There is a simple formula when it comes to calculating the area of a triangle, when given the base length of the triangle and the perpendicular height. A=\dfrac{1}{2}\times \text{base}\times \text{height} is the formula, which uses the base, b, and height, h. Sometimes the height is referred to as the altitude of the triangle and it must always be perpendicular to the base.

A right-angled triangle with two short sides which can be considered as base b and height h.

For a right-angled triangle - the two short sides can be the base and the height because they are always perpendicular to each other.

The area of this triangle is \dfrac{1}{2}bh.

A triangle with side lengths a, b, c, and angle C.

If the triangle is not right-angled we don't have the perpendicular height, however we can still find the area, as long as we know two sides and the angle between them.

A triangle with side lengths a, b, c, height h, and angle C.

The area is still half the base b times the height. But what is the height? It isn't a or c in this case, but we can use a and the angle C to find it.

Here we have the perpendicular height h. If we can find this we can calculate the area of the triangle. It divides the triangle into 2 right-angled triangles.

A triangle with side lengths a, b, h, and angle C.

One of the right-angled triangles has a hypotenuse a and short side h. So we can use trigonometric ratios to find the value of \sin C.\sin C=\dfrac{h}{a}

We can rearrange this to find h:h=a\sin C

Putting this all together with the area formula \text{Area}=\dfrac{1}{2}\text{base}\times \text{height}, we obtain the formula:

\displaystyle \text{Area}	\displaystyle =	\displaystyle \dfrac{1}{2}\times \text{base}\times \text{height}	Area of a triangle formula
	\displaystyle =	\displaystyle \frac{1}{2}\times b\times \left(a\sin C\right)	Substitute the height and base
	\displaystyle =	\displaystyle \frac{1}{2}ab\sin C	Simplify

Examples

Example 1

Calculate the area of the following triangle. Round your answer to two decimal places.

A triangle with side lengths 6 metres and 2.2 metres and with angle of 73 degrees.

Worked Solution

Create a strategy

Use the area rule given by \text{Area }=\dfrac{1}{2}ab\sin C.

Apply the idea

We are given a=6, b=2.2, and C=73\degree.

\displaystyle \text{Area }	\displaystyle =	\displaystyle \dfrac{1}{2}ab\sin C	Write the formula
	\displaystyle =	\displaystyle \frac{1}{2}\times 6\times 2.2\times \sin 73\degree	Substitute the measurements
	\displaystyle \approx	\displaystyle 6.31 \text{ m}^2	Evaluate and round

Example 2

\triangle ABC has an area of 850 \text{ cm}^2. The side BC=65 cm and \angle ACB=31\degree. What is the length of b? Round your answer to the nearest centimetre.

A triangle with side lengths 65 centimetres and b centimetres and angle A C B equal to 31 degrees.

Worked Solution

Create a strategy

Use the area rule given by \text{Area }=\dfrac{1}{2}ab\sin C.

Apply the idea

We are given \text{Area}=850, a=65, and C=31\degree.

\displaystyle \text{Area }	\displaystyle =	\displaystyle \dfrac{1}{2}ab\sin C	Write the formula
\displaystyle 850	\displaystyle =	\displaystyle \frac{1}{2}\times 65\times b\times \sin 31\degree	Substitute the measurements
\displaystyle 850	\displaystyle =	\displaystyle 16.74\times b	Evaluate the multiplication
\displaystyle b	\displaystyle =	\displaystyle \dfrac{850}{16.74 }	Divide both sides by 16.74
	\displaystyle \approx	\displaystyle 51	Evaluate and round

Idea summary

To find the area of a non-right angled triangle we can use the following formula:

\displaystyle \text{Area}=\dfrac{1}{2}ab\sin C

\bm{a}

is the length of one side

\bm{b}

is the length of another side

\bm{C}

is the included angle between a and b

Outcomes

MA5.3-15MG

applies Pythagoras' theorem, trigonometric relationships, the sine rule, the cosine rule and the area rule to solve problems, including problems involving three dimensions

10.08 The area formula

Ideas

The area formula

Examples

Example 1

Example 2

Outcomes

MA5.3-15MG

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