Basic Math Examples

 $\begin{array}{r} h = & 3 \\ l = & 4 \\ w = & 8 \end{array}$

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 $l = 4$ , the width $w = 8$ , and the height $h = 3$ into the formula for surface area of a pyramid.

$4 \cdot 8 + 8 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(3)}^{2}} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $4$ by $8$ .

$32 + 8 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(3)}^{2}} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3.2

Divide $4$ by $2$ .

$32 + 8 \cdot \sqrt{2^{2} + {(3)}^{2}} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3.3

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

$32 + 8 \cdot \sqrt{4 + {(3)}^{2}} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3.4

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

$32 + 8 \cdot \sqrt{4 + 9} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3.5

Add $4$ and $9$ .

$32 + 8 \cdot \sqrt{13} + 4 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(3)}^{2}}$

Step 3.6

Divide $8$ by $2$ .

$32 + 8 \sqrt{13} + 4 \cdot \sqrt{4^{2} + {(3)}^{2}}$

Step 3.7

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

$32 + 8 \sqrt{13} + 4 \cdot \sqrt{16 + {(3)}^{2}}$

Step 3.8

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

$32 + 8 \sqrt{13} + 4 \cdot \sqrt{16 + 9}$

Step 3.9

Add $16$ and $9$ .

$32 + 8 \sqrt{13} + 4 \cdot \sqrt{25}$

Step 3.10

Rewrite $25$ as $5^{2}$ .

$32 + 8 \sqrt{13} + 4 \cdot \sqrt{5^{2}}$

Step 3.11

Pull terms out from under the radical, assuming positive real numbers.

$32 + 8 \sqrt{13} + 4 \cdot 5$

Step 3.12

Multiply $4$ by $5$ .

$32 + 8 \sqrt{13} + 20$

Step 4

Add $32$ and $20$ .

$52 + 8 \sqrt{13}$

Step 5

Calculate the approximate solution to $4$ decimal places.

$80.8444$

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