Quadrangular pyramid

The regular quadrangular pyramid has a base length of 6 cm and a side edge length of 9 centimeters. Calculate its volume and surface area.

Result

V =  95.247 cm3
S =  137.823 cm2

Solution:

a=6 cm s=9 cm  S1=a2=62=36 cm2  u=2 a=2 6=6 2 cm8.4853 cm  h=s2(u/2)2=92(8.4853/2)2=3 7 cm7.9373 cm  V=13 S1 h=13 36 7.9373=36 795.247=95.247 cm3a = 6 \ cm \ \\ s = 9 \ cm \ \\ \ \\ S_{ 1 } = a^2 = 6^2 = 36 \ cm^2 \ \\ \ \\ u = \sqrt{ 2 } \cdot \ a = \sqrt{ 2 } \cdot \ 6 = 6 \ \sqrt{ 2 } \ cm \doteq 8.4853 \ cm \ \\ \ \\ h = \sqrt{ s^2 - (u/2)^2 } = \sqrt{ 9^2 - (8.4853/2)^2 } = 3 \ \sqrt{ 7 } \ cm \doteq 7.9373 \ cm \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ S_{ 1 } \cdot \ h = \dfrac{ 1 }{ 3 } \cdot \ 36 \cdot \ 7.9373 = 36 \ \sqrt{ 7 } \doteq 95.247 = 95.247 \ cm^3
h2=s2(a/2)2=92(6/2)2=6 2 cm8.4853 cm  S2=a h2/2=6 8.4853/2=18 2 cm225.4558 cm2  S=S1+4 S2=36+4 25.4558137.8234=137.823 cm2h_{ 2 } = \sqrt{ s^2 - (a/2)^2 } = \sqrt{ 9^2 - (6/2)^2 } = 6 \ \sqrt{ 2 } \ cm \doteq 8.4853 \ cm \ \\ \ \\ S_{ 2 } = a \cdot \ h_{ 2 }/2 = 6 \cdot \ 8.4853/2 = 18 \ \sqrt{ 2 } \ cm^2 \doteq 25.4558 \ cm^2 \ \\ \ \\ S = S_{ 1 } + 4 \cdot \ S_{ 2 } = 36 + 4 \cdot \ 25.4558 \doteq 137.8234 = 137.823 \ cm^2



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