Pre-Algebra Examples

 $\begin{array}{r} h = & 9 \\ l = & 7 \\ w = & 9 \end{array}$

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

$7 \cdot 9 + 9 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(9)}^{2}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $7$ by $9$ .

$63 + 9 \cdot \sqrt{{(\frac{7}{2})}^{2} + {(9)}^{2}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.2

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

$63 + 9 \cdot \sqrt{\frac{7^{2}}{2^{2}} + {(9)}^{2}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.3

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

$63 + 9 \cdot \sqrt{\frac{49}{2^{2}} + {(9)}^{2}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.4

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

$63 + 9 \cdot \sqrt{\frac{49}{4} + {(9)}^{2}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.5

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

$63 + 9 \cdot \sqrt{\frac{49}{4} + 81} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.6

To write $81$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$63 + 9 \cdot \sqrt{\frac{49}{4} + 81 \cdot \frac{4}{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.7

Combine $81$ and $\frac{4}{4}$ .

$63 + 9 \cdot \sqrt{\frac{49}{4} + \frac{81 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.8

Combine the numerators over the common denominator.

$63 + 9 \cdot \sqrt{\frac{49 + 81 \cdot 4}{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply $81$ by $4$ .

$63 + 9 \cdot \sqrt{\frac{49 + 324}{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.9.2

Add $49$ and $324$ .

$63 + 9 \cdot \sqrt{\frac{373}{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.10

Rewrite $\sqrt{\frac{373}{4}}$ as $\frac{\sqrt{373}}{\sqrt{4}}$ .

$63 + 9 \cdot \frac{\sqrt{373}}{\sqrt{4}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.11

Simplify the denominator.

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Step 3.11.1

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

$63 + 9 \cdot \frac{\sqrt{373}}{\sqrt{2^{2}}} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.11.2

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

$63 + 9 \cdot \frac{\sqrt{373}}{2} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{{(\frac{9}{2})}^{2} + {(9)}^{2}}$

Step 3.12

Apply the product rule to $\frac{9}{2}$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{9^{2}}{2^{2}} + {(9)}^{2}}$

Step 3.13

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

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81}{2^{2}} + {(9)}^{2}}$

Step 3.14

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

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81}{4} + {(9)}^{2}}$

Step 3.15

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

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81}{4} + 81}$

Step 3.16

To write $81$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81}{4} + 81 \cdot \frac{4}{4}}$

Step 3.17

Combine $81$ and $\frac{4}{4}$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81}{4} + \frac{81 \cdot 4}{4}}$

Step 3.18

Combine the numerators over the common denominator.

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81 + 81 \cdot 4}{4}}$

Step 3.19

Simplify the numerator.

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Step 3.19.1

Multiply $81$ by $4$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{81 + 324}{4}}$

Step 3.19.2

Add $81$ and $324$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \sqrt{\frac{405}{4}}$

Step 3.20

Rewrite $\sqrt{\frac{405}{4}}$ as $\frac{\sqrt{405}}{\sqrt{4}}$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{\sqrt{405}}{\sqrt{4}}$

Step 3.21

Simplify the numerator.

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Step 3.21.1

Rewrite $405$ as $9^{2} \cdot 5$ .

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Step 3.21.1.1

Factor $81$ out of $405$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{\sqrt{81 (5)}}{\sqrt{4}}$

Step 3.21.1.2

Rewrite $81$ as $9^{2}$ .

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{\sqrt{9^{2} \cdot 5}}{\sqrt{4}}$

Step 3.21.2

Pull terms out from under the radical.

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{9 \sqrt{5}}{\sqrt{4}}$

Step 3.22

Simplify the denominator.

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Step 3.22.1

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

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{9 \sqrt{5}}{\sqrt{2^{2}}}$

Step 3.22.2

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

$63 + \frac{9 \sqrt{373}}{2} + 7 \cdot \frac{9 \sqrt{5}}{2}$

Step 3.23

Multiply $7 \frac{9 \sqrt{5}}{2}$ .

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Step 3.23.1

Combine $7$ and $\frac{9 \sqrt{5}}{2}$ .

$63 + \frac{9 \sqrt{373}}{2} + \frac{7 (9 \sqrt{5})}{2}$

Step 3.23.2

Multiply $9$ by $7$ .

$63 + \frac{9 \sqrt{373}}{2} + \frac{63 \sqrt{5}}{2}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$220.3456$