Pyramid 8

Calculate the volume and the surface area of a regular quadrangular pyramid with the base side 9 cm and side wall with the base has an angle 75°.

Result

V =  641.27 cm3
S =  516.117 cm2

Solution:

a=9 cm S1=a2=92=81 cm2 s=a/2=9/2=92=4.5 cm h=2 s tan(75 rad)=2 s tan(75 π180 )=2 4.5 tan(75 3.1415926180 )=23.75063 h2=h2+s2=23.75062+4.5224.1732 cm  V=13 S1 h=13 81 23.7506641.2669641.27 cm3a=9 \ \text{cm} \ \\ S_{1}=a^2=9^2=81 \ \text{cm}^2 \ \\ s=a/2=9/2=\dfrac{ 9 }{ 2 }=4.5 \ \text{cm} \ \\ h=\sqrt{ 2 } \cdot \ s \cdot \ \tan ( 75 ^\circ \rightarrow\ \text{rad})=\sqrt{ 2 } \cdot \ s \cdot \ \tan ( 75 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\sqrt{ 2 } \cdot \ 4.5 \cdot \ \tan ( 75 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=23.75063 \ \\ h_{2}=\sqrt{ h^2 + s^2 }=\sqrt{ 23.7506^2 + 4.5^2 } \doteq 24.1732 \ \text{cm} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 81 \cdot \ 23.7506 \doteq 641.2669 \doteq 641.27 \ \text{cm}^3
S2=a h2/2=9 24.1732/2108.7793 cm2  S=S1+4 S2=81+4 108.7793516.1171516.117 cm2S_{2}=a \cdot \ h_{2} / 2=9 \cdot \ 24.1732 / 2 \doteq 108.7793 \ \text{cm}^2 \ \\ \ \\ S=S_{1} + 4 \cdot \ S_{2}=81 + 4 \cdot \ 108.7793 \doteq 516.1171 \doteq 516.117 \ \text{cm}^2



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