Pre-Algebra Examples

 $\begin{array}{r} h = & 5 \\ l = & 6 \\ w = & 8 \end{array}$

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 $l = 6$ , the width $w = 8$ , and the height $h = 5$ into the formula for surface area of a pyramid.

$6 \cdot 8 + 8 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $6$ by $8$ .

$48 + 8 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3.2

Divide $6$ by $2$ .

$48 + 8 \cdot \sqrt{3^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3.3

Raise $3$ to the power of $2$ .

$48 + 8 \cdot \sqrt{9 + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3.4

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

$48 + 8 \cdot \sqrt{9 + 25} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3.5

Add $9$ and $25$ .

$48 + 8 \cdot \sqrt{34} + 6 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(5)}^{2}}$

Step 3.6

Divide $8$ by $2$ .

$48 + 8 \sqrt{34} + 6 \cdot \sqrt{4^{2} + {(5)}^{2}}$

Step 3.7

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

$48 + 8 \sqrt{34} + 6 \cdot \sqrt{16 + {(5)}^{2}}$

Step 3.8

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

$48 + 8 \sqrt{34} + 6 \cdot \sqrt{16 + 25}$

Step 3.9

Add $16$ and $25$ .

$48 + 8 \sqrt{34} + 6 \sqrt{41}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$133.0664$