Regular quadrilateral pyramid

Find the volume and surface of a regular quadrilateral pyramid if the bottom edge is 45 cm long and the pyramid height is 7 cm.

Result

V =  4725 cm3
S =  4145.737 cm2

Solution:

a=45 cm h=7 cm S1=a2=452=2025 cm2 V=S1 h/3=2025 7/3=4725 cm3a=45 \ \text{cm} \ \\ h=7 \ \text{cm} \ \\ S_{1}=a^2=45^2=2025 \ \text{cm}^2 \ \\ V=S_{1} \cdot \ h/3=2025 \cdot \ 7/3=4725 \ \text{cm}^3
h2=h2+(a/2)2=72+(45/2)223.5637 cm S2=a h2/2=45 23.5637/2530.1842 cm2 S=S1+4 S2=2025+4 530.18424145.73694145.737 cm2h_{2}=\sqrt{ h^2+(a/2)^2 }=\sqrt{ 7^2+(45/2)^2 } \doteq 23.5637 \ \text{cm} \ \\ S_{2}=a \cdot \ h_{2} / 2=45 \cdot \ 23.5637 / 2 \doteq 530.1842 \ \text{cm}^2 \ \\ S=S_{1} + 4 \cdot \ S_{2}=2025 + 4 \cdot \ 530.1842 \doteq 4145.7369 \doteq 4145.737 \ \text{cm}^2



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