Pre-Algebra Examples

 $\begin{array}{r} h = & 3.5 \\ l = & 2 \\ w = & 2 \end{array}$

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 $l w$ , 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 $l = 2$ , the width $w = 2$ , and the height $h = 3.5$ into the formula for surface area of a pyramid.

$2 \cdot 2 + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3

Simplify each term.

Tap for more steps...

Step 3.1

Multiply $2$ by $2$ .

$4 + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3.2

Divide $2$ by $2$ .

$4 + 2 \cdot \sqrt{1^{2} + {(3.5)}^{2}} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3.3

One to any power is one.

$4 + 2 \cdot \sqrt{1 + {(3.5)}^{2}} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3.4

Raise $3.5$ to the power of $2$ .

$4 + 2 \cdot \sqrt{1 + 12.25} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3.5

Add $1$ and $12.25$ .

$4 + 2 \cdot \sqrt{13.25} + 2 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(3.5)}^{2}}$

Step 3.6

Divide $2$ by $2$ .

$4 + 2 \sqrt{13.25} + 2 \cdot \sqrt{1^{2} + {(3.5)}^{2}}$

Step 3.7

One to any power is one.

$4 + 2 \sqrt{13.25} + 2 \cdot \sqrt{1 + {(3.5)}^{2}}$

Step 3.8

Raise $3.5$ to the power of $2$ .

$4 + 2 \sqrt{13.25} + 2 \cdot \sqrt{1 + 12.25}$

Step 3.9

Add $1$ and $12.25$ .

$4 + 2 \sqrt{13.25} + 2 \sqrt{13.25}$

Step 4

Add $2 \sqrt{13.25}$ and $2 \sqrt{13.25}$ .

$4 + 4 \sqrt{13.25}$

Step 5

Calculate the approximate solution to $4$ decimal places.

$18.5602$