Pre-Algebra Examples

 $\begin{array}{r} h = & 9 ft \\ l = & 10 ft \\ w = & 12 ft \end{array}$

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 $l = 10$ , the width $w = 12$ , and the height $h = 9$ into the formula for surface area of a pyramid.

$10 \cdot 12 + 12 \cdot \sqrt{{(\frac{10}{2})}^{2} + {(9)}^{2}} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3

Simplify each term.

Tap for more steps...

Step 3.1

Multiply $10$ by $12$ .

$120 + 12 \cdot \sqrt{{(\frac{10}{2})}^{2} + {(9)}^{2}} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3.2

Divide $10$ by $2$ .

$120 + 12 \cdot \sqrt{5^{2} + {(9)}^{2}} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3.3

Raise $5$ to the power of $2$ .

$120 + 12 \cdot \sqrt{25 + {(9)}^{2}} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3.4

Raise $9$ to the power of $2$ .

$120 + 12 \cdot \sqrt{25 + 81} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3.5

Add $25$ and $81$ .

$120 + 12 \cdot \sqrt{106} + 10 \cdot \sqrt{{(\frac{12}{2})}^{2} + {(9)}^{2}}$

Step 3.6

Divide $12$ by $2$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{6^{2} + {(9)}^{2}}$

Step 3.7

Raise $6$ to the power of $2$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{36 + {(9)}^{2}}$

Step 3.8

Raise $9$ to the power of $2$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{36 + 81}$

Step 3.9

Add $36$ and $81$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{117}$

Step 3.10

Rewrite $117$ as $3^{2} \cdot 13$ .

Tap for more steps...

Step 3.10.1

Factor $9$ out of $117$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{9 (13)}$

Step 3.10.2

Rewrite $9$ as $3^{2}$ .

$120 + 12 \sqrt{106} + 10 \cdot \sqrt{3^{2} \cdot 13}$

Step 3.11

Pull terms out from under the radical.

$120 + 12 \sqrt{106} + 10 \cdot (3 \sqrt{13})$

Step 3.12

Multiply $3$ by $10$ .

$120 + 12 \sqrt{106} + 30 \sqrt{13}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$351.7141 {(ft)}^{2}$