Area Rule Worked ExamplesIGCSE Maths 0580

The angle must be between the two sides used. The March 2026 examiner report noted that "A few candidates used the incorrect formula, treating the triangle as a right-angled." — $\tfrac{1}{2} \times 12 \times 8 = 48$ is only correct when the included angle is $90^{\circ}$. A quick bound: the area can never exceed $\tfrac{1}{2}ab = 48$, reached only at $90^{\circ}$ — and $43.5 < 48$. ✓

Rearranging the area rule is straightforward — the difficulty is identifying which side the angle sits between. Substitute back to check: $\tfrac{1}{2}(14)(9.61)\sin 48^{\circ} = 50.0$. ✓

Unlike the ambiguous case in the sine rule, both answers are always valid here — two different triangles with the same two sides genuinely share the same area. Giving only the acute value loses a mark whenever the question says 'two possible values'.

Use the two sides that enclose the angle you found — pairing $A$ with sides $5$ and $7$ would be wrong. Both formulas are on the List of formulas; the June 2025 examiner report noted that some candidates "misquoted formulas such as the volume of a cone, the area of a triangle or the cosine rule." Keep the unrounded angle in your calculator for step 5.

Read the required accuracy — 'nearest square metre' means a whole number, not 3 significant figures. Units are squared for an area; writing $\text{m}$ rather than $\text{m}^2$ can cost a mark.

You never need the length of the diagonal — each triangle already has its two sides and included angle. Note $\sin 100^{\circ}$ is positive even though the angle is obtuse, so no sign adjustment is needed. Carry unrounded values and round only at the end. Do not skip this type: on a quadrilateral-area question the June 2025 examiner report noted that "More than a quarter of candidates did not attempt this part." Splitting along a diagonal into two triangles is always the way in.

Two errors are common here: halving the answer as if the shape were a single triangle ($21\text{ cm}^2$), and believing $\sin 150^{\circ}$ is negative because the angle is obtuse. On the non-calculator paper, the June 2025 examiner report advised that "it is useful to know the trigonometric ratios for the angles 0°, 30°, 45°, 60° and 90°, with the exception of tan90°." Using the other angle of the parallelogram, $30^{\circ}$ at $B$, gives the same area — a useful check. The height $3.5\text{ cm}$ is less than the slanted side $7\text{ cm}$, as it must be. ✓

A frequent error in (b) is to work out $\tfrac{1}{2} \times 3 \times 7 \times \sin 64^{\circ} = 9.44$, treating the shaded region as a triangle with sides $CA$ and $DB$ — but those two sides do not meet at a $64^{\circ}$ angle. In (a) the method mark needs the full sides $8$ and $11$. In (c) the height must be less than both $OA = 8$ and $OB = 11$; $7.62$ passes that check. ✓ The March 2025 examiner report stressed that "intermediate values within longer methods should not be rounded with only the final answer rounded to the appropriate degree of accuracy." — using the rounded $AB = 10.4$ in (c) gives $7.60$, not $7.62$.

In (a) the mark for the exact value is lost if you write $\sin 45^{\circ} = 0.707$ — a decimal cannot 'show' an exact result. The March 2026 examiner report reminded candidates that "When candidates are required to show a given result, they should show every step in their method". In (b) a common slip is to forget the square root and give $r = 70.7$. In (c), $r$ is the distance to a vertex, not the side length, so the perimeter is not $8r$. Sense check: $PQ = 6.44$ is shorter than $r = 8.41$, as it should be for an angle of only $45^{\circ}$ at $O$. ✓

On a similar quadrilateral, the June 2025 examiner report noted that "Some candidates lost accuracy because they used 8 or 8.1 for the length of AB" — a rounded length carried from an earlier part. Here, using $QS = 9.8$ gives a total of $70.26\ldots$, which rounds to $70.3$, not $70.4$. Store the diagonal in your calculator memory and reuse it.

On a similar tunnel question, the March 2025 examiner report found that "A significant number of candidates did not recognise the need to calculate an angle to work with sectors and triangles", and that "some lost accuracy by rounding prematurely and a few omitted to add the area of the triangle." Leaving out the triangle here gives $539\text{ m}^3$. Sense check: the answer must be less than the volume of the whole cylinder, $\pi \times 4^2 \times 15 = 754\text{ m}^3$. ✓

On a question subtracting a sector from a triangle, the June 2025 examiner report noted that "When attempting to find the area of the sector, some used the wrong radius". Here the radius is the sprinkler's reach, $12\text{ m}$, not a side of the bed: using $18$ or $22$ makes the sector larger than the triangle, which gives a negative answer. The triangle needs $\tfrac{1}{2}ab\sin C$ — treating it as right-angled gives $\tfrac{1}{2} \times 18 \times 22 = 198$ and the wrong answer $118\text{ m}^2$.