Exercises: Square Roots

First you have to calculate the area of the circle so that you know how big the area of the square has to be.

You are given the radius $r=13.6cm$.

Calculate the area using the formula.

Determining the side length of the square

You already know:

$A_{\mathrm{circle}}=A_{\mathrm{square}}=581.07c{m}^2$

Now use the formula for the area of a square:

You get two solutions:

In the context of the square, only one of the solutions makes sense, namely the positive one.

So the correct solution is $s=24.11cm$.

The method of nested intervals is a way to approximately compute square roots.

As a first step, you have to find an interval in which the value of $\sqrt{7}$ is definitely included. For this, it is useful to know the nearest smaller and larger square numbers.

Step 1

You already know the values of:

$\sqrt{4}$ and $\sqrt{9}$

This gives you the following estimates:

$\begin{array}{ccccc}{2}^2& <& 7& <& 3^2\\ 4& <& 7& <& 9\end{array}$

Taking the square root, this implies:

So the value of $\sqrt{7}$ is contained in the interval:

The next steps always follow the same algorithm:

First you find the mean value of the old interval boundaries. This mean value will become one of the new interval boundaries. The other new interval boundary will be taken from the old interval.

In order to decide whether the new value is the upper or lower interval boundary, you proceed as follows: Take the square of the mean value and compare it with 7.

• If its square is greater than 7, then the mean value becomes the new upper boundary.

• If its square is greater than 7, then the mean value becomes the new lower boundary.

Step 2

Calculate the mean value of the interval boundaries of $I_0$.

${\displaystyle \frac{2+3}{2}}=2.5$

Calculate the square of $2.5$ and compare it with $7$.

${2.5}^2=6.25$

$6.25<7<9$

${2.5}^2<7<3^2$

Now define the new interval.

Step 3

$I_1=(2.5,3)$

Calculate the mean value of the interval boundaries of $I_1$.

Calculate the square of this mean value and compare it with $7$.

$\begin{array}{ccccc}6.25& <& 7& <& 7.56\\ {2.5}^2& <& 7& <& {2.75}^2\end{array}$

Now define the new interval.

Step 4

Calculate the mean value of the interval boundaries of $I_2$.

Calculate the square of $2.625$ and compare it with $7$.

$\begin{array}{ccccc}6.89& <& 7& <& 7.56\\ {2.625}^2& <& 7& <& {2.75}^2\end{array}$

Now define the new interval.

Step 5

Calculate the mean value of the interval boundaries of $I_3$.

Calculate the square of this mean value and compare it with $7$.

Now define the new interval.

After these five steps, we know that

So the value of $\sqrt{7}$ is between $2.625$ and $2.6875$.

Final Estimate

Depending on what precision you need, you can repeat the procedure as often as you need Here, the problem setting tells us to terminate after 5 steps. So the final interval where you locate $\sqrt{7}$ is $I_4=(2.625,2.6875)$. A good estimate for $\sqrt{7}$ is now given by the "middle of the final interval", i.e., the mean value of its boundaries:

So the final estimate for $\sqrt{7}$ is $2.65625$.

For comparison: The precise value of $\sqrt{7}$ is $2.64575\dots$

Heron's Method

Heron's method is an algorithm that allows you to estimate the value of square roots.

You try to find a square whose area is the number under the root (which is $7$, here). The side length is then the square (which is $\sqrt{7}$, here).

To find the square, you start with a rectangle of area $A=7$ and change its shape to be more "square-like" step by step.

Step 1

Find the first rectangle by picking two side lengths $a$ and $b$, such that the area of the rectangle is

$A=a\cdot b$

In our case, $A=7$ so it follows to choose

$a_0=7$

$b_0=1$

In the figure below you can see the initial rectangle.

The next steps always follow the same pattern:

• First you calculate the mean value of the old side lengths, which will be one of your new side lengths, called $a$.

• Then you calculate the other side length $b$ by dividing $\frac{A}{a}$ with $A$ being the fixed area.

You can find a detailed explanation of these steps in this applet.

Step 2

Calculate the mean value of $a_0$ and $b_0$.

$a_1={\displaystyle \frac{a_0+b_0}{2}}={\displaystyle \frac{7+1}{2}}=4$

Then calculate $b_1$.

