Pre-Algebra Examples

 $\begin{array}{r} h = & 5 \\ l = & 6 \\ w = & 1 \end{array}$

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 $w = 1$ , and the height $h = 5$ into the formula for surface area of a pyramid.

$6 \cdot 1 + 1 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $6$ by $1$ .

$6 + 1 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.2

Multiply $\sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}}$ by $1$ .

$6 + \sqrt{{(\frac{6}{2})}^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.3

Divide $6$ by $2$ .

$6 + \sqrt{3^{2} + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.4

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

$6 + \sqrt{9 + {(5)}^{2}} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.5

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

$6 + \sqrt{9 + 25} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.6

Add $9$ and $25$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(5)}^{2}}$

Step 3.7

Apply the product rule to $\frac{1}{2}$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1^{2}}{2^{2}} + {(5)}^{2}}$

Step 3.8

One to any power is one.

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1}{2^{2}} + {(5)}^{2}}$

Step 3.9

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

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1}{4} + {(5)}^{2}}$

Step 3.10

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

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1}{4} + 25}$

Step 3.11

To write $25$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1}{4} + 25 \cdot \frac{4}{4}}$

Step 3.12

Combine $25$ and $\frac{4}{4}$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1}{4} + \frac{25 \cdot 4}{4}}$

Step 3.13

Combine the numerators over the common denominator.

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1 + 25 \cdot 4}{4}}$

Step 3.14

Simplify the numerator.

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Step 3.14.1

Multiply $25$ by $4$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{1 + 100}{4}}$

Step 3.14.2

Add $1$ and $100$ .

$6 + \sqrt{34} + 6 \cdot \sqrt{\frac{101}{4}}$

Step 3.15

Rewrite $\sqrt{\frac{101}{4}}$ as $\frac{\sqrt{101}}{\sqrt{4}}$ .

$6 + \sqrt{34} + 6 \cdot \frac{\sqrt{101}}{\sqrt{4}}$

Step 3.16

Simplify the denominator.

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Step 3.16.1

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

$6 + \sqrt{34} + 6 \cdot \frac{\sqrt{101}}{\sqrt{2^{2}}}$

Step 3.16.2

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

$6 + \sqrt{34} + 6 \cdot \frac{\sqrt{101}}{2}$

Step 3.17

Cancel the common factor of $2$ .

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Step 3.17.1

Factor $2$ out of $6$ .

$6 + \sqrt{34} + 2 (3) \cdot \frac{\sqrt{101}}{2}$

Step 3.17.2

Cancel the common factor.

$6 + \sqrt{34} + 2 \cdot 3 \cdot \frac{\sqrt{101}}{2}$

Step 3.17.3

Rewrite the expression.

$6 + \sqrt{34} + 3 \cdot \sqrt{101}$

$6 + \sqrt{34} + 3 \sqrt{101}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$41.9806$

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