# Area of a Triangle

Math Analysis - Law of Sines - Area of oblique triangle

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• The area of a triangle given two sides and an included angle is given as $A = \frac{1}{2} a b s i n C$ where a and b are the sides and angle C is the included angle.

Example: Given the triangle below, find the area.

Solution:
$A = \frac{1}{2} a b s i n C$
$A = \frac{1}{2} \left(7\right) \left(9\right) s i n 30 \mathrm{de} g r e e s$
$A = \frac{1}{2} \left(63\right) \left(\frac{1}{2}\right)$
$A = \frac{63}{4}$ square units

• Heron's formula allows you to evaluate the area of a triangle knowing the length of its three sides.
The area $A$ of a triangle with sides of lengths $a , b$ and $c$ is given by:

A=sqrt(sp×(sp-a)×(sp-b)×(sp-c))

Where $s p$ is the semiperimeter:

$s p = \frac{a + b + c}{2}$

For example; consider the triangle:

The area of this triangle is A=(base×height)/2
So: A=(4×3)/2=6
Using Heron's formula:
$s p = \frac{3 + 4 + 5}{2} = 6$
And:
A=sqrt(6×(6-5)×(6-4)×(6-3))=6

The demonstration of Heron's formula can be found in textbooks of geometry or maths or in many websites. If you need it have a look at:
http://en.m.wikipedia.org/wiki/Heron%27s_formula

• You can use it whenever you know the lengths of all three sides of a triangle.

I hope that this was helpful.

## Questions

• · 2 months ago
• · 2 months ago