1. Square 8.5 by multiplying it by itself:

   $$8.5^2=8.5\times 8.5=72.25$$

2. Find $\sqrt{72.25}$ by asking, “What number squared gives 72.25?”

   $$\sqrt{72.25}=8.5$$

A right-angled triangle has shorter sides of 7.2 cm and 9.6 cm. Find the hypotenuse.

1. Let the hypotenuse be $x$.

2. Use Pythagoras, adding the squares of the shorter sides:

   $$x^2=7.2^2+9.6^2$$

3. Square and add:

   $$x^2=51.84+92.16=144$$

4. Square-root to find $x$:

   $$x=\sqrt{144}=12$$

5. The hypotenuse is 12 cm.

A right-angled triangle has hypotenuse 15 cm and one shorter side 8 cm. Find the other shorter side correct to 1 decimal place.

1. Let the missing shorter side be $x$.

2. Start with Pythagoras:

   $$15^2=8^2+x^2$$

3. Rearrange by subtracting $8^2$:

   $$x^2=15^2-8^2$$

4. Square and subtract:

   $$x^2=225-64=161$$

5. Square-root to find $x$:

   $$x=\sqrt{161}=12.688\dots$$

6. Rounded to 1 decimal place, the missing side is 12.7 cm.

A calculation gives a length of $18.3579\dots$ metres. Round it to 3 significant figures.

1. The first three significant figures are 1, 8 and 3.

2. Look at the next digit, which is 5, so round the 3 up to 4.

3. The rounded answer is 18.4 m.

A rectangle is 16 cm long and 9 cm wide. Find the length of its diagonal correct to 1 decimal place.

1. The length, width and diagonal form a right-angled triangle.

2. The diagonal is the hypotenuse, so add the squares:

   $$d^2=16^2+9^2$$

3. Square and add:

   $$d^2=256+81=337$$

4. Square-root:

   $$d=\sqrt{337}=18.357\dots$$

5. Rounded to 1 decimal place, the diagonal is 18.4 cm.

An isosceles triangle has equal sides of 13 cm and a base of 10 cm. Find its perpendicular height.

1. The height splits the base into two equal parts, so each half is 5 cm.

2. Use one half of the triangle. The 13 cm side is the hypotenuse.

3. Let the height be $h$:

   $$h^2=13^2-5^2$$

4. Square and subtract:

   $$h^2=169-25=144$$

5. Square-root:

   $$h=\sqrt{144}=12$$

6. The perpendicular height is 12 cm.

Two right-angled triangles share a side. In the first triangle, the hypotenuse is 20 m and one shorter side is 12 m. In the second triangle, the shared side and a side of 9 m form the shorter sides. Find the final hypotenuse correct to 3 significant figures.

1. First find the shared side. Let it be $s$.

2. In the first triangle, subtract because 20 m is the hypotenuse:

   $$s^2=20^2-12^2$$

3. Calculate $s$:

   $$s^2=400-144=256$$

4. Square-root:

   $$s=\sqrt{256}=16$$

5. Now use the second triangle. Let the final hypotenuse be $x$:

   $$x^2=16^2+9^2$$

6. Calculate and square-root:

   $$x=\sqrt{337}=18.357\dots$$

7. Correct to 3 significant figures, the final length is 18.4 m.

A ladder reaches 2.4 m up a wall. The base of the ladder is 80 cm from the wall. Find the length of the ladder.

1. Convert 80 cm to metres so the units match:

   $$80cm=0.8m$$

2. The wall and ground make a right angle. The ladder is the hypotenuse.

3. Let the ladder length be $L$:

   $$L^2=2.4^2+0.8^2$$

4. Square and add:

   $$L^2=5.76+0.64=6.4$$

5. Square-root:

   $$L=\sqrt{6.4}=2.529\dots$$

6. The ladder is about 2.53 m long.

A screen has diagonal 65 inches. Its length and width are in the ratio 16:9. Find the length and width correct to 1 decimal place.

1. Write the length as $16k$ and the width as $9k$.

2. Use Pythagoras with the diagonal as the hypotenuse:

   $$65^2=(16k{)}^2+(9k{)}^2$$

3. Square the ratio parts:

   $$4225=256k^2+81k^2$$

4. Combine like terms:

   $$4225=337k^2$$

5. Solve for $k$:

   $$k=\sqrt{\frac{4225}{337}}=3.542\dots$$

6. Find the two dimensions:

   $$16k=56.7\dots ,9k=31.8\dots$$

7. The length is 56.7 inches and the width is 31.9 inches, correct to 1 decimal place.