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 $r=13.6cmr=\; 13.6cm$.

Calculate the area using the formula.

Determining the side length of the square

You already know:

$A_{\mathrm{c}\mathrm{i}\mathrm{r}\mathrm{c}\mathrm{l}\mathrm{e}}=A_{\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{e}}=581.07c{m}^{2}A\_\{\backslash rm\; circle\}\; =\; A\_\{\backslash rm\; square\}\; =\; 581.07cm^2$

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 $s=24.11cms\; =\; 24.11cm$.

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 $\sqrt{7}\backslash sqrt\{7\}$ 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:

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

This gives you the following estimates:

Taking the square root, this implies:

So the value of $\sqrt{7}\backslash sqrt\{7\}$ 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 $I_{0}I\_0$.

${\displaystyle \frac{2+3}{2}}=2.5\backslash dfrac\{2+3\}\{2\}=2\{.\}5$

Calculate the square of $2.52.5$ and compare it with $77$.

$2.5^2=6.252\{.\}5^2\; =\; 6\{.\}25$

Now define the new interval.

Step 3

$I_1=(2.5,3)I\_1\; =\; (2\{.\}5,\; 3)$

Calculate the mean value of the interval boundaries of $I_{1}I\_1$.

Calculate the square of this mean value and compare it with $77$.

Now define the new interval.

Step 4

Calculate the mean value of the interval boundaries of $I_{2}I\_2$.

Calculate the square of $2.6252.625$ and compare it with $77$.

Now define the new interval.

Step 5

Calculate the mean value of the interval boundaries of $I_{3}I\_3$.

Calculate the square of this mean value and compare it with $77$.

Now define the new interval.

After these five steps, we know that

So the value of $\sqrt{7}\backslash sqrt\{7\}$ is between $2.6252.625$ and $2.68752.6875$.

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 $\sqrt{7}\backslash sqrt\{7\}$ is $I_4=(2.625,2.6875)I\_4\; =\; (\; 2\{.\}625,\; 2\{.\}6875\; )$. A good estimate for $\sqrt{7}\backslash sqrt\{7\}$ is now given by the "middle of the final interval", i.e., the mean value of its boundaries:

So the final estimate for $\sqrt{7}\backslash sqrt\{7\}$ is $2.656252\{.\}65625$.

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

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 $77$, here). The side length is then the square (which is $\sqrt{7}\backslash sqrt\{7\}$, here).

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

Step 1

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

$A=a\cdot bA=a\backslash cdot\; b$

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

$a_0=7a\_0=7$

$b_0=1b\_0=1$

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 $aa$.

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

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

Step 2

Calculate the mean value of $a_{0}a\_0$ and $b_{0}b\_0$.

$a_1={\displaystyle \frac{a_0+b_0}{2}}={\displaystyle \frac{7+1}{2}}=4a\_1\; =\; \backslash dfrac\{a\_0+b\_0\}\{2\}\; =\; \backslash dfrac\{7+1\}\{2\}=4$

Then calculate $b_{1}b\_1$.

$b_1={\displaystyle \frac{A}{a_1}}={\displaystyle \frac{\text{radicand}}{a_1}}={\displaystyle \frac{7}{4}}=1.75b\_1\; =\; \backslash dfrac\{A\}\{a\_1\}\; =\; \backslash dfrac\{\backslash text\{radicand\}\}\{a\_1\}\; =\; \backslash dfrac\{7\}\{4\}\; =\; 1.75$

Your new side lengths are

$a_1=4a\_1=4$

$b_1=1.75b\_1=1.75$

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 $a_{1}a\_1$ and $b_{1}b\_1$.

$a_2={\displaystyle \frac{a_1+b_1}{2}}={\displaystyle \frac{4+1.75}{2}}=2.875a\_2\; =\; \backslash dfrac\{a\_1+b\_1\}\{2\}\; =\; \backslash dfrac\{4+1.75\}\{2\}=2.875$

Then calculate $b_{2}b\_2$.

$b_2={\displaystyle \frac{A}{a_2}}={\displaystyle \frac{\text{radicand}}{a_2}}={\displaystyle \frac{7}{2.875}}=2.4348b\_2\; =\; \backslash dfrac\{A\}\{a\_2\}\; =\; \backslash dfrac\{\backslash text\{radicand\}\}\{a\_2\}\backslash \backslash \; \backslash ;\; \backslash ;\; \backslash ;\; =\; \backslash dfrac\{7\}\{2.875\}\; =\; 2.4348$

Your new side lengths are

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

Step 4

Calculate the mean value of $a_{2}a\_2$ and $b_{2}b\_2$.

$a_3={\displaystyle \frac{a_2+b_2}{2}}={\displaystyle \frac{2.875+2.4348}{2}}=2.6548a\_3\; =\; \backslash dfrac\{a\_2+b\_2\}\{2\}\; =\; \backslash dfrac\{2.875+2.4348\}\{2\}=2.6548$

Then calculate $b_{3}b\_3$.

$b_3={\displaystyle \frac{A}{a_3}}={\displaystyle \frac{\text{radicand}}{a_3}}={\displaystyle \frac{7}{2.6548}}=2.6367b\_3\; =\; \backslash dfrac\{A\}\{a\_3\}\; =\; \backslash dfrac\{\backslash text\{radicand\}\}\{a\_3\}\backslash \backslash \; \backslash ;\; \backslash ;\; \backslash ;\; =\; \backslash dfrac\{7\}\{2.6548\}\; =\; 2.6367$

Your new side lengths are $a_3=2.6548a\_3=2.6548$ and $b_3=2.6367b\_3=2.6367$.

