Formulas

An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation
2x+1=72x+1=7
has the unique solution
x=3x=3
because when we substitute 3 for
xx
in the equation, we obtain the true statement
2(3)+1=72\left(3\right)+1=7
.

A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area
AA
of a circle in terms of the radius
rr
of the circle:
A=πr2A=\pi {r}^{2}
. For any value of
rr
, the area
AA
can be found by evaluating the expression
πr2\pi {r}^{2}
.

Example 11: Using a Formula

A right circular cylinder with radius
rr
and height
hh
has the surface area
SS
(in square units) given by the formula
S=2πr(r+h)S=2\pi r\left(r+h\right)
. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of
π\pi
.

A right circular cylinder with an arrow extending from the center of the top circle outward to the edge, labeled: r. Another arrow beside the image going from top to bottom, labeled: h. Figure 3. Right circular cylinder


Solution

Evaluate the expression
2πr(r+h)2\pi r\left(r+h\right)
for
r=6r=6
and
h=9h=9
.

S=2πr(r+h)=2π(6)[(6)+(9)]=2π(6)(15)=180π\begin{matrix} S\qquad&=2\pi r\left(r+h\right) \\ \qquad& =2\pi\left(6\right)[\left(6\right)+\left(9\right)] \\ \qquad& =2\pi\left(6\right)\left(15\right) \\ \qquad& =180\pi\end{matrix}
The surface area is
180π180\pi
square inches.

Try It 11

/ An art frame with a piece of artwork in the center. The frame has a width of 8 centimeters. The artwork itself has a length of 32 centimeters and a width of 24 centimeters. Figure 4


A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm2) is found to be
A=(L+16)(W+16)−L⋅WA=\left(L+16\right)\left(W+16\right)-L\cdot W
. Find the area of a matte for a photograph with length 32 cm and width 24 cm.

Solution

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