Pre-Algebra Examples

 $\begin{array}{r} h = & 8 ft \\ l = & 16 ft \\ w = & 1 \end{array}$

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 $l = 16$ , the width $w = 1$ , and the height $h = 8$ into the formula for surface area of a pyramid.

$16 \cdot 1 + 1 \cdot \sqrt{{(\frac{16}{2})}^{2} + {(8)}^{2}} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $16$ by $1$ .

$16 + 1 \cdot \sqrt{{(\frac{16}{2})}^{2} + {(8)}^{2}} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.2

Multiply $\sqrt{{(\frac{16}{2})}^{2} + {(8)}^{2}}$ by $1$ .

$16 + \sqrt{{(\frac{16}{2})}^{2} + {(8)}^{2}} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.3

Divide $16$ by $2$ .

$16 + \sqrt{8^{2} + {(8)}^{2}} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.4

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

$16 + \sqrt{64 + {(8)}^{2}} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.5

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

$16 + \sqrt{64 + 64} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.6

Add $64$ and $64$ .

$16 + \sqrt{128} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.7

Rewrite $128$ as $8^{2} \cdot 2$ .

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Step 3.7.1

Factor $64$ out of $128$ .

$16 + \sqrt{64 (2)} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.7.2

Rewrite $64$ as $8^{2}$ .

$16 + \sqrt{8^{2} \cdot 2} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.8

Pull terms out from under the radical.

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{{(\frac{1}{2})}^{2} + {(8)}^{2}}$

Step 3.9

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

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1^{2}}{2^{2}} + {(8)}^{2}}$

Step 3.10

One to any power is one.

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1}{2^{2}} + {(8)}^{2}}$

Step 3.11

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

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1}{4} + {(8)}^{2}}$

Step 3.12

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

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1}{4} + 64}$

Step 3.13

To write $64$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1}{4} + 64 \cdot \frac{4}{4}}$

Step 3.14

Combine $64$ and $\frac{4}{4}$ .

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1}{4} + \frac{64 \cdot 4}{4}}$

Step 3.15

Combine the numerators over the common denominator.

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1 + 64 \cdot 4}{4}}$

Step 3.16

Simplify the numerator.

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Step 3.16.1

Multiply $64$ by $4$ .

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{1 + 256}{4}}$

Step 3.16.2

Add $1$ and $256$ .

$16 + 8 \sqrt{2} + 16 \cdot \sqrt{\frac{257}{4}}$

Step 3.17

Rewrite $\sqrt{\frac{257}{4}}$ as $\frac{\sqrt{257}}{\sqrt{4}}$ .

$16 + 8 \sqrt{2} + 16 \cdot \frac{\sqrt{257}}{\sqrt{4}}$

Step 3.18

Simplify the denominator.

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Step 3.18.1

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

$16 + 8 \sqrt{2} + 16 \cdot \frac{\sqrt{257}}{\sqrt{2^{2}}}$

Step 3.18.2

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

$16 + 8 \sqrt{2} + 16 \cdot \frac{\sqrt{257}}{2}$

Step 3.19

Cancel the common factor of $2$ .

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Step 3.19.1

Factor $2$ out of $16$ .

$16 + 8 \sqrt{2} + 2 (8) \cdot \frac{\sqrt{257}}{2}$

Step 3.19.2

Cancel the common factor.

$16 + 8 \sqrt{2} + 2 \cdot 8 \cdot \frac{\sqrt{257}}{2}$

Step 3.19.3

Rewrite the expression.

$16 + 8 \sqrt{2} + 8 \cdot \sqrt{257}$

$16 + 8 \sqrt{2} + 8 \sqrt{257}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$155.5635 {(ft)}^{2}$