__How to find the area of triangles without perpendicular base or height__

__How to find the area of triangles without perpendicular base or height__

From the diagram shown above, we learnt in past chapters that we can find the area of a triangle using the formula:

**Area of triangle = ½ x base x height**

This is applicable for the right-angled triangle ABC, since the length of its base and height can be found in a straightforward manner i.e. parallel to the x and y-axis.

However, for triangle DEF, the length of its base and height are not as clear-cut as for triangle ABC. So, how do we find the area of triangle DEF? In this chapter, we will be learning a **new formula** to calculate the area of any triangles when we are **given the coordinates of all its vertices**. Let’s begin!

From the diagram above, you can see the triangle **ABC** such that **A (x1, y1), B (x2, y2) **and** C (x3, y3)** are in an anticlockwise direction.

We can see that: Area of triangle ABC = Area of trapezium **BMLA** + area of trapezium **ALNC** – area of trapezium **BMNC**

= [½ (y1 + y2) (x1 – x2)] + [½ (y1 + y3) (x3 – x1)] – [½ (y2 + y3) (x3 – x2)]

= ½ [(x1y1 – x2y1 + x1y2 – x2y2) + (x3y1 – x1y1 + x3y3 – x1y3) – (x3y2 – x2y2 + x3y3 – x2y3)]

= ½ (x1y2 + x2y3 + x3y1 – x2y1 – x3y2 – x1y3)

= ½ [(x1y2 + x2y3 + x3y1) – (x2y1 + x3y2 + x1y3)]

If you find the above formula complex, don't worry! The above formula can be simplified as:

Where the products taken in the direction pointed by the **black arrows** are given positive signs,

**i.e. x1y2 + x2y3 + x3y1**

and the products taken in the direction pointed by the **blue arrows** are given negative signs.

**i.e. -x2y1 - x3y2 - x1y3 or – (x2y1 + x3y2 + x1y3)**

Therefore, if **A (x1, y1), B (x2, y2) **and** C (x3, y3)** are in an **anticlockwise **direction, then the formula above holds true.

**Note:** This formula is only applicable when the vertices are taken in the **anticlockwise** direction. If the vertices are taken to be in a **clockwise** direction, then the resultant value obtained will hence be **negative**.

Let’s take a look at the application of this formula in the example question shown below:

· **Find the area of the triangle with vertices A (2, 7), B (1, -2) and C (-3, -5).**

**Solution:**

Area of triangle ABC = ½ {[(2) (-5) + (-3) (-2) + (1) (7)] – [(7) (-3) + (-5) (1) + (-2) (2)]}

= ½ [(-10 + 6 + 7) – (-21 – 5 – 4)]

= ½ [(3) – (-30)]

= 16.5 units^2

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