Pre-Algebra Examples

 $\begin{array}{r} h = & 12 \\ r = & 4 \end{array}$

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 $(r = 4)$ , the height $(h = 12)$ , and an approximation for $π (3.14)$ into the formula.

$(3.14) {(4)}^{2} + 3.14 (4) \sqrt{{(4)}^{2} + {(12)}^{2}}$

Simplify each term.

Tap for more steps...

Remove parentheses around $4$ .

$3.14 \cdot 4^{2} + 3.14 (4) \sqrt{{(4)}^{2} + {(12)}^{2}}$

Raise $4$ to the power of $2$ to get $16$ .

$3.14 \cdot 16 + 3.14 (4) \sqrt{{(4)}^{2} + {(12)}^{2}}$

Multiply $3.14$ by $16$ to get $50.24$ .

$50.24 + 3.14 (4) \sqrt{{(4)}^{2} + {(12)}^{2}}$

Multiply $3.14$ by $4$ to get $12.56$ .

$50.24 + 12.56 \sqrt{{(4)}^{2} + {(12)}^{2}}$

Remove parentheses around $4$ .

$50.24 + 12.56 \sqrt{4^{2} + {(12)}^{2}}$

Raise $4$ to the power of $2$ to get $16$ .

$50.24 + 12.56 \sqrt{16 + {(12)}^{2}}$

Remove parentheses around $12$ .

$50.24 + 12.56 \sqrt{16 + 12^{2}}$

Raise $12$ to the power of $2$ to get $144$ .

$50.24 + 12.56 \sqrt{16 + 144}$

Add $16$ and $144$ to get $160$ .

$50.24 + 12.56 \sqrt{160}$

Rewrite $160$ as $4^{2} \cdot 10$ .

Tap for more steps...

Factor $16$ out of $160$ .

$50.24 + 12.56 \sqrt{16 (10)}$

Rewrite $16$ as $4^{2}$ .

$50.24 + 12.56 \sqrt{4^{2} \cdot 10}$

Pull terms out from under the radical.

$50.24 + 12.56 (4 \sqrt{10})$

Simplify $12.56 (4 \sqrt{10})$ .

Tap for more steps...

Multiply $4$ by $12.56$ to get $50.24$ .

$50.24 + 50.24 \sqrt{10}$

Multiply $50.24$ by $\sqrt{10}$ to get $158.87282964$ .

$50.24 + 158.87282964$

Add $50.24$ and $158.87282964$ to get $209.11282964$ .

$209.11282964$

Enter YOUR Problem