Pre-Algebra Examples

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\begin{matrix} h = & 5 l = & 7 w = & 2 \end{matrix}

$\begin{array}{r} h = & 5 \\ l = & 7 \\ w = & 2 \end{array}$

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

$l w$ , and

s_{l}

$s_{l}$ and

s_{w}

$s_{w}$ 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}

$(l e n g t h) \cdot (w i d t h) + (w i d t h) \cdot s_{l} + (l e n g t h) \cdot s_{w}$

Step 2

Substitute the values of the length

l = 7

$l = 7$ , the width

w = 2

$w = 2$ , and the height

h = 5

$h = 5$ into the formula for surface area of a pyramid.

7 \cdot 2 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$7 \cdot 2 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply

7

$7$ by

2

$2$ .

14 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.2

Apply the product rule to

\frac{7}{2}

$\frac{7}{2}$ .

14 + 2 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.3

Raise

7

$7$ to the power of

2

$2$ .

14 + 2 \cdot \sqrt{\frac{49}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.4

Raise

2

$2$ to the power of

2

$2$ .

14 + 2 \cdot \sqrt{\frac{49}{4} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49}{4} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.5

Raise

5

$5$ to the power of

2

$2$ .

14 + 2 \cdot \sqrt{\frac{49}{4} + 25} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49}{4} + 25} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.6

To write

25

$25$ as a fraction with a common denominator, multiply by

\frac{4}{4}

$\frac{4}{4}$ .

14 + 2 \cdot \sqrt{\frac{49}{4} + 25 \cdot \frac{4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49}{4} + 25 \cdot \frac{4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.7

Combine

25

$25$ and

\frac{4}{4}

$\frac{4}{4}$ .

14 + 2 \cdot \sqrt{\frac{49}{4} + \frac{25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49}{4} + \frac{25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.8

Combine the numerators over the common denominator.

14 + 2 \cdot \sqrt{\frac{49 + 25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49 + 25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply

25

$25$ by

4

$4$ .

14 + 2 \cdot \sqrt{\frac{49 + 100}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{49 + 100}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.9.2

Add

49

$49$ and

100

$100$ .

14 + 2 \cdot \sqrt{\frac{149}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{149}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

14 + 2 \cdot \sqrt{\frac{149}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \sqrt{\frac{149}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.10

Rewrite

\sqrt{\frac{149}{4}}

$\sqrt{\frac{149}{4}}$ as

\frac{\sqrt{149}}{\sqrt{4}}

$\frac{\sqrt{149}}{\sqrt{4}}$ .

14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.11

Simplify the denominator.

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Step 3.11.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{2^{2}}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{2^{2}}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.11.2

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

14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}

$14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.12

Cancel the common factor of

2

$2$ .

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Step 3.12.1

Cancel the common factor.

$14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.12.2

Rewrite the expression.

$14 + \sqrt{149} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.13

Divide

$2$ by

$2$ .

$14 + \sqrt{149} + 7 \cdot \sqrt{1^{2} + {(5)}^{2}}$

Step 3.14

One to any power is one.

$14 + \sqrt{149} + 7 \cdot \sqrt{1 + {(5)}^{2}}$

Step 3.15

Raise

$5$ to the power of

$2$ .

$14 + \sqrt{149} + 7 \cdot \sqrt{1 + 25}$

Step 3.16

Add

$1$ and

$25$ .

$14 + \sqrt{149} + 7 \sqrt{26}$

Step 4

Calculate the approximate solution to

$4$ decimal places.

$61.8997$

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