$b_1={\displaystyle \frac{A}{a_1}}={\displaystyle \frac{\text{radicand}}{a_1}}={\displaystyle \frac{7}{4}}=1.75$

Your new side lengths are

$a_1=4$

$b_1=1.75$

In the figure below you can see the original rectangle (red) and the new one (blue), which has now become a bit more square-like.

Step 3

Calculate the mean value of $a_1$ and $b_1$.

$a_2={\displaystyle \frac{a_1+b_1}{2}}={\displaystyle \frac{4+1.75}{2}}=2.875$

Then calculate $b_2$.

$\begin{array}{l}{b}_2={\displaystyle \frac{A}{a_2}}={\displaystyle \frac{\text{radicand}}{a_2}}\\ ={\displaystyle \frac{7}{2.875}}=2.4348\end{array}$

Your new side lengths are

In the figure below you can see the new rectangle in orange.

Step 4

Calculate the mean value of $a_2$ and $b_2$.

$a_3={\displaystyle \frac{a_2+b_2}{2}}={\displaystyle \frac{2.875+2.4348}{2}}=2.6548$

Then calculate $b_3$.

$\begin{array}{l}{b}_3={\displaystyle \frac{A}{a_3}}={\displaystyle \frac{\text{radicand}}{a_3}}\\ ={\displaystyle \frac{7}{2.6548}}=2.6367\end{array}$

Your new side lengths are $a_3=2.6548$ and $b_3=2.6367$.

You may notice these two values are not that different anymore.

In the figurebelow you can see the change in the rectangles. The last one (purple) is already almost a square.

Final Estimate

You could do as many further steps as you wish in order to get a more precise result. Now, the problem setting tells us to do exactly four steps, so we will stop here.

Since $a_3=2.6548$ and $b_3=2.6367$, the value of $\sqrt{7}$ must be between these two numbers. So a good estimate for $\sqrt{7}$ is given by the mean value of the two numbers:

For comparison: The precise value of $\sqrt{7}$ is $2.64575\dots$

$37$ lies between the square numbers $36=6^2$ and $49=7^2$.

So $\sqrt{37}$ lies between $\sqrt{36}=6$ and $\sqrt{49}=7$.

For the calculation of the side length of the square you need the area $A_{\mathrm{square}}$ of the square. The rectangle and the square have the same area, so you may calculate the area of the rectangle $A_{\mathrm{rectangle}}$, first:

$\begin{array}{rcl}{A}_{\mathrm{rectangle}}& =& \mathrm{length}\cdot \mathrm{width}\\ & & \\ & =& c\cdot b\\ & & \\ & =& 8\mathrm{cm}\cdot 2\mathrm{cm}\\ & & \\ & =& 16{\mathrm{cm}}^2\end{array}$

The area of the square $A_{\mathrm{square}}$ is $16{\mathrm{cm}}^2$. With this information, you may calculate the side length $a$ of the square:

$\begin{array}{rcl}\mathrm{length}\cdot \mathrm{width}& =& A_{\mathrm{square}}\\ & & \\ a\cdot a& =& 16{\mathrm{cm}}^2\\ & & \\ a^2& =& 16{\mathrm{cm}}^2\\ & & \\ \sqrt{a^2}& =& \sqrt{16{\mathrm{cm}}^2}\\ & & \\ a& =& 4\mathrm{cm}\end{array}$

The side lengths of the square thus have a length of $4\mathrm{cm}$.

The total picture should have a size of $63{\mathrm{cm}}^2$. The area of a single mosaic tile can be calculated by dividing the area of the complete cat picture by the number of tiles.

$\begin{array}{rcl}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}& =& 7\cdot 9\\ & & \\ & =& 63\end{array}$

The cat picture is made up of $63$ mosaic tiles.

$\begin{array}{rcl}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}& =& \mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}÷\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}\\ & & \\ & =& 63{\mathrm{cm}}^2÷63\\ & & \\ & =& 1{\mathrm{cm}}^2\end{array}$

A single tile has the size of $1{\mathrm{cm}}^2$.

In order to compute the side length, use the respective formula for the area. Since the tiles are square-shaped, the length is equal to the width.

