Answer:
Volume = 120.7 units cubed
Lateral Area = 110.9 units squared
Surface Area = 174.9 units squared
Step-by-step explanation:
The formula for the volume of a square-based pyramid is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Volume of a Square-based Pyramid}}\\\\V =\dfrac{1}{3}s^2\sqrt{l^2-\dfrac{s^2}{2}}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$V$ is the volume.}\\\phantom{ww}\bullet\;\textsf{$s$ is the length of the base edge.}\\\phantom{ww}\bullet\;\textsf{$l$ is the length of the lateral edge.}\end{array}}[/tex]
in this case, all the edges of the pyramid have length 8, so:
Substitute these values into the formula and solve for volume:
[tex]V =\dfrac{1}{3}\cdot 8^2\sqrt{8^2-\dfrac{8^2}{2}}\\\\\\\\V =\dfrac{1}{3}\cdot 64\sqrt{64-32}\\\\\\\\V =\dfrac{64\sqrt{32}}{3}\\\\\\V=120.67955732...\\\\\\V=120.7\; \sf units\;cubed[/tex]
Therefore, the volume of the pyramid is 120.7 units cubed, rounded to the nearest tenth.
[tex]\dotfill[/tex]
The lateral area of a pyramid is the total surface area of all the triangular faces of the pyramid, excluding the base area.
Given that all the edges of the pyramid have length 8, each triangular face is an equilateral triangle.
The formula for the area of an equilateral triangle with side length 's' is:
[tex]A=\dfrac{\sqrt{3}}{4}s^2[/tex]
Therefore, the lateral area (LA) of the pyramid is:
[tex]LA=4 \cdot \dfrac{\sqrt{3}}{4}\cdot 8^2\\\\\\LA=\sqrt{3}\cdot 64\\\\\\LA=64\sqrt{3}\\\\\\LA=110.85125168...\\\\\\LA=110.9\; \sf units\;squared[/tex]
So, the lateral area of the pyramid is 110.9 units squared, rounded to the nearest tenth.
[tex]\dotfill[/tex]
The total surface area of a pyramid is composed of the lateral area and the area of the base. Given that all the edges of the pyramid have length 8, the base is a square with side length s = 8. Therefore:
[tex]SA=LA+s^2\\\\\\SA=64\sqrt{3}+8^2\\\\\\SA=64\sqrt{3}+64\\\\\\SA=174.851251684...\\\\\\SA=174.9 \; \sf units\;squared[/tex]
So, the total surface area of the pyramid is 174.9 units squared, rounded to the nearest tenth.