Pre-Algebra Examples

 $\begin{array}{r} h = & 4 \\ r = & 2 \end{array}$

Step 1

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

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

Step 2

Substitute the values of the length $r = 2$ and height $h = 4$ into the formula. Pi $π$ is approximately equal to $3.14$ .

$(π \cdot 2) (2 + \sqrt{{(2)}^{2} + {(4)}^{2}})$

Step 3

Move $2$ to the left of $π$ .

$2 \cdot π (2 + \sqrt{{(2)}^{2} + {(4)}^{2}})$

Step 4

Simplify each term.

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

$2 π (2 + \sqrt{4 + {(4)}^{2}})$

Raise $4$ to the power of $2$ .

$2 π (2 + \sqrt{4 + 16})$

Add $4$ and $16$ .

$2 π (2 + \sqrt{20})$

Rewrite $20$ as $2^{2} \cdot 5$ .

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Factor $4$ out of $20$ .

$2 π (2 + \sqrt{4 (5)})$

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

$2 π (2 + \sqrt{2^{2} \cdot 5})$

Pull terms out from under the radical.

$2 π (2 + 2 \sqrt{5})$

Step 5

Simplify by multiplying through.

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Apply the distributive property.

$2 π \cdot 2 + 2 π (2 \sqrt{5})$

Multiply.

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Multiply $2$ by $2$ .

$4 π + 2 π (2 \sqrt{5})$

Multiply $2$ by $2$ .

$4 π + 4 π \sqrt{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$4 π + 4 π \sqrt{5}$

Decimal Form:

$40.66562953 \dots$

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