Pre-Algebra Examples

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

The surface area of a cone is equal to the area of the base plus the area of the cone.

$π \cdot {(r a d i u s)}^{2} + π \cdot ((r a d i u s) \sqrt{r a d i u s^{2} + h e i g h t^{2}})$

Substitute the values of the length $(r = 5)$ , the height $(h = 5)$ , and an approximation for $π (3.14)$ into the formula.

$(3.14) {(5)}^{2} + 3.14 (5) \sqrt{{(5)}^{2} + {(5)}^{2}}$

Simplify each term.

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Remove parentheses around $5$ .

$3.14 \cdot 5^{2} + 3.14 (5) \sqrt{{(5)}^{2} + {(5)}^{2}}$

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

$3.14 \cdot 25 + 3.14 (5) \sqrt{{(5)}^{2} + {(5)}^{2}}$

Multiply $3.14$ by $25$ .

$78.5 + 3.14 (5) \sqrt{{(5)}^{2} + {(5)}^{2}}$

Multiply $3.14$ by $5$ .

$78.5 + 15.7 \sqrt{{(5)}^{2} + {(5)}^{2}}$

Remove parentheses around $5$ .

$78.5 + 15.7 \sqrt{5^{2} + {(5)}^{2}}$

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

$78.5 + 15.7 \sqrt{25 + {(5)}^{2}}$

Remove parentheses around $5$ .

$78.5 + 15.7 \sqrt{25 + 5^{2}}$

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

$78.5 + 15.7 \sqrt{25 + 25}$

Add $25$ and $25$ .

$78.5 + 15.7 \sqrt{50}$

Rewrite $50$ as $5^{2} \cdot 2$ .

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Factor $25$ out of $50$ .

$78.5 + 15.7 \sqrt{25 (2)}$

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

$78.5 + 15.7 \sqrt{5^{2} \cdot 2}$

Pull terms out from under the radical.

$78.5 + 15.7 (5 \sqrt{2})$

Simplify $15.7 (5 \sqrt{2})$ .

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Multiply $5$ by $15.7$ .

$78.5 + 78.5 \sqrt{2}$

Multiply $78.5$ by $\sqrt{2}$ .

$78.5 + 111.01576464$

Add $78.5$ and $111.01576464$ .

$189.51576464$

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