Find the Surface Area pyramid (8)(8)(10)

 $\begin{array}{r} h = & 8 \\ l = & 10 \\ w = & 8 \end{array}$

Step 1

The surface area of a pyramid is equal to the sum of the areas of each side of the pyramid. The base of the pyramid has area $l w$ , and $s_{l}$ and $s_{w}$ represent the slant height on the length and slant height on the width.

$(l e n g t h) \cdot (w i d t h) + (w i d t h) \cdot s_{l} + (l e n g t h) \cdot s_{w}$

Step 2

Substitute the values of the length $l = 10$ , the width $w = 8$ , and the height $h = 8$ into the formula for surface area of a pyramid.

$10 \cdot 8 + 8 \cdot \sqrt{{(\frac{10}{2})}^{2} + {(8)}^{2}} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Step 3

Simplify each term.

Multiply $10$ by $8$ .

$80 + 8 \cdot \sqrt{{(\frac{10}{2})}^{2} + {(8)}^{2}} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Divide $10$ by $2$ .

$80 + 8 \cdot \sqrt{5^{2} + {(8)}^{2}} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Raise $5$ to the power of $2$ .

$80 + 8 \cdot \sqrt{25 + {(8)}^{2}} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Raise $8$ to the power of $2$ .

$80 + 8 \cdot \sqrt{25 + 64} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Add $25$ and $64$ .

$80 + 8 \cdot \sqrt{89} + 10 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(8)}^{2}}$

Divide $8$ by $2$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{4^{2} + {(8)}^{2}}$

Raise $4$ to the power of $2$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{16 + {(8)}^{2}}$

Raise $8$ to the power of $2$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{16 + 64}$

Add $16$ and $64$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{80}$

Rewrite $80$ as $4^{2} \cdot 5$ .

Factor $16$ out of $80$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{16 (5)}$

Rewrite $16$ as $4^{2}$ .

$80 + 8 \sqrt{89} + 10 \cdot \sqrt{4^{2} \cdot 5}$

Pull terms out from under the radical.

$80 + 8 \sqrt{89} + 10 \cdot (4 \sqrt{5})$

Multiply $4$ by $10$ .

$80 + 8 \sqrt{89} + 40 \sqrt{5}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$244.9146$

Find the Surface Area pyramid (8)(8)(10)

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