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 $a_3=2.6548a\_3=2.6548$ and $b_3=2.6367b\_3=2.6367$, the value of $\sqrt{7}\backslash sqrt\{7\}$ must be between these two numbers. So a good estimate for $\sqrt{7}\backslash sqrt\{7\}$ is given by the mean value of the two numbers:

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

$3737$ lies between the square numbers $36=6^{2}36\; =\; 6^2$ and $49=7^{2}49=7^2$.

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

For the calculation of the side length of the square you need the area $A_{\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{e}}A\_\{\backslash rm\; square\}$ of the square. The rectangle and the square have the same area, so you may calculate the area of the rectangle $A_{\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}}A\_\{\backslash rm\; rectangle\}$, first:

The area of the square $A_{\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{e}}A\_\{\backslash rm\; square\}$ is $16{\mathrm{c}\mathrm{m}}^{2}16\backslash ;\backslash mathrm\{cm\}^2$. With this information, you may calculate the side length $aa$ of the square:

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

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

The cat picture is made up of $6363$ mosaic tiles.

A single tile has the size of $1{\mathrm{c}\mathrm{m}}^{2}1\backslash ;\{\backslash rm\; cm\}^2$.

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.

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

First, you replace 102 by 100.

Here, $100^2=100\cdot 100=10000100^2\; =\; 100\; \backslash cdot100=10000$

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

$102^{2}102^2$ is therefore slightly greater than $1000010000$ and $\sqrt{100}\backslash sqrt\{100\}$ is slightly greater than $1010$.

What we are looking for is the result of $\sqrt{56}\backslash sqrt\{56\}$, i.e. a number that, if you multiply by itself, becomes $5656$. Now, $5656$ is not a square number. But there are square numbers that are close to $5656$.

$49=7\cdot 749\; =\; 7\; \backslash cdot\; 7$

$$$64=8\cdot 864\; =\; 8\; \backslash cdot\; 8$

$5656$ lies between the square numbers $4949$ and $6464$, which means that the value of $\sqrt{56}\backslash sqrt\{56\}$ must lie between $77$ and $88$.

The square numbers $4949$ and $6464$ are approximately equidistant from $5656$. For this reason, the value of the root is close to $7.57.5$.

$\sqrt{56}\approx 7.5\backslash sqrt\{56\}\backslash approx7.5$

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:

$a=4\text{cm}\Rightarrow A_{\mathrm{\square }}=(4\text{cm}{)}^2=16{\text{cm}}^{2}a=4\backslash text\{cm\}\; \backslash Rightarrow\; A\_\backslash square=(4\backslash text\{cm\})^2=16\backslash text\{cm\}^2$

$a=1\text{m}\Rightarrow A_{\mathrm{\square }}=(1\text{m}{)}^2=1{\text{m}}^{2}a=1\backslash text\{m\}\; \backslash Rightarrow\; A\_\backslash square=\; (1\backslash text\{m\})^2=1\backslash text\{m\}^2$

$a=7\text{cm}\Rightarrow A_{\mathrm{\square }}=(7\text{cm}{)}^2=49{\text{cm}}^{2}a=7\backslash text\{cm\}\; \backslash Rightarrow\; A\_\backslash square=(7\backslash text\{cm\})^2=49\backslash text\{cm\}^2$

$a=10\text{mm}\Rightarrow A_{\mathrm{\square }}=(10\text{mm}{)}^2=100{\text{mm}}^{2}a=10\backslash text\{mm\}\; \backslash Rightarrow\; A\_\backslash square=(10\backslash text\{mm\})^2=100\backslash text\{mm\}^2$

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 $4949$?" $\to \backslash rightarrow$ $7\cdot 7=497\backslash cdot7=49$, so the side length you are looking for is $7\text{m}7\backslash text\{m\}$.

$A_{\mathrm{\square }}=49{\text{m}}^2\Rightarrow 7\cdot 7=49\Rightarrow a=7\text{m}A\_\backslash square\; =\; 49\backslash text\{m\}^2\; \backslash Rightarrow\; 7\backslash cdot\; 7\; =49\; \backslash \backslash \; \backslash Rightarrow\; a=7\backslash text\{m\}$

$A_{\mathrm{\square }}=81{\text{cm}}^2\Rightarrow 9\cdot 9=81\Rightarrow a=9\text{cm}A\_\backslash square\; =\; 81\backslash text\{cm\}^2\; \backslash Rightarrow\; 9\backslash cdot\; 9\; =81\; \backslash \backslash \; \backslash Rightarrow\; a=9\backslash text\{cm\}$

$A_{\mathrm{\square }}=121{\text{dm}}^2\Rightarrow 11\cdot 11=121\Rightarrow a=11\text{dm}A\_\backslash square\; =\; 121\backslash text\{dm\}^2\; \backslash Rightarrow\; 11\backslash cdot\; 11\; =121\; \backslash \backslash \; \backslash Rightarrow\; a=11\backslash text\{dm\}$

$A_{\mathrm{\square }}=25{\text{cm}}^2\Rightarrow 5\cdot 5=25\Rightarrow a=5\text{cm}A\_\backslash square\; =\; 25\backslash text\{cm\}^2\; \backslash Rightarrow\; 5\backslash cdot\; 5\; =25\; \backslash \backslash \; \backslash Rightarrow\; a=5\backslash text\{cm\}$

Solution

Calculating of the Side Length of the $\text{Coloured Square}\backslash text\{\backslash color\{009999\}\{Coloured\; Square\}\}$

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

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

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

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

Determining the Perimeter of the Rectangle

The formula for the perimeter of a rectangle with length $ll$ and the width $bb$ 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:

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

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

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., $\sqrt{20}\backslash sqrt\{20\}$ lies between the natural numbers $44$ and $55$.

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

So, the square root of 20 is approximately 4.5.