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danielmaduroh
The question is to find the surface area of the square pyramid:

This pyramid encloses a volume V, so it has five surfaces, namely:

*One square base.
*four triangular surfaces.

1. Surface area of the square base:

All four sides are equal, so:

As = LxL = (8.4yd)(8.4yd) = 70.56[tex]yd^2[/tex]

2. Area of triangular surfaces:

Given that the angle β=60°, we have four equilateral triangles which have three equal sides each, therefore:

At = [tex] 4(\frac{base.height}{2}) [/tex] = [tex]2L.Ap[/tex]

Now we need to calculate the apothem Ap.

Using trigonometry, we can calculate Ap using the sine function:

sinβ = [tex] \frac{opposite}{hypotenuse} [/tex] = [tex] \frac{Ap}{a} [/tex] 

Given that the triangles are equilateral, then a = L

∴ Ap = (a)(sinβ) = 8.4sin(60°) = (8.4)([tex] \frac{ \sqrt{3} }{2} [/tex]) = 7.275yd

So:

[tex]At = 2(8.4)(7.275) = 122.22 yd^{2}[/tex]

Therefore, the surface area of the pyramid is:

A = As + At = 70.56 + 122.22 = 192.78 [tex] yd^{2} [/tex]


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LilNil

The answer is 192.8 yd²

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