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. According to our trigonometric ratios from 3.02, the value of $\sin C$sinC is the opposite side, $h$h, divided by the hypotenuse, $b$b. This means the height $h$h is equal to $b\sin C$bsinC, and putting this all together we now have a formula for the area of any triangle!
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.
Round your answer to two decimal places.
Question 2
The diagram shows a triangular paddock with measurements as shown.
Find the area.
Round your answer to the nearest square metre.
What is the area in hectares?
Round your answer to two decimal places.
Question 3
$\triangle ABC$△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.