Hexagonal pyramid

Regular hexagonal pyramid has dimensions: length edge of the base a = 1.8 dm and the height of the pyramid = 2.4 dm. Calculate the surface area and volume of a pyramid.

Correct result:

S =  23.872 dm2
V =  6.734 dm3

Solution:

a=1.8 dm v=2.4 dm h1=a⋅ 3/2=1.8⋅ 3/2≐1.5588 dm h2=v2+h12=2.42+1.55882≐2.8618 dm  S1=a⋅ h1/2=1.8⋅ 1.5588/2≐1.403 dm2 S2=6⋅ S1=6⋅ 1.403≐8.4178 dm2 S3=a⋅ h2/2=1.8⋅ 2.8618/2≐2.5756 dm2 S=S2+6⋅ S3=8.4178+6⋅ 2.5756=23.872 dm2a=1.8 \ \text{dm} \ \\ v=2.4 \ \text{dm} \ \\ h_{1}=a \cdot \ \sqrt{ 3 }/2=1.8 \cdot \ \sqrt{ 3 }/2 \doteq 1.5588 \ \text{dm} \ \\ h_{2}=\sqrt{ v^2+h_{1}^2 }=\sqrt{ 2.4^2+1.5588^2 } \doteq 2.8618 \ \text{dm} \ \\ \ \\ S_{1}=a \cdot \ h_{1}/2=1.8 \cdot \ 1.5588/2 \doteq 1.403 \ \text{dm}^2 \ \\ S_{2}=6 \cdot \ S_{1}=6 \cdot \ 1.403 \doteq 8.4178 \ \text{dm}^2 \ \\ S_{3}=a \cdot \ h_{2}/2=1.8 \cdot \ 2.8618/2 \doteq 2.5756 \ \text{dm}^2 \ \\ S=S_{2} + 6 \cdot \ S_{3}=8.4178 + 6 \cdot \ 2.5756=23.872 \ \text{dm}^2
V=13⋅ S2⋅ v=13⋅ 8.4178⋅ 2.4=6.734 dm3V=\dfrac{ 1 }{ 3 } \cdot \ S_{2} \cdot \ v=\dfrac{ 1 }{ 3 } \cdot \ 8.4178 \cdot \ 2.4=6.734 \ \text{dm}^3

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