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 sides of length $a$ and hypotenuse $h$ , we use the Pythagoras theorem:

\begin{aligned} h^2 &= a^2 + a^2 = 2a^2 \\ h &= \sqrt{2a^2} = a\sqrt{2} \end{aligned}

To find decimal bounds for the hypotenuse, we square $h$ to get $h^2 = 2a^2$ , find the two consecutive integers bounding it, and then test the squares of tenths (one decimal place) to sandwich the value between two consecutive tenths.

(i) Find the length of the hypotenuse and its bounds for side length 3.

Step 1 · Calculate Hypotenuse and Find Bounds for $a = 3$

Diagram 1

Length of hypotenuse: $h = 3\sqrt{2}$

Squaring the hypotenuse:

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

Since $4^2 = 16$ and $5^2 = 25$ , we have $16 < 18 < 25$ , so: $4 < 3\sqrt{2} < 5$

Testing squares of tenths between $4$ and $5$ :

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

Since $17.64 < 18 < 18.49$ , taking square roots gives: $4.2 < 3\sqrt{2} < 4.3$

Answer

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

(ii) Find the length of the hypotenuse and its bounds for side length 4.

Step 1 · Calculate Hypotenuse and Find Bounds for $a = 4$

Length of hypotenuse: $h = 4\sqrt{2}$

Squaring the hypotenuse:

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

Since $5^2 = 25$ and $6^2 = 36$ , we have $25 < 32 < 36$ , so: $5 < 4\sqrt{2} < 6$

Testing squares of tenths 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 $31.36 < 32 < 32.49$ , taking square roots gives: $5.6 < 4\sqrt{2} < 5.7$

Answer

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

(iii) Find the length of the hypotenuse and its bounds for side length 6.

Step 1 · Calculate Hypotenuse and Find Bounds for $a = 6$

Length of hypotenuse: $h = 6\sqrt{2}$

Squaring the hypotenuse:

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

Since $8^2 = 64$ and $9^2 = 81$ , we have $64 < 72 < 81$ , so: $8 < 6\sqrt{2} < 9$

Testing squares of tenths 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 $70.56 < 72 < 72.25$ , taking square roots gives: $8.4 < 6\sqrt{2} < 8.5$

Answer

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

(iv) Find the length of the hypotenuse and its bounds for side length 8.

Step 1 · Calculate Hypotenuse and Find Bounds for $a = 8$

Length of hypotenuse: $h = 8\sqrt{2}$

Squaring the hypotenuse:

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

Since $11^2 = 121$ and $12^2 = 144$ , we have $121 < 128 < 144$ , so: $11 < 8\sqrt{2} < 12$

Testing squares of tenths 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 $127.69 < 128 < 129.96$ , taking square roots gives: $11.3 < 8\sqrt{2} < 11.4$

Answer

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

(v) Find the length of the hypotenuse and its bounds for side length 9.

Step 1 · Calculate Hypotenuse and Find Bounds for $a = 9$

Length of hypotenuse: $h = 9\sqrt{2}$

Squaring the hypotenuse:

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

Since $12^2 = 144$ and $13^2 = 169$ , we have $144 < 162 < 169$ , so: $12 < 9\sqrt{2} < 13$

Testing squares of tenths 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 $161.29 < 162 < 163.84$ , taking square roots gives: $12.7 < 9\sqrt{2} < 12.8$

Answer

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

Common Mistakes

Squaring Error: Forgetting to square both the integer and the root when calculating $(a\sqrt{2})^2 = a^2 \times 2$ , mistakenly writing it as $2a$ or $a^2 + 2$ .
Incorrect Bounds: Choosing decimal values whose squares do not strictly straddle the target value (e.g., stopping before finding the upper bound that exceeds $h^2$ ).

Question 2

More questions in FIO