Pre-Algebra Examples

 $\begin{array}{r} h = & 12 m \\ l = & 7 m \\ w = & 14 m \end{array}$

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 $l = 7$ , the width $w = 14$ , and the height $h = 12$ into the formula for surface area of a pyramid.

$7 \cdot 14 + 14 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(12)}^{2}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $7$ by $14$ .

$98 + 14 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(12)}^{2}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.2

Apply the product rule to $\frac{7}{2}$ .

$98 + 14 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(12)}^{2}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.3

Raise $7$ to the power of $2$ .

$98 + 14 \cdot \sqrt{\frac{49}{2^{2}} + {(12)}^{2}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.4

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

$98 + 14 \cdot \sqrt{\frac{49}{4} + {(12)}^{2}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.5

Raise $12$ to the power of $2$ .

$98 + 14 \cdot \sqrt{\frac{49}{4} + 144} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.6

To write $144$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$98 + 14 \cdot \sqrt{\frac{49}{4} + 144 \cdot \frac{4}{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.7

Combine $144$ and $\frac{4}{4}$ .

$98 + 14 \cdot \sqrt{\frac{49}{4} + \frac{144 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.8

Combine the numerators over the common denominator.

$98 + 14 \cdot \sqrt{\frac{49 + 144 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply $144$ by $4$ .

$98 + 14 \cdot \sqrt{\frac{49 + 576}{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.9.2

Add $49$ and $576$ .

$98 + 14 \cdot \sqrt{\frac{625}{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.10

Rewrite $\sqrt{\frac{625}{4}}$ as $\frac{\sqrt{625}}{\sqrt{4}}$ .

$98 + 14 \cdot \frac{\sqrt{625}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.11

Simplify the numerator.

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Step 3.11.1

Rewrite $625$ as $25^{2}$ .

$98 + 14 \cdot \frac{\sqrt{25^{2}}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.11.2

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

$98 + 14 \cdot \frac{25}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.12

Simplify the denominator.

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Step 3.12.1

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

$98 + 14 \cdot \frac{25}{\sqrt{2^{2}}} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.12.2

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

$98 + 14 \cdot \frac{25}{2} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.13

Cancel the common factor of $2$ .

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Step 3.13.1

Factor $2$ out of $14$ .

$98 + 2 (7) \cdot \frac{25}{2} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.13.2

Cancel the common factor.

$98 + 2 \cdot 7 \cdot \frac{25}{2} + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.13.3

Rewrite the expression.

$98 + 7 \cdot 25 + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.14

Multiply $7$ by $25$ .

$98 + 175 + 7 \cdot \sqrt{{(\frac{14}{2})}^{2} + {(12)}^{2}}$

Step 3.15

Divide $14$ by $2$ .

$98 + 175 + 7 \cdot \sqrt{7^{2} + {(12)}^{2}}$

Step 3.16

Raise $7$ to the power of $2$ .

$98 + 175 + 7 \cdot \sqrt{49 + {(12)}^{2}}$

Step 3.17

Raise $12$ to the power of $2$ .

$98 + 175 + 7 \cdot \sqrt{49 + 144}$

Step 3.18

Add $49$ and $144$ .

$98 + 175 + 7 \sqrt{193}$

Step 4

Add $98$ and $175$ .

$273 + 7 \sqrt{193}$

Step 5

Calculate the approximate solution to $4$ decimal places.

$370.2471 m^{2}$