$\begin{array}{rcl}\mathrm{length}\cdot \mathrm{width}& =& \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\\ & & \\ a\cdot a& =& \mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\\ & & \\ a^2& =& 1{\mathrm{cm}}^2\\ & & \\ \sqrt{a^2}& =& \sqrt{1{\mathrm{cm}}^2}\\ & & \\ a& =& 1\mathrm{cm}\end{array}$

For the cat picture, Tina needs mosaic tiles with a side length of $1\mathrm{cm}$.

First, you replace 102 by 100.

Here, ${100}^2=100\cdot 100=10000$

and $\sqrt{100}=\sqrt{{10}^2}=10$

${102}^2$ is therefore slightly greater than $10000$ and $\sqrt{100}$ is slightly greater than $10$.

What we are looking for is the result of $\sqrt{56}$, i.e. a number that, if you multiply by itself, becomes $56$. Now, $56$ is not a square number. But there are square numbers that are close to $56$.

$49=7\cdot 7$

$$$64=8\cdot 8$

$56$ lies between the square numbers $49$ and $64$, which means that the value of $\sqrt{56}$ must lie between $7$ and $8$.

The square numbers $49$ and $64$ are approximately equidistant from $56$. For this reason, the value of the root is close to $7.5$.

$\sqrt{56}\approx 7.5$

In this exercise, for example, you can first calculate the missing area values and then the missing side lengths.

Calculating the areas

You can determine the area of a square usingthe formula:

Now use the formula to calculate the missing area values:

$a=4\text{cm}\Rightarrow A_{\square }=(4\text{cm}{)}^2=16{\text{cm}}^2$

$a=1\text{m}\Rightarrow A_{\square }=(1\text{m}{)}^2=1{\text{m}}^2$

$a=7\text{cm}\Rightarrow A_{\square }=(7\text{cm}{)}^2=49{\text{cm}}^2$

$a=10\text{mm}\Rightarrow A_{\square }=(10\text{mm}{)}^2=100{\text{mm}}^2$

Calculating of the side lengths

To get the missing side lengths, you need to find out, which number, when multiplied by itself, becomes the corresponding area.

So, for example, "Which number times, if multiplied by itself, becomes $49$?" $\to$ $7\cdot 7=49$, so the side length you are looking for is $7\text{m}$.

$\begin{array}{l}{A}_{\square }=49{\text{m}}^2\Rightarrow 7\cdot 7=49\\ \Rightarrow a=7\text{m}\end{array}$

$\begin{array}{l}{A}_{\square }=81{\text{cm}}^2\Rightarrow 9\cdot 9=81\\ \Rightarrow a=9\text{cm}\end{array}$

$\begin{array}{l}{A}_{\square }=121{\text{dm}}^2\Rightarrow 11\cdot 11=121\\ \Rightarrow a=11\text{dm}\end{array}$

$\begin{array}{l}{A}_{\square }=25{\text{cm}}^2\Rightarrow 5\cdot 5=25\\ \Rightarrow a=5\text{cm}\end{array}$

Solution

Calculating of the Side Length of the $\text{C}\text{o}\text{l}\text{o}\text{u}\text{r}\text{e}\text{d}\text{S}\text{q}\text{u}\text{a}\text{r}\text{e}$

The formula for the area of a square with side length $a$ reads:

You know that the area of the coloured square has to be $9{\text{ cm}}^2$.

In order to get the side length, you have to find out, which side length becomes $9{\text{ cm}}^2$, when multiplied by itself.

The coloured square has a side length of $a=3\text{ cm}$.

Determining the Perimeter of the Rectangle

The formula for the perimeter of a rectangle with length $l$ and the width $b$ is:

In the figure you can see that the length of the rectangle is 6 times the side length of the square and the width of the rectangle is 4 times the side length of the square.

Now you can substitute the length and the width into the formula for the perimeter of the rectangle. This gives you the solution:

$P_{\text{rectangle}}=2\cdot l+2\cdot b=2\cdot 18\text{ cm}+2\cdot 12\text{ cm}=60\text{ cm}$

The perimeter of the rectangle is thus $60\text{ cm}$.

The next square numbers with respect to 20 are:

The number 20 lies between the square numbers 16 and 25, so you know about the square root of 20 that

I.e., $\sqrt{20}$ lies between the natural numbers $4$ and $5$.

Since the square numbers 16 and 25 have approximately the same distance to 20, you know that the number $\sqrt{20}$ is approximately in the middle of 4 and 5.

So, the square root of 20 is approximately 4.5.