Pre-Algebra Examples

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

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

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

Substitute the values of the length $r = 3$ and height $h = 3$ into the formula. Pi $π$ is approximately equal to $3.14$ .

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

Move $3$ to the left of $π$ .

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

Simplify each term.

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

$3 π (3 + \sqrt{9 + {(3)}^{2}})$

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

$3 π (3 + \sqrt{9 + 9})$

Add $9$ and $9$ .

$3 π (3 + \sqrt{18})$

Rewrite $18$ as $3^{2} \cdot 2$ .

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Factor $9$ out of $18$ .

$3 π (3 + \sqrt{9 (2)})$

Rewrite $9$ as $3^{2}$ .

$3 π (3 + \sqrt{3^{2} \cdot 2})$

Pull terms out from under the radical.

$3 π (3 + 3 \sqrt{2})$

Simplify by multiplying through.

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Apply the distributive property.

$3 π \cdot 3 + 3 π (3 \sqrt{2})$

Multiply.

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Multiply $3$ by $3$ .

$9 π + 3 π (3 \sqrt{2})$

Multiply $3$ by $3$ .

$9 π + 9 π \sqrt{2}$

The result can be shown in multiple forms.

Exact Form:

$9 π + 9 π \sqrt{2}$

Decimal Form:

$68.26028032 \dots$

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