If w = 3 in and h = 4 in, what is the surface area of the right square pyramid? *Note: Picture not drawn to scale.

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If w = 3 in and h = 4 in, what is the surface area of the right square pyramid? *Note: Picture not drawn to scale.

Mathematics
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60 in2 111 in2 66 in2 96 in2 I was thinking C 66
Since you have been given w and h, you need to find a way to express s in terms of the other two known variables. This is the only way you will be able to get a numerical answer. The only trig relation you can make in this situation (since there are no angles) is to use the Pythagorean theorem. In this case, that would imply h^2 + w^2 = s^2 This implies that s = +/-sqrt(h^2 + w^2) But since we are dealing with distances and we don't care about the negative answer, we can simply say that s = sqrt(h^2 + w^2)

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so ?
Now, the surface area of the pyramid is going to be the sum of the area of the four triangles and the sum of the square base. The area of each triangle is going to be it's base times height which is Area of triangle = 2w*s = 2w*sqrt(h^2+w^2) And obviously, the square part is going to be Area of square = (2w)^2 = 4w^2 So, all you have to do is plug in the values.
so would it be C
do its D got it :D
Yeah it is D. My triangle area should be ws not 2ws, but I'm sure you figured that out.

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