Question 2

We can find the length of the hypotenuse of a right-angled triangle using the Pythagorean theorem.

Step 1 — Finding the Hypotenuse Formula

The Pythagorean theorem tells us that in a right-angled triangle, the square of the longest side (called the hypotenuse) is equal to the sum of the squares of the other two sides (called legs). An isosceles right triangle has two equal legs and one right angle.

Let $a$ be the length of the two equal sides. Let $h$ be the length of the hypotenuse.

Using the Pythagorean theorem:

$h^2 = a^2 + a^2$

$h^2 = 2a^2$

To find $h$, we take the square root of both sides:

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

$h = a\sqrt{2}$

$\boxed{\text{Hypotenuse} = a\sqrt{2}}$

Step 2 — Calculating and Bounding for $a=3$

We use the formula $h = a\sqrt{2}$ to find the hypotenuse when $a=3$.

$h = 3\sqrt{2}$

To find the bounds, we first square the value of $h$:

$h^2 = (3\sqrt{2})^2$

$h^2 = 3^2 \times (\sqrt{2})^2$

$h^2 = 9 \times 2$

$h^2 = 18$

Now, we find two consecutive whole numbers whose squares are just below and just above 18.

We know that $4^2 = 16$ and $5^2 = 25$. Since $16 < 18 < 25$, we can say:

$\sqrt{16} < \sqrt{18} < \sqrt{25}$

$4 < 3\sqrt{2} < 5$

To find bounds with one decimal point, we check the squares of numbers between 4 and 5 that have one decimal place.

$4.1^2 = 16.81$

$4.2^2 = 17.64$

$4.3^2 = 18.49$

Since $17.64 < 18 < 18.49$, the hypotenuse $3\sqrt{2}$ lies between $\sqrt{17.64}$ and $\sqrt{18.49}$.

$\boxed{4.2 < 3\sqrt{2} < 4.3}$

Step 3 — Calculating and Bounding for $a=4$

We use the formula $h = a\sqrt{2}$ to find the hypotenuse when $a=4$.

$h = 4\sqrt{2}$

To find the bounds, we square the value of $h$:

$h^2 = (4\sqrt{2})^2$

$h^2 = 4^2 \times (\sqrt{2})^2$

$h^2 = 16 \times 2$

$h^2 = 32$

Now, we find two consecutive whole numbers whose squares are just below and just above 32.

We know that $5^2 = 25$ and $6^2 = 36$. Since $25 < 32 < 36$, we can say:

$\sqrt{25} < \sqrt{32} < \sqrt{36}$

$5 < 4\sqrt{2} < 6$

To find bounds with one decimal point, we check the squares of numbers between 5 and 6 that have one decimal place.

$5.1^2 = 26.01$

$5.2^2 = 27.04$

$5.3^2 = 28.09$

$5.4^2 = 29.16$

$5.5^2 = 30.25$

$5.6^2 = 31.36$

$5.7^2 = 32.49$

Since $31.36 < 32 < 32.49$, the hypotenuse $4\sqrt{2}$ lies between $\sqrt{31.36}$ and $\sqrt{32.49}$.

$\boxed{5.6 < 4\sqrt{2} < 5.7}$

Step 4 — Calculating and Bounding for $a=6$

We use the formula $h = a\sqrt{2}$ to find the hypotenuse when $a=6$.

$h = 6\sqrt{2}$

To find the bounds, we square the value of $h$:

$h^2 = (6\sqrt{2})^2$

$h^2 = 6^2 \times (\sqrt{2})^2$

$h^2 = 36 \times 2$

$h^2 = 72$

Now, we find two consecutive whole numbers whose squares are just below and just above 72.

We know that $8^2 = 64$ and $9^2 = 81$. Since $64 < 72 < 81$, we can say:

$\sqrt{64} < \sqrt{72} < \sqrt{81}$

$8 < 6\sqrt{2} < 9$

To find bounds with one decimal point, we check the squares of numbers between 8 and 9 that have one decimal place.

$8.1^2 = 65.61$

$8.2^2 = 67.24$

$8.3^2 = 68.89$

$8.4^2 = 70.56$

$8.5^2 = 72.25$

Since $70.56 < 72 < 72.25$, the hypotenuse $6\sqrt{2}$ lies between $\sqrt{70.56}$ and $\sqrt{72.25}$.

$\boxed{8.4 < 6\sqrt{2} < 8.5}$

Step 5 — Calculating and Bounding for $a=8$

We use the formula $h = a\sqrt{2}$ to find the hypotenuse when $a=8$.

$h = 8\sqrt{2}$

To find the bounds, we square the value of $h$:

$h^2 = (8\sqrt{2})^2$

$h^2 = 8^2 \times (\sqrt{2})^2$

$h^2 = 64 \times 2$

$h^2 = 128$

Now, we find two consecutive whole numbers whose squares are just below and just above 128.

We know that $11^2 = 121$ and $12^2 = 144$. Since $121 < 128 < 144$, we can say:

$\sqrt{121} < \sqrt{128} < \sqrt{144}$

$11 < 8\sqrt{2} < 12$

To find bounds with one decimal point, we check the squares of numbers between 11 and 12 that have one decimal place.

$11.1^2 = 123.21$

$11.2^2 = 125.44$

$11.3^2 = 127.69$

$11.4^2 = 129.96$

Since $127.69 < 128 < 129.96$, the hypotenuse $8\sqrt{2}$ lies between $\sqrt{127.69}$ and $\sqrt{129.96}$.

$\boxed{11.3 < 8\sqrt{2} < 11.4}$

Step 6 — Calculating and Bounding for $a=9$

We use the formula $h = a\sqrt{2}$ to find the hypotenuse when $a=9$.

$h = 9\sqrt{2}$

To find the bounds, we square the value of $h$:

$h^2 = (9\sqrt{2})^2$

$h^2 = 9^2 \times (\sqrt{2})^2$

$h^2 = 81 \times 2$

$h^2 = 162$

Now, we find two consecutive whole numbers whose squares are just below and just above 162.

We know that $12^2 = 144$ and $13^2 = 169$. Since $144 < 162 < 169$, we can say:

$\sqrt{144} < \sqrt{162} < \sqrt{169}$

$12 < 9\sqrt{2} < 13$

To find bounds with one decimal point, we check the squares of numbers between 12 and 13 that have one decimal place.

$12.1^2 = 146.41$

$12.2^2 = 148.84$

$12.3^2 = 151.29$

$12.4^2 = 153.76$

$12.5^2 = 156.25$

$12.6^2 = 158.76$

$12.7^2 = 161.29$

$12.8^2 = 163.84$

Since $161.29 < 162 < 163.84$, the hypotenuse $9\sqrt{2}$ lies between $\sqrt{161.29}$ and $\sqrt{163.84}$.

$\boxed{12.7 < 9\sqrt{2} < 12.8}$

Answer

(i) The hypotenuse is $3\sqrt{2}$, and its length is between 4.2 and 4.3. (ii) The hypotenuse is $4\sqrt{2}$, and its length is between 5.6 and 5.7. (iii) The hypotenuse is $6\sqrt{2}$, and its length is between 8.4 and 8.5. (iv) The hypotenuse is $8\sqrt{2}$, and its length is between 11.3 and 11.4. (v) The hypotenuse is $9\sqrt{2}$, and its length is between 12.7 and 12.8.