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:

$a=6 \ \text{cm} \ \\ s=9 \ \text{cm} \ \\ \ \\ S_{1}=a^2=6^2=36 \ \text{cm}^2 \ \\ \ \\ u=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 6 \doteq 6 \ \sqrt{ 2 } \ \text{cm} \doteq 8.4853 \ \text{cm} \ \\ \ \\ h=\sqrt{ s^2 - (u/2)^2 }=\sqrt{ 9^2 - (8.4853/2)^2 } \doteq 3 \ \sqrt{ 7 } \ \text{cm} \doteq 7.9373 \ \text{cm} \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 36 \cdot \ 7.9373 \doteq 36 \ \sqrt{ 7 } \doteq 95.247 \doteq 95.247 \ \text{cm}^3$
$h_{2}=\sqrt{ s^2 - (a/2)^2 }=\sqrt{ 9^2 - (6/2)^2 } \doteq 6 \ \sqrt{ 2 } \ \text{cm} \doteq 8.4853 \ \text{cm} \ \\ \ \\ S_{2}=a \cdot \ h_{2}/2=6 \cdot \ 8.4853/2 \doteq 18 \ \sqrt{ 2 } \ \text{cm}^2 \doteq 25.4558 \ \text{cm}^2 \ \\ \ \\ S=S_{1} + 4 \cdot \ S_{2}=36 + 4 \cdot \ 25.4558 \doteq 137.8234 \doteq 137.823 \ \text{cm}^2$

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