.4 \ dm \ \\ b=3.2 \ \text{dm} \ \\ h=0.4 \ m=0.4 \cdot \ 10 \ dm=4 \ dm \ \\ \ \\ S_{1}=a \cdot \ b=2.4 \cdot \ 3.2=\dfrac{ 192 }{ 25 }=7.68 \ \text{dm}^2 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 7.68 \cdot \ 4=\dfrac{ 256 }{ 25 }=10.24 \ \text{dm}^3
s2=h2+(a/2)2=42+(2.4/2)24.1761 dm s3=h2+(b/2)2=42+(3.2/2)24.3081 dm  S2=a s32=2.4 4.308125.1698 dm2 S3=b s22=3.2 4.176126.6818 dm2  S=S1+2 S2+2 S3=7.68+2 5.1698+2 6.681831.383131.383 dm2s_{2}=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 4^2 + (2.4/2)^2 } \doteq 4.1761 \ \text{dm} \ \\ s_{3}=\sqrt{ h^2 + (b/2)^2 }=\sqrt{ 4^2 + (3.2/2)^2 } \doteq 4.3081 \ \text{dm} \ \\ \ \\ S_{2}=\dfrac{ a \cdot \ s_{3} }{ 2 }=\dfrac{ 2.4 \cdot \ 4.3081 }{ 2 } \doteq 5.1698 \ \text{dm}^2 \ \\ S_{3}=\dfrac{ b \cdot \ s_{2} }{ 2 }=\dfrac{ 3.2 \cdot \ 4.1761 }{ 2 } \doteq 6.6818 \ \text{dm}^2 \ \\ \ \\ S=S_{1} + 2 \cdot \ S_{2} + 2 \cdot \ S_{3}=7.68 + 2 \cdot \ 5.1698 + 2 \cdot \ 6.6818 \doteq 31.3831 \doteq 31.383 \ \text{dm}^2



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