Victorian Curriculum Year 10A - 2020 Edition
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7.07 The area formula

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. $A=\frac{1}{2}\times\text{base}\times\text{height}$A=12×base×height is the formula, which uses the base, $b$b, and height, $h$h. Sometimes the height is referred to as the altitude of the triangle and it must always be perpendicular to the base.

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 $\frac{1}{2}bh$12bh

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:

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

Here we have added the side $h$h. This is the perpendicular height of the triangle. If we can find this we can calculate the area of the triangle. It is perpendicular to the base and divides the triangle into $2$2 right-angled triangles.

One of the right-angled triangles within our larger triangle, has a hypotenuse $a$a and short side $h$h. This is now a right-angled triangle so we can use trigonometric ratios to find the value of $\sin C$sinC. It is the opposite side, $h$h, divided by the hypotenuse, $a$a.

$\sin C=\frac{h}{a}$sinC=ha

We can rearrange to find the value of $h$h,

$h=a\sin C$h=asinC

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

$\text{Area}$Area $=$= $\frac{1}{2}\times\text{base}\times\text{height}$12×base×height

General formula for triangle

  $=$= $\frac{1}{2}\times b\times\left(a\sin C\right)$12×b×(asinC)

Substitute for triangle where two sides and an enclosed angle are known

  $=$= $\frac{1}{2}ab\sin C$12absinC

Simplify equation



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.

Round your answer to two decimal places.

Question 2

$\triangle ABC$ABC has an area of $850$850 cm2. The side $BC=65$BC=65 cm and $\angle ACB=31^\circ$ACB=31°.

What is the length of $b$b?

Round your answer to the nearest centimetre.


A triangular-shaped field has sides of length $25$25 m, $29$29 m and $36$36 m.

  1. Find the value of the angle $x$x opposite the side that is $25$25 m long.

    Round your answer to two decimal places.

  2. Hence or otherwise, find the area of the field to two decimal places.

  3. Bob has been hired to plough the field and to erect fencing around its perimeter.

    If he charges $\$3$$3 per square metre for ploughing and $\$7$$7 per metre for fencing, how much does he charge in total?

    Round your answer to two decimal places.



VCMMG367 (10a)

Establish the sine, cosine and area rules for any triangle and solve related problems.

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