Basic Math Examples

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\begin{matrix} h = & 3 l = & 4 w = & 8 \end{matrix}

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

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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