Area of a Triangle
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Theory
The area rule, \(A = \dfrac{1}{2}ab\sin C\), gives the area of any triangle from two sides and the angle between them, with no height needed. This Year 10A maths page covers exact areas, working backwards to a side or an angle, triangles given by three sides, and areas of quadrilaterals and regular polygons.
The area rule says the area of a triangle is half the product of two sides times the sine of the included angle, the angle between those two sides.
It comes from \(A = \dfrac{1}{2} \times \text{base} \times \text{height}\): with base \(a\), the height is \(b\sin C\). It works for acute and obtuse angles, because \(\sin(180^\circ - C) = \sin C\).
Since \(\sin C \le 1\), the largest area two sides can make is \(\dfrac{1}{2}ab\), when the angle between them is \(90^\circ\).
Working backwards from an area gives \(\sin C\). In a triangle that usually means two possible angles, \(C\) and \(180^\circ - C\), and both make a triangle with the same area.
How to use the area rule
- Identify two sides and the angle between them. If the angle you know is not between them, find another side or angle first.
- Substitute into \(A = \dfrac{1}{2}ab\sin C\). Use exact values if an exact answer is asked for.
- Answer with square units, rounded as asked. Use the unrounded area for costs.
Working backwards and composite shapes
- Unknown side: substitute the area and solve the linear equation. If the sides are \(x\) and \(x + 4\), you get a quadratic: expand, factorise and reject negative lengths.
- Unknown angle: solve for \(\sin C\), then give \(C\) and \(180^\circ - C\) unless the diagram shows which one.
- Quadrilaterals: split along a diagonal into two triangles. Find the diagonal with the cosine rule if you need it, then add the two areas. A parallelogram is two congruent triangles.
- Regular polygons: join the centre to each vertex and multiply the area of one triangle by \(n\) (Example 4).
- Largest area: the area \(\dfrac{1}{2}ab\sin\theta\) is greatest when \(\sin\theta = 1\), that is \(\theta = 90^\circ\).
(i) Find its area, correct to the nearest square metre.
(ii) Turf costs \(\$12.50\) per square metre. Find the cost of turfing the park, to the nearest dollar.
Two sides and the angle between them: use the area rule.
| \(A\) | \(=\) | \(\dfrac{1}{2}ab\sin C\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times 46 \times 38 \times \sin 74^\circ 15'\) |
| \(\) | \(=\) | \(874\sin 74^\circ 15'\) |
| \(\) | \(=\) | \(841.185\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{841}\text{ m}^2\) |
(ii) Use the unrounded area:
| \(\text{cost}\) | \(=\) | \(841.185\ldots \times 12.50\) |
| \(\) | \(\approx\) | \(\mathbf{\$10\,515}\) |
(i) Find the exact area of a triangle with sides \(6\) cm and \(8\) cm and an included angle of \(135^\circ\).
(ii) Another triangle has sides \(7\) cm and \(12\) cm and an area of \(33.6\) cm\(^2\). Find the two possible sizes of the angle \(\theta\) between these sides, correct to one decimal place.
(i) \(\sin 135^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}}\).
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 6 \times 8 \times \dfrac{1}{\sqrt{2}}\) |
| \(\) | \(=\) | \(\dfrac{24}{\sqrt{2}} = \mathbf{12\sqrt{2}}\text{ cm}^2\) |
(ii) Substitute the area and solve for \(\sin\theta\).
| \(\dfrac{1}{2} \times 7 \times 12 \times \sin\theta\) | \(=\) | \(33.6\) |
| \(42\sin\theta\) | \(=\) | \(33.6\) |
| \(\sin\theta\) | \(=\) | \(0.8\) |
| \(\theta\) | \(=\) | \(53.130\ldots^\circ\) or \(180^\circ - 53.130\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{53.1^\circ}\text{ or }\mathbf{126.9^\circ}\) |
Both angles give a real triangle with the same area, because \(\sin(180^\circ - \theta) = \sin\theta\).
The area rule needs an angle. Find any angle with the cosine rule, here \(\theta\) between the \(7\) cm and \(9\) cm sides.
| \(\cos\theta\) | \(=\) | \(\dfrac{7^2 + 9^2 - 12^2}{2 \times 7 \times 9} = -\dfrac{14}{126}\) |
| \(\theta\) | \(=\) | \(96.379\ldots^\circ\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 7 \times 9 \times \sin 96.379\ldots^\circ\) |
| \(\) | \(=\) | \(31.304\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{31.3}\text{ cm}^2\) |
Use the sides that form \(\theta\) in the area rule. Keep \(\theta\) unrounded in the calculator.
Join the centre \(O\) to each vertex. This makes \(5\) identical triangles, each with two sides of \(8\) cm and an angle of \(\dfrac{360^\circ}{5} = 72^\circ\) at \(O\).
| \(A_1\) | \(=\) | \(\dfrac{1}{2} \times 8 \times 8 \times \sin 72^\circ = 30.433\ldots\) |
| \(A\) | \(=\) | \(5 \times 30.433\ldots\) |
| \(\) | \(=\) | \(152.169\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{152.2}\text{ cm}^2\) |
In general, a regular \(n\)-gon in a circle of radius \(r\) has area \(\dfrac{n}{2}r^2\sin\left(\dfrac{360^\circ}{n}\right)\).
Common pitfalls
Frequently asked questions
What is the area rule for a triangle?
The area of a triangle is half of two sides times the sine of the angle between them: \(A = \dfrac{1}{2}ab\sin C\). It works for any triangle, not just right-angled ones.
How do you find the area of a triangle without the height?
If you know two sides and the angle between them, use A = one half a b sin C. If you know all three sides, use the cosine rule to find an angle first, then the area rule.
How do you find an angle from the area of a triangle?
Substitute the area and the two sides into A = one half a b sin C and solve for sin C. Then find C with inverse sine, and remember that 180 degrees minus C has the same sine, so it may also be an answer.
Does the area rule work for obtuse angles?
Yes. The sine of an obtuse angle is positive and \(\sin(180^\circ - C) = \sin C\), so \(\sin 130^\circ = \sin 50^\circ\). Substitute the obtuse angle directly.
When is the area of a triangle with two fixed sides the largest?
The area one half a b sin theta is largest when sin theta equals 1, which happens when the angle between the two sides is 90 degrees. The largest area is then one half a b.