Find the surface area of a square pyramid whose base edge is 6cm and whose slant edge is 5cm

so let's notice, the base is a 6x6 square, and triangular faces have a base of 6 and an altitude/height of 5. So we can just get the area of the square and the triangles and sum them up and that's the area of the pyramid.

[tex]\bf \stackrel{\textit{triangles' area}}{4\left[ \cfrac{1}{2}(6)(5) \right]}+\stackrel{\textit{square's area}}{(6\cdot 6)}\implies 60+36\implies 96[/tex]

Answer Link

carlosego carlosego · Answer 2 · 2019-05-23T00:25:32+02:00

For this case we have that by definition, the surface area of a regular pyramid, is given by:

[tex]SA = \frac {1} {2} p * l + B[/tex]

Where:

p: Represents the perimeter of the base

l: The inclination height

B: The area of the base

Now, since the base is square we have:

[tex]B = 6 ^ 2 = 36 \ cm ^ 2\\p = 6 + 6 + 6 + 6 = 24 \ cm\\l = 5 \ cm[/tex]

Then, replacing the values:

[tex]SA = \frac {1} {2} 24 * 5 + 36\\SA = 60 + 36\\SA = 96 \ cm ^ 2[/tex]

ANswer

[tex]96 \ cm ^ 2[/tex]