Ontario 10 Applied (MFM2P) Formulas, substitution and equations II
Lesson We know that we calculate the area of a rectangle by multiplying the length by the width.

However, we can also write this statement algebraically. If we let the area be $A$A, the length be $l$l and the width be $b$b, then we can write the formula for the area of a rectangle as:

$A=l\times b$A=l×b or even just $A=lb$A=lb

Algebraic equations are a shorthand way of writing mathematical relationships and there are lots of examples of this in geometrical measurement. We can write algebraic equations for perimeter, area, volume and surface area, just to name a few.

Even though we are usually solving to find the subject of the equation, sometimes we may be asked to find one of the other variables. To do this, we need to change the subject of an equation to get the variable we are trying to find by itself on one side of the equation.

## Solving measurement equations

The process for solving measurement equations is just the same as solving regular equations:

1. Write the formula out.
2. Substitute in the values for the given variables.
3. Simplify and solve.

Let's see this process in action by looking at some examples.

#### Worked examples

##### Example 1

The perimeter of a square with side a is given by formula $P=4a$P=4a. Given that $a=5$a=5 cm find $P$P.

Think: We need to substitute in the known value of $a$a to find the value of $P$P.

Do:

 $P$P $=$= $4a$4a $=$= $4\times5$4×5 $=$= $20$20 cm

##### Example 2

$P=2\left(a+b\right)$P=2(a+b) is the formula that describes the perimeter of the rectangle. Find $b$b if $a=2$a=2 and $P=22$P=22.

Think: This time we have a value for $a$a and a value for $P$P. We'll need to rearrange the equation to make $b$b the subject.

Do:

 $P$P $=$= $2\left(a+b\right)$2(a+b) $22$22 $=$= $2\left(2+b\right)$2(2+b) $11$11 $=$= $2+b$2+b $9$9 $=$= $b$b $b$b $=$= $9$9 units

##### Example 3

Find the circumference of the circle shown, correct to $2$2 decimal places. Think: What is the formula for the circumference of a circle?

Do:

 $C$C $=$= $2\pi r$2πr $=$= $2\times\pi\times8$2×π×8 $=$= $16\pi$16π $=$= $50.2654$50.2654... $=$= $50.27$50.27 cm (2 d.p.)

#### Practice questions

##### Question 1

The surface area of a rectangular prism is given by formula $S=2\left(lw+wh+lh\right)$S=2(lw+wh+lh), where $l$l , $w$w and $h$h are the dimensions of the prism.

Given that a rectangular prism has a length of $8$8 cm, a width of $7$7 cm and a height of $9$9 cm, find its surface area.

##### Question 2

The volume of a cylinder is given by $V=\pi r^2h$V=πr2h.

Find the volume of the cylinder shown, rounding your answer to two decimal places. ### Outcomes

#### 10P.LR1.02

Determine the value of a variable in the first degree, using a formula