Julia is training for the Pedal for Heart and Stroke charity ride. Participants can ride a 10 km, 25 km or a 50 km route. Julia has decided to ride the 25 km route. She looks at the wheels on her bicycle and wonders if the new tires will last for all the times they will have to go around to complete this ride. Julia knows the tire warranty guarantees 25 000 rotations.

If her bicycle tire has a diameter of 61 cm, does the tire warranty cover enough rotations to get her to the end of the 25 km route?

In this concept, you will learn to approximate solutions to equations involving irrational numbers.

### Irrational Numbers

An **irrational number** is one that cannot be written as the ratio of two integers and is a non-terminating, non-repeating decimal. Most scientific calculators will display non-terminating decimals to nine places after the decimal point. If you were to use all of these decimal places when solving an equation involving irrational numbers, then your answer would be more accurate than if you rounded the decimal to the nearest hundredth. However, your answer would still be an approximation and not an accurate answer.

It is acceptable to round irrational numbers to the nearest hundredth or thousandth when solving equations. The answer to the equation should not include an equal sign but rather the symbol

which means “is approximately equal to.” If the value for the variable is “approximately equal to” use this same symbol when verifying the answer using the original equation. Both sides of the equation may not be equal but their values will be very close to being equal.### Examples

#### Example 1

Earlier, you were given a problem about Julia and her 25 km bicycle ride. Julia needs to find out how many rotations her tires will make over the 25 km ride to determine if the warranty will last until the end of the ride.

First, calculate the circumference of the bicycle tire.

The diameter of the tire is 61 cm. The circumference of the tire can be found using the formula:

\begin{align*}\begin{array}{rcl}
C &=& \pi d\\
C &=& (3.14)(61)\\
C &\approx& 191.45 \ cm
\end{array}\end{align*}

Next, convert the circumference of the tire from centimeters to meters using proportions.

\begin{align*}\begin{array}{rcl} \frac{100 \ cm}{191.54 \ cm} &=& \frac{1 \ m}{x}\\ \\ \frac{100 \ \cancel{cm}}{191.54 \ \cancel{cm}} &=& \frac{1 \ m}{x}\\ \\ 100 x &=& 191.54\\ \\ \frac{\overset{{1}}{\cancel{100}} x}{\cancel{100}} &=& \frac{191.54}{100}\\ \\ x &\approx& 1.9154 \ m \end{array}\end{align*}

Next, convert the circumference of the tire from meters to kilometers using proportions.

\begin{align*}\begin{array}{rcl} \frac{1000 \ m}{1.9154 \ m} &=& \frac{1 \ km}{x}\\ \\ \frac{1000 \ \cancel{m}}{1.9154 \ \cancel{m}} &=& \frac{1 \ km}{x}\\ \\ 1000 x &=& 1.9154\\ \\ \frac{\overset{{1}}{\cancel{1000}} x}{\cancel{1000}} &=& \frac{1.9154}{1000}\\ \\ x &\approx& 0.0019154 \ km \end{array}\end{align*}

Then, divide the length of the race by the circumference of the tire to determine the number of rotations. Round you answer to the nearest whole number.

\begin{align*}\begin{array}{rcl} \text{Number of rotations} &=& \frac{25 \ km}{0.0019154 \ km}\\ \\ \text{Number of rotations} &=& \frac{25 \ \cancel{km}}{0.0019154 \ \cancel{km}}\\ \\ \text{Number of rotations} &=& 13,052.104\\ \\ \text{Number of rotations} & \approx & 13,052 \end{array}\end{align*}

The answer is approximately equal to 13,052 rotations.

The tire warranty guarantees 25 000 rotations. Julia and her tires will make it to the end of the ride.

#### Example 2

You can calculate the area of a circle by multiplying the radius of the circle (the distance from the center to the edge) by π (pi), which is an irrational number approximately equal to 3.14159265359. Calculations would be inconveniently complex if you used that many decimal places, so pi is commonly rounded to 3.14.

Calculate the approximate area of a circle with a radius of 5 inches. Use Area = pi times radius squared, and estimate pi to two decimal places.

\begin{align*}A= & \pi \times r^2\\\end{align*}

First, substitute 5 in place of the \begin{align*}r\end{align*} in the equation.

\begin{align*} A = & \pi \times (5)^2\end{align*}

Next, simplify the radius squared. Remember that 'squared' means 'multiplied by itself'.

\begin{align*}A = & \pi \times 25\end{align*}

Then, approximate pi as 3.14 and change the equals sign to 'approximately equal'.

\begin{align*}A \approx & 3.14 \times 25\end{align*}

Finally, multiply.

\begin{align*}A \approx 78.5\end{align*}

The area of the circle is approximately 78.5 square inches.

#### Example 3

Find the approximate volume of a pop can with a radius of 2 inches and height of 5 inches. Use the formula where *V* is the volume, *r* is the radius, and *h* is the height of the cylinder. Approximate pi as 3.14. The volume will be in inches cubed.

First, substitute 2 for *r*, and 5 for *h* into the formula.

\begin{align*}V= \pi \times 2^2 \times 5\end{align*}

Next, simplify the numbers.

\begin{align*}V= & \pi \times 2^2 \times 5\\ V = & \pi \times 4 \times 5\\ V = & \pi \times 20\\\end{align*}

Then, approximate pi as 3.14, and multiply.

\begin{align*}V = & \pi \times 20\\ V \approx & 3.14 \times 20\\ V \approx & 62.8\\\end{align*}

The volume of the can is approximately 62.8 inches cubed.

#### Example 4

You can calculate the diagonal of a square by multiplying the length of one of the sides by

, which can be approximated as 1.414. What is the approximate measure of the diagonal of a square with sides 6 inches long?First, write the formula for the diagonal of a square, using *s* for the length of a side, and *d* for the diagonal.

\begin{align*}d= \sqrt2 \times s\end{align*}

Next, substitute 6 for *s* and 1.414 for the square root of 2.

\begin{align*}d= & \sqrt2 \times s\\ d = & \sqrt2 \times 6\\ d \approx & 1.414 \times 6\\\end{align*}

Finally, multiply.

\begin{align*}d \approx & 1.414 \times 6\\ d \approx & 8.484\end{align*}

The diagonal of the square is approximately 8.484 inches.

### Review

Approximate the solution for each equation involving irrational numbers.

- \begin{align*}a : a = 3\pi \end{align*}
- \begin{align*}x : x = 8\pi \end{align*}
- \begin{align*}b:b=9\pi \end{align*}
- \begin{align*}c:c=12\pi \end{align*}
- \begin{align*}a:a=2\pi \end{align*}
- \begin{align*}y:y=6\pi \end{align*}
- \begin{align*}b:b=7\pi \end{align*}
- \begin{align*}d:d=12\pi -6\end{align*}
- \begin{align*}a:a=14\pi -9\end{align*}
- \begin{align*}x:x=11\pi -5\end{align*}
- \begin{align*}\sqrt{2} +5 = x\end{align*}
- \begin{align*}8=\sqrt{2} + x\end{align*}
- \begin{align*}t=\pi -5.3\end{align*}
- \begin{align*}\sqrt{h} = \sqrt{6} - \frac{3}{4}\end{align*}
- Mrs. DeFazio wrote the following equation on the board: \begin{align*}w=\sqrt{11}-2^2\end{align*} What is the value of ‘w’ in Mrs. DeFazio’s equation?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.5.