The length of a side of the base of a regular square pyramid is 7 in., and the pyramid's slant height is 16 in. Find the surface area of the pyramid.

[tex]\boxed{\begin{array}{l}\underline{\textsf{Surface Area of a Square-based Pyramid}}\\\\A = s^2 + 2sh\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A$ is the surface area.}\\\phantom{ww}\bullet\;\textsf{$s$ is the length of a side of the base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the slant height.}\end{array}}[/tex]

In this case:

[tex]s = 7\;\sf in[/tex]
[tex]h = 16\;\sf in[/tex]

Substitute the given values into the formula:

[tex]A = (7)^2 + 2 \cdot 7 \cdot 16[/tex]

[tex]A = 49 + 224[/tex]

[tex]A = 273\; \sf in^2[/tex]

So, the surface area of the square-based pyramid is 273 in².

msm555 msm555 · Answer 2 · 2024-01-24T06:12:35-05:00

Answer:

[tex]273 \, \textsf{in}^2[/tex]

Step-by-step explanation:

The surface area ([tex]A[/tex]) of a regular square pyramid can be found using the formula:

[tex] A = B + \dfrac{1}{2}\cdot P\cdot l [/tex]

where:

[tex]B[/tex] is the area of the base,
[tex]P[/tex] is the perimeter of the base, and
[tex]l[/tex] is the slant height.

For a square pyramid, the base is a square, so [tex]B = s^2[/tex] where [tex]s[/tex] is the length of a side of the base.

The perimeter of the base ([tex]P[/tex]) of a square is given by:

[tex]P = 4s[/tex]

Now, let's substitute the given values into the formula:

[tex] \sf B = s^2 [/tex]
[tex] \sf P = 4s [/tex]
[tex] \sf A = s^2 + \dfrac{1}{2}(4s)l [/tex]

Given:

[tex]s = 7 [/tex] inches (length of a side of the base),
[tex]l = 16[/tex] inches (slant height).

Substitute these values into the formula:

[tex] A = 7^2 + \dfrac{1}{2}(4 \times 7 \times 16) [/tex]

[tex] A = 49 + 224 [/tex]

[tex] A = 273 [/tex]

So, the surface area of the pyramid is [tex]273 \, \textsf{in}^2[/tex].