Pre-Algebra Examples

 $\begin{array}{r} h = & 15 \\ l = & 20 \\ w = & 8 \end{array}$

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 $l = 20$ , the width $w = 8$ , and the height $h = 15$ into the formula for surface area of a pyramid.

$20 \cdot 8 + 8 \cdot \sqrt{{(\frac{20}{2})}^{2} + {(15)}^{2}} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3

Simplify each term.

Tap for more steps...

Step 3.1

Multiply $20$ by $8$ .

$160 + 8 \cdot \sqrt{{(\frac{20}{2})}^{2} + {(15)}^{2}} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.2

Divide $20$ by $2$ .

$160 + 8 \cdot \sqrt{10^{2} + {(15)}^{2}} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.3

Raise $10$ to the power of $2$ .

$160 + 8 \cdot \sqrt{100 + {(15)}^{2}} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.4

Raise $15$ to the power of $2$ .

$160 + 8 \cdot \sqrt{100 + 225} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.5

Add $100$ and $225$ .

$160 + 8 \cdot \sqrt{325} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.6

Rewrite $325$ as $5^{2} \cdot 13$ .

Tap for more steps...

Step 3.6.1

Factor $25$ out of $325$ .

$160 + 8 \cdot \sqrt{25 (13)} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.6.2

Rewrite $25$ as $5^{2}$ .

$160 + 8 \cdot \sqrt{5^{2} \cdot 13} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.7

Pull terms out from under the radical.

$160 + 8 \cdot (5 \sqrt{13}) + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.8

Multiply $5$ by $8$ .

$160 + 40 \sqrt{13} + 20 \cdot \sqrt{{(\frac{8}{2})}^{2} + {(15)}^{2}}$

Step 3.9

Divide $8$ by $2$ .

$160 + 40 \sqrt{13} + 20 \cdot \sqrt{4^{2} + {(15)}^{2}}$

Step 3.10

Raise $4$ to the power of $2$ .

$160 + 40 \sqrt{13} + 20 \cdot \sqrt{16 + {(15)}^{2}}$

Step 3.11

Raise $15$ to the power of $2$ .

$160 + 40 \sqrt{13} + 20 \cdot \sqrt{16 + 225}$

Step 3.12

Add $16$ and $225$ .

$160 + 40 \sqrt{13} + 20 \sqrt{241}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$614.7055$