Pre-Algebra Examples Measurement Area and Volume

Find the Surface Area

 $\begin{array}{r} h = & 5 \\ l = & 7 \\ w = & 2 \end{array}$

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 $l = 7$ , the width $w = 2$ , and the height $h = 5$ into the formula for surface area of a pyramid.

$7 \cdot 2 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $7$ by $2$ .

$14 + 2 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.2

Apply the product rule to $\frac{7}{2}$ .

$14 + 2 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.3

Raise $7$ to the power of $2$ .

$14 + 2 \cdot \sqrt{\frac{49}{2^{2}} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.4

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

$14 + 2 \cdot \sqrt{\frac{49}{4} + {(5)}^{2}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.5

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

$14 + 2 \cdot \sqrt{\frac{49}{4} + 25} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.6

To write $25$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$14 + 2 \cdot \sqrt{\frac{49}{4} + 25 \cdot \frac{4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.7

Combine $25$ and $\frac{4}{4}$ .

$14 + 2 \cdot \sqrt{\frac{49}{4} + \frac{25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.8

Combine the numerators over the common denominator.

$14 + 2 \cdot \sqrt{\frac{49 + 25 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply $25$ by $4$ .

$14 + 2 \cdot \sqrt{\frac{49 + 100}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.9.2

Add $49$ and $100$ .

$14 + 2 \cdot \sqrt{\frac{149}{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.10

Rewrite $\sqrt{\frac{149}{4}}$ as $\frac{\sqrt{149}}{\sqrt{4}}$ .

$14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.11

Simplify the denominator.

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Step 3.11.1

Rewrite $4$ as $2^{2}$ .

$14 + 2 \cdot \frac{\sqrt{149}}{\sqrt{2^{2}}} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.11.2

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

$14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.12

Cancel the common factor of $2$ .

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Step 3.12.1

Cancel the common factor.

$14 + 2 \cdot \frac{\sqrt{149}}{2} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.12.2

Rewrite the expression.

$14 + \sqrt{149} + 7 \cdot \sqrt{{(\frac{2}{2})}^{2} + {(5)}^{2}}$

Step 3.13

Divide $2$ by $2$ .

$14 + \sqrt{149} + 7 \cdot \sqrt{1^{2} + {(5)}^{2}}$

Step 3.14

One to any power is one.

$14 + \sqrt{149} + 7 \cdot \sqrt{1 + {(5)}^{2}}$

Step 3.15

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

$14 + \sqrt{149} + 7 \cdot \sqrt{1 + 25}$

Step 3.16

Add $1$ and $25$ .

$14 + \sqrt{149} + 7 \sqrt{26}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$61.8997$

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