Find the Surface Area pyramid (5)(5)(5)



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

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

$w = 5$ , and the height

$h = 5$ into the formula for surface area of a pyramid.

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

Step 3

Simplify each term.

Multiply

$5$ by

$5$ .

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

Apply the product rule to

$\frac{5}{2}$ .

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

Raise

$5$ to the power of

$2$ .

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

Raise

$2$ to the power of

$2$ .

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

Raise

$5$ to the power of

$2$ .

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

To write

$25$ as a fraction with a common denominator, multiply by

$\frac{4}{4}$ .

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

Combine

$25$ and

$\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply

$25$ by

$4$ .

$25 + 5 \cdot \sqrt{\frac{25 + 100}{4}} + 5 \cdot \sqrt{{(\frac{5}{2})}^{2} + {(5)}^{2}}$

Add

$25$ and

$100$ .

$25 + 5 \cdot \sqrt{\frac{125}{4}} + 5 \cdot \sqrt{{(\frac{5}{2})}^{2} + {(5)}^{2}}$

Rewrite

$\sqrt{\frac{125}{4}}$ as

$\frac{\sqrt{125}}{\sqrt{4}}$ .

$25 + 5 \cdot \frac{\sqrt{125}}{\sqrt{4}} + 5 \cdot \sqrt{{(\frac{5}{2})}^{2} + {(5)}^{2}}$

Simplify the numerator.

Rewrite

$125$ as

$5^{2} \cdot 5$ .

Factor

$25$ out of

$125$ .

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

Rewrite

$25$ as

$5^{2}$ .

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

Pull terms out from under the radical.

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

Simplify the denominator.

Rewrite

$4$ as

$2^{2}$ .

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

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

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

Multiply

$5 \frac{5 \sqrt{5}}{2}$ .

Combine

$5$ and

$\frac{5 \sqrt{5}}{2}$ .

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

Multiply

$5$ by

$5$ .

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

Apply the product rule to

$\frac{5}{2}$ .

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

Raise

$5$ to the power of

$2$ .

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

Raise

$2$ to the power of

$2$ .

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

Raise

$5$ to the power of

$2$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{25}{4} + 25}$

To write

$25$ as a fraction with a common denominator, multiply by

$\frac{4}{4}$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{25}{4} + 25 \cdot \frac{4}{4}}$

Combine

$25$ and

$\frac{4}{4}$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{25}{4} + \frac{25 \cdot 4}{4}}$

Combine the numerators over the common denominator.

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{25 + 25 \cdot 4}{4}}$

Simplify the numerator.

Multiply

$25$ by

$4$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{25 + 100}{4}}$

Add

$25$ and

$100$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \sqrt{\frac{125}{4}}$

Rewrite

$\sqrt{\frac{125}{4}}$ as

$\frac{\sqrt{125}}{\sqrt{4}}$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{\sqrt{125}}{\sqrt{4}}$

Simplify the numerator.

Rewrite

$125$ as

$5^{2} \cdot 5$ .

Factor

$25$ out of

$125$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{\sqrt{25 (5)}}{\sqrt{4}}$

Rewrite

$25$ as

$5^{2}$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{\sqrt{5^{2} \cdot 5}}{\sqrt{4}}$

Pull terms out from under the radical.

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{5 \sqrt{5}}{\sqrt{4}}$

Simplify the denominator.

Rewrite

$4$ as

$2^{2}$ .

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{5 \sqrt{5}}{\sqrt{2^{2}}}$

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

$25 + \frac{25 \sqrt{5}}{2} + 5 \cdot \frac{5 \sqrt{5}}{2}$

Multiply

$5 \frac{5 \sqrt{5}}{2}$ .

Combine

$5$ and

$\frac{5 \sqrt{5}}{2}$ .

$25 + \frac{25 \sqrt{5}}{2} + \frac{5 (5 \sqrt{5})}{2}$

Multiply

$5$ by

$5$ .

$25 + \frac{25 \sqrt{5}}{2} + \frac{25 \sqrt{5}}{2}$

Step 4

Simplify terms.

Combine the numerators over the common denominator.

$25 + \frac{25 \sqrt{5} + 25 \sqrt{5}}{2}$

Add

$25 \sqrt{5}$ and

$25 \sqrt{5}$ .

$25 + \frac{50 \sqrt{5}}{2}$

Cancel the common factor of

$50$ and

$2$ .

Factor

$2$ out of

$50 \sqrt{5}$ .

$25 + \frac{2 (25 \sqrt{5})}{2}$

Cancel the common factors.

Factor

$2$ out of

$2$ .

$25 + \frac{2 (25 \sqrt{5})}{2 (1)}$

Cancel the common factor.

$25 + \frac{2 (25 \sqrt{5})}{2 \cdot 1}$

Rewrite the expression.

$25 + \frac{25 \sqrt{5}}{1}$

Divide

$25 \sqrt{5}$ by

$1$ .

$25 + 25 \sqrt{5}$

Step 5

Calculate the approximate solution to

$4$ decimal places.

$80.9017$

Find the Surface Area pyramid (5)(5)(5)

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