# 7.05 Area of a non-right-angled triangle

Lesson

We know how to find the area of right-angled triangles - we just multiply the two short sides together, and halve the result:

The area of this triangle is $\frac{1}{2}ab$12ab, half the area of the rectangle.

If the triangle is not right-angled we can still find the area, as long as we know two sides and the angle between them:

The area of this triangle is the base $a$a times the height, and then halved, just like for right-angled triangles. But what is the height? It isn't $b$b in this case, but we can use $b$b and the angle $C$C to find it.

Here we have made a small right-angled triangle within our larger triangle, with hypotenuse $b$b and short side $h$h, the height of our large triangle. Using trigonometric ratios, the value of $\sin C$sinC is the opposite side, $h$h, divided by the hypotenuse, $b$b.

So $\sin C=\frac{h}{b}$sinC=hb and hence, $h=b\sin C$h=bsinC.

Putting this all together with the area formula $Area=\frac{1}{2}base\times height$Area=12base×height, we obtain the formula:

Sine area rule

If a triangle has sides of length $a$a and $b$b, and the angle between these sides is $C$C, then:

$\text{Area }=\frac{1}{2}ab\sin C$Area =12absinC

#### Practice questions

##### Question 1

Calculate the area of the following triangle.

##### Question 2

$\triangle ABC$ABC has an area of $520$520 cm2. The side $BC=48$BC=48 cm and $\angle ACB=35^\circ$ACB=35°.

What is the length of $b$b?

## Further applications

Many shapes can be divided into triangles and if we have sufficient information we can apply the sine area rule to solve problems involving areas.

#### Worked examples

##### Example 1

Find the area of a regular pentagon inscribed in a circle of radius $8$8 cm.

Pictured left is a regular pentagon inscribed in a circle. Drawing lines out from the centre to the corners of the pentagon we will form 5 congruent triangles. Each triangle will be isosceles with two sides equal to the radius. The central angle can be found by dividing a full rotation by 5:

 $\theta$θ $=$= $\frac{360^\circ}{5}$360°5​ $=$= $72^\circ$72°

We now have enough information to use our area rule.

 Area $=$= $5\times\frac{1}{2}ab\sin C$5×12​absinC $=$= $5\times\frac{1}{2}\times8\times8\times\sin\left(72^\circ\right)$5×12​×8×8×sin(72°) $=$= $152$152$cm^2$cm2 (to the nearest $cm^2$cm2)
##### Example 2

A rhombus has area $50$50$cm^2$cm2 and its acute angle is $30^\circ$30°. What is the side length of the rhombus?

A rhombus can be split into two isosceles triangles with side length $x$x cm and angle between the sides $30^\circ$30°.

Hence, the area could be found using the formula:

 Area $=$= $2\times\frac{1}{2}ab\sin C$2×12​absinC $=$= $x^2\times\sin\left(30^\circ\right)$x2×sin(30°)

Substituting our known area and rearrange for $x$x:

 $50$50 $=$= $x^2\times\sin\left(30^\circ\right)$x2×sin(30°) $50$50 $=$= $x^2\times\frac{1}{2}$x2×12​ $100$100 $=$= $x^2$x2 $x$x $=$= $10$10, since $x$x is positive.

The side length of the rhombus is $10$10 cm.

#### Practice questions

##### Question 3

A parallelogram has two adjacent sides of length $12$12 cm and $7$7 cm respectively, with an included angle that measures $101^\circ$101°.

Find the area of the parallelogram.

##### Question 4

A hexagon has sides with length $10$10 cm.

Find the area of the hexagon.

### Outcomes

#### 4.2.1.3

establish and use the sine (ambiguous case is required) and cosine rules and the formula 𝐴rea=1/2 bc sin𝐴 for the area of a triangle

#### 4.2.1.4

construct mathematical models using the sine and cosine rules in two- and three-dimensional contexts (including bearings in two-dimensional context) and use the model to solve problems; verify and evaluate the usefulness of the model using qualitative statements and quantitative analysis