Pre-Algebra Examples



\begin{array}{r} h = & 5 \\ r = & 1 \end{array}

$\begin{array}{r} h = & 5 \\ r = & 1 \end{array}$

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}})

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

Step 2

Substitute the values of the length

r = 1

$r = 1$ and height

h = 5

$h = 5$ into the formula. Pi

π

$π$ is approximately equal to

3.14

$3.14$ .

(π \cdot 1) (1 + \sqrt{{(1)}^{2} + {(5)}^{2}})

$(π \cdot 1) (1 + \sqrt{{(1)}^{2} + {(5)}^{2}})$

Step 3

Multiply

π

$π$ by

1

$1$ .

π (1 + \sqrt{{(1)}^{2} + {(5)}^{2}})

$π (1 + \sqrt{{(1)}^{2} + {(5)}^{2}})$

Step 4

Simplify each term.

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Step 4.1

One to any power is one.

π (1 + \sqrt{1 + {(5)}^{2}})

$π (1 + \sqrt{1 + {(5)}^{2}})$

Step 4.2

Raise

5

$5$ to the power of

2

$2$ .

π (1 + \sqrt{1 + 25})

$π (1 + \sqrt{1 + 25})$

Step 4.3

Add

1

$1$ and

25

$25$ .

π (1 + \sqrt{26})

$π (1 + \sqrt{26})$

π (1 + \sqrt{26})

$π (1 + \sqrt{26})$

Step 5

Simplify by multiplying through.

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Step 5.1

Apply the distributive property.

π \cdot 1 + π \sqrt{26}

$π \cdot 1 + π \sqrt{26}$

Step 5.2

Multiply

π

$π$ by

1

$1$ .

π + π \sqrt{26}

$π + π \sqrt{26}$

π + π \sqrt{26}

$π + π \sqrt{26}$

Step 6

The result can be shown in multiple forms.

Exact Form:

π + π \sqrt{26}

$π + π \sqrt{26}$

Decimal Form:

19.16063489 \dots

$19.16063489 \dots$

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