National 5 Maths
Area of a Triangle

In this example, the vertex letters A, B and C exactly match the formula. So:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2} \ \normalsize ab\ sin\ C \\[6pt] &=& \small\frac{1}{2}\normalsize \times 9 \times 15\ sin\ 60^\circ \\[6pt] &\approx& 58.5\ \small\textsf{mm²} \\[12pt] \end{eqnarray} $$

This question would be suitable for Paper 1. We have been told sin B, not B itself.

In this example, the vertex letters A, B and C don't match the formula. So we can rewrite the formula using the different letters.

The SQA has never combined area of a triangle with the surds topic, but there is no reason why they shouldn't!

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ac\ sin\ B\:\ \small\textsf{(same formula really!)}\normalsize\\[6pt] &=& \small\frac{1}{2}\normalsize \times 3 \times 5\ \times \small \frac{1}{\sqrt{2}}\normalsize \\[6pt] &=& \frac{15}{2 \sqrt{2}} \\[6pt] &=& \frac{15}{2 \sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} \ \small\textsf{(rationalising)}\normalsize\\[6pt] &=& \frac{15\sqrt{2}}{4}\ \small\textsf{cm²} \end{eqnarray} $$

This question relies on our knowledge that equilateral triangles contain three 60° angles.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C \\[6pt] &=& \small\frac{1}{2}\normalsize \times 5 \times 5 \times sin\ 60^\circ \\[6pt] &\approx& 10.8\ \small\textsf{cm²} \end{eqnarray} $$

This question relies on our knowledge that a regular hexagon consists of 6 equilateral triangles.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& 6 \times \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& 6 \times \small\frac{1}{2}\normalsize \times 10 \times 10 \times sin\ 60^\circ \\[6pt] &=& 259.807... \\[6pt] &\approx& 260\ \small\textsf{cm²}\ \small\textsf{(correct to 3 s.f.)}\normalsize \end{eqnarray} $$

This question requires us to work backwards. For convenience, let's call the unknown angle C°.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] 50 &=& \small\frac{1}{2}\normalsize \times 12\times 10\ sin\ C\\[6pt] 100 &=& 120\small \ \normalsize sin\ C \\[6pt] sin\ C&=& \small\frac{100}{120}\normalsize \\[6pt] C&=& sin^{-1}\ \small\ \frac{100}{120}\normalsize \\[6pt] &\approx& 56.4^\circ\ \end{eqnarray} $$

This is almost the same as the previous example, except now the angle is obtuse, not acute.

The working is exactly the same, all the way down to \(C \approx 56.4^\circ\), which cannot be correct as \(C\) is obtuse.

However, the final step is simple if your knowledge of trig equations is solid. \(C\) is the 2nd quadrant (obtuse) angle related to \(56.4^\circ\), so it's \(180-56.4^\circ=123.6^\circ\).

A hexagon comprises six equilateral triangles, which contain 60° angles.

Each diagonal of the hexagon is 40 cm so each side of the triangles is 20 cm.

Now we can find the area of each triangle:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& \small\frac{1}{2}\normalsize \times 20\times 20\ sin\ 60^\circ\\[6pt] &\approx& 173.205... \end{eqnarray} $$

So the area of the hexagon is \(6\times 173.205... \approx\ 1039.2\) cm²

The sides DE and EF enclose the angle E°, so we can find the area as follows:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\times\small \ \normalsize \small\textsf{DE}\normalsize\times\small\textsf{EF}\normalsize\times sin\,E\\[6pt] &=& \small\frac{1}{2}\normalsize \times 8\times 12\times\small\frac{2}{3}\\[6pt] &=& \left(\small\frac{1}{2}\normalsize\times\small\frac{2}{3}\right) \times \left(8\times 12\right)\\[6pt] &=& \small\frac{1}{3}\normalsize \times 96\\[6pt] &=& 32\ \small\textsf{cm²} \end{eqnarray} $$

This is a straightfoward 2-mark use of the area formula.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& \small\frac{1}{2}\normalsize \times 45\times 70\ sin\ 129^\circ\\[6pt] &=& 1224.0048...\\[6pt] &\approx& 1224.0\ \small\textsf{cm²} \end{eqnarray} $$

This 4-mark question first needs SOH CAH TOA in triangle ABC to find the sine of the angle at A.

$$ \begin{eqnarray} sin\,A &=& \small\frac{8}{18}\normalsize\\[6pt] &=& \small\frac{4}{9}\normalsize \end{eqnarray} $$

Note that we don't need to find angle A itself, because the area formula uses sin A°, not A°.

Now we use the area formula for triangle ADE, as follows:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\times\small\textsf{AD}\normalsize\times\small\textsf{AE}\normalsize\times sin\,A\\[8pt] 160&=& \small\frac{1}{2}\normalsize \times 24\times \small\textsf{AE}\normalsize\times\small\frac{4}{9}\\[8pt] 160&=& \small\frac{2}{9}\normalsize \times 24\times \small\textsf{AE}\normalsize\\[8pt] 160\times 9&=& 2 \times 24\times \small\textsf{AE}\normalsize\\[8pt] 1440 &=& 48\times\small\textsf{AE}\normalsize\\[8pt] \small\textsf{AE}\normalsize &=& \small\frac{1440}{48}\normalsize\\[8pt] &=& 30\ \small\textsf{cm} \end{eqnarray} $$