Question

Surface area of a square pyramid with side length 6 yd and slant height 7 yd

Answer 1

Answer:

127.38

Step-by-step explanation:

The surface area formula for a square pyramid is:

A = a^2 + 2a * sqrt[ (a^2 / 4 ) + h^2 ]

So we input the base height as the a value, and the slant as H)

A = 6^2 + 2(6) * sqrt[ (6^2 / 4 ) + 7^2 ]

When using the exponents we get:

A = 36 + 2(6) * sqrt[ (36) / 4 ) + 49 ]

Then multiply/divide:

A = 36 + 12 * sqrt[ (9) + 49 ]

Then add the value in the sqrt.

A = 36 + 12 * sqrt[ 58 ]

Now, if we find the sqrt of 58, we get:     sqrt[ 58 ] ≅ 7.615

A = 36 + 12 * 7.615

When multiplying we get:

A = 36 + 91.38

Finally, after adding, the surface area of the square pyramid is:

A = 127.38

Answer 2

Check the picture below.

so the pyramid is really 4 triangular faces with a base of 6 and a height of 7, and a 6x6 square at the bottom.

$\stackrel{\textit{\LARGE Areas}}{\stackrel{\textit{four triangular faces}}{4\left[\cfrac{1}{2}(\stackrel{b}{6})(\stackrel{h}{7}) \right]}~~ + ~~\stackrel{square}{(6)(6)}}\implies 84~~ + ~~36\implies 120~yd^2$