Basic Math Examples

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\begin{matrix} h = & 5 l = & 6 w = & 1 \end{matrix}

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

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

$l w$ , 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 = 6

$l = 6$ , the width

w = 1

$w = 1$ , and the height

h = 5

$h = 5$ 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}}

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

Step 3

Simplify each term.

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Step 3.1

Multiply

6

$6$ by

1

$1$ .

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

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

Step 3.2

Multiply

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

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

1

$1$ .

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

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

Step 3.3

Divide

6

$6$ by

2

$2$ .

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

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

Step 3.4

Raise

3

$3$ to the power of

2

$2$ .

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

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

Step 3.5

Raise

5

$5$ to the power of

2

$2$ .

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

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

Step 3.6

Add

9

$9$ and

25

$25$ .

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

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

Step 3.7

Apply the product rule to

\frac{1}{2}

$\frac{1}{2}$ .

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

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

Step 3.9

Raise

2

$2$ to the power of

2

$2$ .

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

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

Step 3.10

Raise

5

$5$ to the power of

2

$2$ .

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

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

Step 3.11

To write

25

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

\frac{4}{4}

$\frac{4}{4}$ .

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

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

Step 3.12

Combine

25

$25$ and

\frac{4}{4}

$\frac{4}{4}$ .

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

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

Step 3.14

Simplify the numerator.

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Step 3.14.1

Multiply

25

$25$ by

4

$4$ .

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

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

Step 3.14.2

Add

1

$1$ and

100

$100$ .

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

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

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

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

Step 3.15

Rewrite

\sqrt{\frac{101}{4}}

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

\frac{\sqrt{101}}{\sqrt{4}}

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

6 + \sqrt{34} + 6 \cdot \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

$4$ as

2^{2}

$2^{2}$ .

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

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

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

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

Step 3.17

Cancel the common factor of

2

$2$ .

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Step 3.17.1

Factor

2

$2$ out of

6

$6$ .

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

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

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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