Pre-Algebra Examples

 $\begin{array}{r} h = & 5 \\ r = & 3 \end{array}$

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 $(r = 3)$ , the height $(h = 5)$ , and an approximation for $π (3.14)$ into the formula.

$(3.14) {(3)}^{2} + 3.14 (3) \sqrt{{(3)}^{2} + {(5)}^{2}}$

Simplify each term.

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Remove parentheses.

$3.14 \cdot 3^{2} + 3.14 (3) \sqrt{{(3)}^{2} + {(5)}^{2}}$

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

$3.14 \cdot 9 + 3.14 (3) \sqrt{{(3)}^{2} + {(5)}^{2}}$

Multiply $3.14$ by $9$ .

$28.26 + 3.14 (3) \sqrt{{(3)}^{2} + {(5)}^{2}}$

Multiply $3.14$ by $3$ .

$28.26 + 9.42 \sqrt{{(3)}^{2} + {(5)}^{2}}$

Remove parentheses.

$28.26 + 9.42 \sqrt{3^{2} + {(5)}^{2}}$

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

$28.26 + 9.42 \sqrt{9 + {(5)}^{2}}$

Remove parentheses.

$28.26 + 9.42 \sqrt{9 + 5^{2}}$

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

$28.26 + 9.42 \sqrt{9 + 25}$

Add $9$ and $25$ .

$28.26 + 9.42 \sqrt{34}$

Multiply $9.42$ by $\sqrt{34}$ .

$28.26 + 54.92756684$

Add $28.26$ and $54.92756684$ .

$83.18756684$

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