Pre-Algebra Examples

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\begin{array}{r} h = & 3 \\ r = & 4 \end{array}

$\begin{array}{r} h = & 3 \\ r = & 4 \end{array}$

Step 1

The surface area of a cylinder is equal to the sum of the areas of the bases each having an area of

π \cdot r^{2}

$π \cdot r^{2}$ , plus the area of the side. The area of the side is equal to the area of a rectangle with length

2 π r

$2 π r$ (the circumference of the base) times the height.

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

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

Step 2

Substitute the values of the radius

r = 4

$r = 4$ and height

h = 3

$h = 3$ into the formula. Pi

π

$π$ is approximately equal to

3.14

$3.14$ .

2 (π) {(4)}^{2} + 2 (π) (4) (3)

$2 (π) {(4)}^{2} + 2 (π) (4) (3)$

Step 3

Simplify each term.

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Step 3.1

Raise

4

$4$ to the power of

2

$2$ .

2 π \cdot 16 + 2 (π) (4) (3)

$2 π \cdot 16 + 2 (π) (4) (3)$

Step 3.2

Multiply

16

$16$ by

2

$2$ .

32 π + 2 (π) (4) (3)

$32 π + 2 (π) (4) (3)$

Step 3.3

Multiply

4

$4$ by

2

$2$ .

32 π + 8 π \cdot 3

$32 π + 8 π \cdot 3$

Step 3.4

Multiply

3

$3$ by

8

$8$ .

32 π + 24 π

$32 π + 24 π$

32 π + 24 π

$32 π + 24 π$

Step 4

Add

32 π

$32 π$ and

24 π

$24 π$ .

56 π

$56 π$

Step 5

The result can be shown in multiple forms.

Exact Form:

56 π

$56 π$

Decimal Form:

175.92918860 \dots

$175.92918860 \dots$

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