Basic Math Examples



\begin{array}{r} h = & 4 \\ l = & 6 \\ w = & 4 \end{array}

$\begin{array}{r} h = & 4 \\ l = & 6 \\ w = & 4 \end{array}$

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 = 4

$w = 4$ , and the height

h = 4

$h = 4$ into the formula for surface area of a pyramid.

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

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

Step 3

Simplify each term.

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Step 3.1

Multiply

6

$6$ by

4

$4$ .

24 + 4 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{{(\frac{6}{2})}^{2} + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.2

Divide

6

$6$ by

2

$2$ .

24 + 4 \cdot \sqrt{3^{2} + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{3^{2} + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.3

Raise

3

$3$ to the power of

2

$2$ .

24 + 4 \cdot \sqrt{9 + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{9 + {(4)}^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.4

Raise

4

$4$ to the power of

2

$2$ .

24 + 4 \cdot \sqrt{9 + 16} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{9 + 16} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.5

Add

9

$9$ and

16

$16$ .

24 + 4 \cdot \sqrt{25} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{25} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.6

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

24 + 4 \cdot \sqrt{5^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot \sqrt{5^{2}} + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.7

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

24 + 4 \cdot 5 + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 4 \cdot 5 + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.8

Multiply

4

$4$ by

5

$5$ .

24 + 20 + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}

$24 + 20 + 6 \cdot \sqrt{{(\frac{4}{2})}^{2} + {(4)}^{2}}$

Step 3.9

Divide

4

$4$ by

2

$2$ .

24 + 20 + 6 \cdot \sqrt{2^{2} + {(4)}^{2}}

$24 + 20 + 6 \cdot \sqrt{2^{2} + {(4)}^{2}}$

Step 3.10

Raise

2

$2$ to the power of

2

$2$ .

24 + 20 + 6 \cdot \sqrt{4 + {(4)}^{2}}

$24 + 20 + 6 \cdot \sqrt{4 + {(4)}^{2}}$

Step 3.11

Raise

4

$4$ to the power of

2

$2$ .

24 + 20 + 6 \cdot \sqrt{4 + 16}

$24 + 20 + 6 \cdot \sqrt{4 + 16}$

Step 3.12

Add

4

$4$ and

16

$16$ .

24 + 20 + 6 \cdot \sqrt{20}

$24 + 20 + 6 \cdot \sqrt{20}$

Step 3.13

Rewrite

20

$20$ as

2^{2} \cdot 5

$2^{2} \cdot 5$ .

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Step 3.13.1

Factor

4

$4$ out of

20

$20$ .

24 + 20 + 6 \cdot \sqrt{4 (5)}

$24 + 20 + 6 \cdot \sqrt{4 (5)}$

Step 3.13.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

24 + 20 + 6 \cdot \sqrt{2^{2} \cdot 5}

$24 + 20 + 6 \cdot \sqrt{2^{2} \cdot 5}$

24 + 20 + 6 \cdot \sqrt{2^{2} \cdot 5}

$24 + 20 + 6 \cdot \sqrt{2^{2} \cdot 5}$

Step 3.14

Pull terms out from under the radical.

24 + 20 + 6 \cdot (2 \sqrt{5})

$24 + 20 + 6 \cdot (2 \sqrt{5})$

Step 3.15

Multiply

2

$2$ by

6

$6$ .

24 + 20 + 12 \sqrt{5}

$24 + 20 + 12 \sqrt{5}$

24 + 20 + 12 \sqrt{5}

$24 + 20 + 12 \sqrt{5}$

Step 4

Add

24

$24$ and

20

$20$ .

44 + 12 \sqrt{5}

$44 + 12 \sqrt{5}$

Step 5

Calculate the approximate solution to

4

$4$ decimal places.

70.8328

$70.8328$