Let \(x\) represent the length of the square, in centimetres.

The diagonal splits the square into two right-angled triangles. The perpendicular sides are both \(x\) cm and the hypotenuse is 10 cm. So:

$$ \begin{eqnarray} x^2+x^2 &=& 10^2 \\[6pt] 2x^2 &=& 100\\[6pt] x^2 &=& 50 \end{eqnarray} $$

There is no need to square root here, because the area of the square is \(x^2\small.\)

So the area is 50 cm².

Perpendicular means at right angles.

Since 2019, the SQA has been very fussy about how answers to Converse of Pythagoras questions are set out. One way to make sure you get all the marks is to use a table, like this, with the square of the longest side in one column and the sum of the squares of the two shorter sides in the other column.

$$ \begin{array} {c|c} 52^2 & 28^2+45^2 \\ =2704 & =784+2025 \\ & =2809 \\ \end{array} $$

\(52^2\neq 28^2+45^2\small,\) so by the Converse of Pythagoras' Theorem, this is not a right-angled triangle. Therefore, the two shorter sides are not perpendicular.

If C is due north of B, then the angle at B is 90° and so Pythagoras' Theorem applies. Let's check if it really does:

$$ \begin{array} {c|c} 97^2 & 65^2 + 72^2 \\ =9409 & =4225+5184 \\ & =9409 \\ \end{array} $$

\(97^2=65^2+72^2\small,\) so by the Converse of Pythagoras' Theorem, \(\angle\)ABC \(= 90^\circ\small.\)

Ceebury is therefore due north of Beeville.

Perpendicular means at right angles, so we need to check whether or not Pythagoras' Theorem applies.

$$ \begin{array} {c|c} 8^2 & 7^2 + 4^2 \\ =64 & =49+16 \\ & =65 \\ \end{array} $$

\(8^2\neq 7^2+4^2\small,\) so by the Converse of Pythagoras' Theorem, \(\angle\)ACB \(\neq 90^\circ\small.\)

The wall is not perpendicular to the ground.

Space diagonal means the distance through the cuboid from one corner to the furthest away corner.

Pythagoras' Theorem extends easily into three dimensions:

$$ \begin{eqnarray} \small\textsf{AH}\normalsize ^2&=&\small\textsf{AB}\normalsize ^2+\small\textsf{BC}\normalsize ^2+\small\textsf{CH}\normalsize ^2\\[6pt] &=& 12^2+7^2+5^2\\[6pt] &=& 144+49+25\\[6pt] &=& 218\\[6pt] \small\textsf{AH}\normalsize &=& \sqrt{218}\\[6pt] &=& 14.7648...\\[6pt] &\approx& 14.8\,\small\textsf{cm (to 3 sig.fig.)} \end{eqnarray} $$

We use 3D Pythagoras to calculate the length of the space diagonal PM.

$$ \begin{eqnarray} \small\textsf{PM}\normalsize ^2&=& 40^2+40^2+70^2\\[6pt] &=& 1600+1600+4900\\[6pt] &=& 8100\\[6pt] \small\textsf{PM}\normalsize &=& \sqrt{8100}\\[6pt] &=& 90\ \small\textsf{cm} \end{eqnarray} $$

Now we compare the length of the umbrella to the length of the space diagonal.

\(85 \lt 90\) so Chris is correct that the umbrella will fit into the locker.

Side note: We think this question was quite badly flawed. It didn't just ask whether the umbrella will fit into the locker. It actually asked whether the umbrella will fit “from corner P to corner M”, which it will not, because it is too short to reach all the way from M up to P. So we would have answered: “Chris is not correct, because the umbrella is shorter than the space diagonal, so if one end is at M, the other end will not reach P.” We still don't know if we would have lost a mark for reading the wording of the question precisely!

First, we draw in a radius (PC or QC) to make a right-angled triangle. Either triangle, PAC or QAC, will work.

Let's choose to draw the line PC and use triangle PAC.

PC = BC = 15 cm (as they are radii) so AC = 27 – 15 = 12 cm.

Now we use Pythagoras to calculate AP:

$$ \begin{eqnarray} \small\textsf{AP}\normalsize ^2&=& \small\textsf{PC}\normalsize ^2-\small\textsf{AC}\normalsize ^2\\[6pt] &=& 15^2-12^2\\[6pt] &=& 225-144\\[6pt] &=& 81\\[6pt] \small\textsf{AP}\normalsize &=& \sqrt{81}\\[6pt] &=& 9\,\small\textsf{cm} \end{eqnarray} $$

Finally, PQ = 2AP, so PQ = 18 cm.

First, we need to construct our right angle triangle.

Let C be the mid-point of the chord ML, and think about the right-angled triangle OCL.

OL is a radius, so OL = 1.2 metres.

CL is half of ML, so CL = 1.8 ÷ 2 = 0.9 metres.

Now we use Pythagoras to calculate OC:

$$ \begin{eqnarray} \small\textsf{OC}\normalsize ^2&=& \small\textsf{OL}\normalsize ^2-\small\textsf{CL}\normalsize ^2\\[6pt] &=& 1.2^2-0.9^2\\[6pt] &=& 1.44-0.81\\[6pt] &=& 0.63\\[6pt] \small\textsf{OC}\normalsize &=& \sqrt{0.63}\\[6pt] &=& 0.79372...\\[6pt] &\approx& 0.79\,\small\textsf{metres} \end{eqnarray} $$

Finally, the depth of the milk is a radius plus OC, so it is 1.2 + 0.79 = 1.99 metres (accurate to 2 d.p.)

The two shapes are identical, so we will find the height of just one of them, and then double that.

Draw a vertical line from C₁ down to the chord. Name the mid-point of the chord M.

AM = AB ÷ 4 = 12 cm and AC₁ = 14 cm.

Now we use Pythagoras to find C₁M.

$$ \begin{eqnarray} \small\textsf{C₁M}\normalsize ^2&=& \small\textsf{AC₁}\normalsize ^2-\small\textsf{AM}\normalsize ^2\\[6pt] &=& 14^2-12^2\\[6pt] &=& 196-144\\[6pt] &=& 52\\[6pt] \small\textsf{OM}\normalsize &=& \sqrt{52}\\[6pt] &=& 7.2111... \end{eqnarray} $$

The height of the left-hand shape is therefore 14 + 7.211... = 21.211... cm, and the total height of the logo is double this, so it is 42.4 cm (correct to 1 d.p.)

Let M represent the mid-point of the chord AB, and think about the right-angled triangle AOM.

OA is a radius, so OA = 13 cm.

AM is half of AB, so AM = 10 cm.

Now we use Pythagoras to calculate OM:

$$ \begin{eqnarray} \small\textsf{OM}\normalsize ^2&=& \small\textsf{OA}\normalsize ^2-\small\textsf{AM}\normalsize ^2\\[6pt] &=& 13^2-10^2\\[6pt] &=& 169-100\\[6pt] &=& 69\\[6pt] \small\textsf{OM}\normalsize &=& \sqrt{69}\\[6pt] &=& 8.30662...\\[6pt] &\approx& 8.3\,\small\textsf{cm} \end{eqnarray} $$

Finally, the width of the shape is a radius plus OM, so it is 13 + 8.3 = 21.3 cm (accurate to 1 d.p.)