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 $b$bis 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
When we are using the lengths of 2 sides ($a$a and $b$b) and the angle between them $C$C.
$\text{Area }=\frac{1}{2}\times\text{Side 1 }\times\text{Side 2 }\times\sin\theta$Area =12×Side 1 ×Side 2 ×sinθ.
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. The formula is most commonly written as follows:
$Area=\frac{1}{2}ab\sin C$Area=12absinC
Where $a$a and $b$b are the known side lengths, and $C$C is the given angle between them, as per the diagram above.
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 sides, (no perpendicular heights or angles necessary).
$A=\sqrt{s(s-a)(s-b)(s-c)}$A=√s(s−a)(s−b)(s−c), where s is known as the semiperimeter of the triangle and is calculated using $s=\frac{a+b+c}{2}$s=a+b+c2
Worked example
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.
First we need to identify $s$s, the semiperimeter, which is $s=\frac{a+b+c}{2}=\frac{2+4+3}{2}=4.5$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 square units.
Practice questions
question 4
Find the area of the triangle.
When the area is known
When the area of the triangle is known, we can use algebra and inverse operations to find unknown side lengths or angles.
Practice questions
question 5
$\triangle ABC$△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 6
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.