There are several different ways in which we can find the area of a triangle. The situation and the information we are presented with will determine which formula we will use.
Base and perpendicular height
When the base ($b$b) and perpendicular height ($h$h) are known then we can use the familiar formula, half base times height.
$A=\frac{1}{2}bh$A=12bh
Where $b$b is the length of the base and $h$h is the perpendicular height.
Practice questions
question 1
Find the area of the triangle shown.
Two sides and the included angle
We can also find the area of a triangle when 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. 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:
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
Note: It really doesn't matter what you call the sides as long as you have two sides and the included angle. It's worth noting that we often label the sides with lower case letters, and the angles directly opposite the sides with a capital of the same letter.
Practice questions
Question 2
Calculate the area of the following triangle.
Round your answer to two decimal places.
Question 3
Calculate the area of the following triangle.
Round your answer to the nearest square centimetre.
All three sides
When we know the length of all three sides of the triangle we can use Heron's Formula to calculate the area. Heron's formula requires you to know the lengths of all $3$3 sides, (no perpendicular heights or angles necessary).
$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$A=√s(s−a)(s−b)(s−c)
Where $s$s is known as the semi-perimeter of the triangle and is calculated using $s=\frac{a+b+c}{2}$s=a+b+c2.
Worked example
Example 1
Use Heron's formula to find the area of triangle $ABC$ABC, if $AB=3,BC=2$AB=3,BC=2 and $CA=4$CA=4.
Think: First we need to calculate $s$s, the semi-perimeter, then we can apply Heron's formula.
Do:
Find $s$s:
$s=\frac{a+b+c}{2}=\frac{2+4+3}{2}=4.5$s=a+b+c2=2+4+32=4.5
Now we can use Heron's Formula:
| $A$A | $=$= | $\sqrt{s(s-a)(s-b)(s-c)}$√s(s−a)(s−b)(s−c) |
| $A$A | $=$= | $\sqrt{4.5(4.5-2)(4.5-4)(4.5-3)}$√4.5(4.5−2)(4.5−4)(4.5−3) |
| $A$A | $=$= | $\sqrt{4.5(2.5)(0.5)(1.5)}$√4.5(2.5)(0.5)(1.5) |
| $A$A | $=$= | $2.9$2.9 (to 1dp) |
So the area of the triangle is $2.9$2.9 units2.
Practice questions
question 4
Find the area of the triangle.
Further applications
Many shapes can be divided into triangles and if we have sufficient information we can apply one of our area formulae to solve problems involving areas. We can also work backwards when the area of the triangle is known, using algebra and inverse operations to find unknown side lengths or angles.
Worked examples
Example 2
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$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$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$5×12absinC |
| $=$= | $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 3
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$2×12absinC |
| $=$= | $x^2\times\sin\left(30^\circ\right)$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.
Round your answer to two decimal places.
question 4
$\triangle ABC$△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?
Round your answer to the nearest centimetre.
Given $\triangle ABC$△ABC with length of side $AC$AC labeled as $b$b cm and side $BC$BC labeled as $48$48 cm and their interior $\angle ACB$∠ACB that measures $35$35º. $\angle ACB$∠ACB is an included angle of the two given sides. Both side AC and BC are adjacent to the given included angle.
question 5
We want to find the area of a trapezium with parallel sides of length $19$19 cm and $28$28 cm, and non-parallel sides of length $10$10 cm and $17$17 cm.
We want to break this trapezium up into a parallelogram and a triangle, where the parallelogram has sides of length $10$10 cm and $19$19 cm.
Find the area of the triangle.
Hence solve for $h$h, the height (in centimetres) of the trapezium.
Hence find the area of the trapezium.