Squares and Square Roots | FIO

Question 2

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.

(i) 3

(ii) 4

(iii) 6

(iv) 8

(v) 9

Solution

Understand the Question

For an isosceles right triangle with equal legs of length $a$ , the hypotenuse $h$ is found using the Pythagoras theorem:

h = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}

To find bounds on $h$ $h$ with at least one digit after the decimal point:
1. Calculate $h^2 = 2a^2$ .
2. Find two consecutive whole numbers whose squares bracket $2a^2$ .
3. Test consecutive numbers with one decimal place ( $x.y$ and $x.y+0.1$ ) whose squares lie immediately below and above $2a^2$ .

(i) 3

Step 1 · Calculate Hypotenuse and Find Decimal Bounds

For $a = 3$ , the hypotenuse is $h = 3\sqrt{2}$ . Diagram 1

Squaring the hypotenuse:

\begin{aligned} h^2 &= (3\sqrt{2})^2 \\ &= 3^2 \times (\sqrt{2})^2 \\ &= 9 \times 2 \\ &= 18 \end{aligned}

Finding integer bounds: Since $4^2 = 16$ and $5^2 = 25$ , we have $16 < 18 < 25$ :

\begin{aligned} \sqrt{16} &< \sqrt{18} < \sqrt{25} \\ 4 &< 3\sqrt{2} < 5 \end{aligned}

Testing squares with one decimal place between $4$ and $5$ :

\begin{aligned} 4.1^2 &= 16.81 \\ 4.2^2 &= 17.64 \\ 4.3^2 &= 18.49 \end{aligned}

Since $17.64 < 18 < 18.49$ :

4.2 < 3\sqrt{2} < 4.3

Answer

(i) Hypotenuse $= 3\sqrt{2}$ , with bounds $4.2 < 3\sqrt{2} < 4.3$

(ii) 4

Step 1 · Calculate Hypotenuse and Find Decimal Bounds

For $a = 4$ , the hypotenuse is $h = 4\sqrt{2}$ .

Squaring the hypotenuse:

\begin{aligned} h^2 &= (4\sqrt{2})^2 \\ &= 4^2 \times (\sqrt{2})^2 \\ &= 16 \times 2 \\ &= 32 \end{aligned}

Finding integer bounds: Since $5^2 = 25$ and $6^2 = 36$ , we have $25 < 32 < 36$ :

\begin{aligned} \sqrt{25} &< \sqrt{32} < \sqrt{36} \\ 5 &< 4\sqrt{2} < 6 \end{aligned}

Testing squares with one decimal place between $5$ and $6$ :

\begin{aligned} 5.1^2 &= 26.01 \\ 5.2^2 &= 27.04 \\ 5.3^2 &= 28.09 \\ 5.4^2 &= 29.16 \\ 5.5^2 &= 30.25 \\ 5.6^2 &= 31.36 \\ 5.7^2 &= 32.49 \end{aligned}

Since $31.36 < 32 < 32.49$ :

5.6 < 4\sqrt{2} < 5.7

Answer

(ii) Hypotenuse $= 4\sqrt{2}$ , with bounds $5.6 < 4\sqrt{2} < 5.7$

(iii) 6

Step 1 · Calculate Hypotenuse and Find Decimal Bounds

For $a = 6$ , the hypotenuse is $h = 6\sqrt{2}$ .

Squaring the hypotenuse:

\begin{aligned} h^2 &= (6\sqrt{2})^2 \\ &= 6^2 \times (\sqrt{2})^2 \\ &= 36 \times 2 \\ &= 72 \end{aligned}

Finding integer bounds: Since $8^2 = 64$ and $9^2 = 81$ , we have $64 < 72 < 81$ :

\begin{aligned} \sqrt{64} &< \sqrt{72} < \sqrt{81} \\ 8 &< 6\sqrt{2} < 9 \end{aligned}

Testing squares with one decimal place between $8$ and $9$ :

\begin{aligned} 8.1^2 &= 65.61 \\ 8.2^2 &= 67.24 \\ 8.3^2 &= 68.89 \\ 8.4^2 &= 70.56 \\ 8.5^2 &= 72.25 \end{aligned}

Since $70.56 < 72 < 72.25$ :

8.4 < 6\sqrt{2} < 8.5

Answer

(iii) Hypotenuse $= 6\sqrt{2}$ , with bounds $8.4 < 6\sqrt{2} < 8.5$

(iv) 8

Step 1 · Calculate Hypotenuse and Find Decimal Bounds

For $a = 8$ , the hypotenuse is $h = 8\sqrt{2}$ .

Squaring the hypotenuse:

\begin{aligned} h^2 &= (8\sqrt{2})^2 \\ &= 8^2 \times (\sqrt{2})^2 \\ &= 64 \times 2 \\ &= 128 \end{aligned}

Finding integer bounds: Since $11^2 = 121$ and $12^2 = 144$ , we have $121 < 128 < 144$ :

\begin{aligned} \sqrt{121} &< \sqrt{128} < \sqrt{144} \\ 11 &< 8\sqrt{2} < 12 \end{aligned}

Testing squares with one decimal place between $11$ and $12$ :

\begin{aligned} 11.1^2 &= 123.21 \\ 11.2^2 &= 125.44 \\ 11.3^2 &= 127.69 \\ 11.4^2 &= 129.96 \end{aligned}

Since $127.69 < 128 < 129.96$ :

11.3 < 8\sqrt{2} < 11.4

Answer

(iv) Hypotenuse $= 8\sqrt{2}$ , with bounds $11.3 < 8\sqrt{2} < 11.4$

(v) 9

Step 1 · Calculate Hypotenuse and Find Decimal Bounds

For $a = 9$ , the hypotenuse is $h = 9\sqrt{2}$ .

Squaring the hypotenuse:

\begin{aligned} h^2 &= (9\sqrt{2})^2 \\ &= 9^2 \times (\sqrt{2})^2 \\ &= 81 \times 2 \\ &= 162 \end{aligned}

Finding integer bounds: Since $12^2 = 144$ and $13^2 = 169$ , we have $144 < 162 < 169$ :

\begin{aligned} \sqrt{144} &< \sqrt{162} < \sqrt{169} \\ 12 &< 9\sqrt{2} < 13 \end{aligned}

Testing squares with one decimal place between $12$ and $13$ :

\begin{aligned} 12.1^2 &= 146.41 \\ 12.2^2 &= 148.84 \\ 12.3^2 &= 151.29 \\ 12.4^2 &= 153.76 \\ 12.5^2 &= 156.25 \\ 12.6^2 &= 158.76 \\ 12.7^2 &= 161.29 \\ 12.8^2 &= 163.84 \end{aligned}

Since $161.29 < 162 < 163.84$ :

12.7 < 9\sqrt{2} < 12.8

Answer

(v) Hypotenuse $= 9\sqrt{2}$ , with bounds $12.7 < 9\sqrt{2} < 12.8$

Common Mistakes

Squaring Error: Miscalculating $(a\sqrt{2})^2$ as $2a$ or $a^2\sqrt{2}$ instead of $a^2 \times 2 = 2a^2$ .
Skipping Integer Bounds: Guessing decimal bounds directly without first determining the two consecutive whole numbers can lead to tedious trial-and-error.

Question 2

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