Basic Math Examples

 $\begin{array}{r} h = & 7 \\ l = & 5 \\ w = & 3 \end{array}$

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 $l = 5$ , the width $w = 3$ , and the height $h = 7$ into the formula for surface area of a pyramid.

$5 \cdot 3 + 3 \cdot \sqrt{{(\frac{5}{2})}^{2} + {(7)}^{2}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3

Simplify each term.

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Step 3.1

Multiply $5$ by $3$ .

$15 + 3 \cdot \sqrt{{(\frac{5}{2})}^{2} + {(7)}^{2}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.2

Apply the product rule to $\frac{5}{2}$ .

$15 + 3 \cdot \sqrt{\frac{5^{2}}{2^{2}} + {(7)}^{2}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.3

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

$15 + 3 \cdot \sqrt{\frac{25}{2^{2}} + {(7)}^{2}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.4

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

$15 + 3 \cdot \sqrt{\frac{25}{4} + {(7)}^{2}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.5

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

$15 + 3 \cdot \sqrt{\frac{25}{4} + 49} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.6

To write $49$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$15 + 3 \cdot \sqrt{\frac{25}{4} + 49 \cdot \frac{4}{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.7

Combine $49$ and $\frac{4}{4}$ .

$15 + 3 \cdot \sqrt{\frac{25}{4} + \frac{49 \cdot 4}{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.8

Combine the numerators over the common denominator.

$15 + 3 \cdot \sqrt{\frac{25 + 49 \cdot 4}{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.9

Simplify the numerator.

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Step 3.9.1

Multiply $49$ by $4$ .

$15 + 3 \cdot \sqrt{\frac{25 + 196}{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.9.2

Add $25$ and $196$ .

$15 + 3 \cdot \sqrt{\frac{221}{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.10

Rewrite $\sqrt{\frac{221}{4}}$ as $\frac{\sqrt{221}}{\sqrt{4}}$ .

$15 + 3 \cdot \frac{\sqrt{221}}{\sqrt{4}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.11

Simplify the denominator.

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Step 3.11.1

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

$15 + 3 \cdot \frac{\sqrt{221}}{\sqrt{2^{2}}} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.11.2

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

$15 + 3 \cdot \frac{\sqrt{221}}{2} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{{(\frac{3}{2})}^{2} + {(7)}^{2}}$

Step 3.12

Apply the product rule to $\frac{3}{2}$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{3^{2}}{2^{2}} + {(7)}^{2}}$

Step 3.13

Raise $3$ to the power of $2$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9}{2^{2}} + {(7)}^{2}}$

Step 3.14

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

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9}{4} + {(7)}^{2}}$

Step 3.15

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

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9}{4} + 49}$

Step 3.16

To write $49$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9}{4} + 49 \cdot \frac{4}{4}}$

Step 3.17

Combine $49$ and $\frac{4}{4}$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9}{4} + \frac{49 \cdot 4}{4}}$

Step 3.18

Combine the numerators over the common denominator.

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9 + 49 \cdot 4}{4}}$

Step 3.19

Simplify the numerator.

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Step 3.19.1

Multiply $49$ by $4$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{9 + 196}{4}}$

Step 3.19.2

Add $9$ and $196$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \sqrt{\frac{205}{4}}$

Step 3.20

Rewrite $\sqrt{\frac{205}{4}}$ as $\frac{\sqrt{205}}{\sqrt{4}}$ .

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \frac{\sqrt{205}}{\sqrt{4}}$

Step 3.21

Simplify the denominator.

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Step 3.21.1

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

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \frac{\sqrt{205}}{\sqrt{2^{2}}}$

Step 3.21.2

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

$15 + \frac{3 \sqrt{221}}{2} + 5 \cdot \frac{\sqrt{205}}{2}$

$15 + \frac{3 \sqrt{221}}{2} + \frac{5 \sqrt{205}}{2}$

Step 4

Calculate the approximate solution to $4$ decimal places.

$73.0937$

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