In Geometry, a triangle is the 3 – sided polygon which has 3 edges and 3 vertices. Area of the triangle is a measure of the space covered by the triangle in the two-dimensional plane. In this article, let us discuss what is the area of a triangle, and different methods used to find the area of a triangle in the coordinate geometry.

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## Methods to Find the Area of a Triangle

Area of a triangle can be found using three different methods. The three different methods are discussed below

### Method 1

When the base and altitude of the triangle are given.

Area of the triangle, A = bh/2 square units

where b and h are base and altitude of the triangle respectively.

## Method 2

When the length of three sides of the triangle are given, the area of a triangle can be found using the Heron’s formula.

Therefore, the area of the triangle is calculated using the equation,

A = \( \sqrt{s(s~-~a)~(s~-~b)~(s~-~c)} \)

Where a,b, c are the side lengths of the triangle and s is half of the perimeter

The value of s is found using the formula

s = \( \frac {a~+~b~+~c}{2} \)

## Method 3

If the vertices of a triangle are given, first we have to find the length of three sides of a triangle. The length can be found using the distance formula

The procedure to find the area of a triangle when the vertices are known in the coordinate plane is given below.

Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) respectively.

From the figure, the area of a triangle PQR, lines such asÂ \( \overleftrightarrow {QA}\) , \( \overleftrightarrow {PB}\) and \( \overleftrightarrow {RC}\) are drawn from Q,P and R respectively perpendicular to x – axis.

Now, three different trapeziums are formed such asÂ PQAB, PBCR and QACRÂ in the coordinate plane.

Now, calculate the area of all the trapeziums.

Therefore, **the area of âˆ†PQR is calculated as, Area of âˆ†PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR]** —-(1)

**Finding Area of a TrapeziumÂ PQABÂ **

We know that, the formula to find the area of a trapezium is

Since, Area of a trapezium = (1/2)Â (sum of the parallel sides)Ã—(distance between them)

Area of trapezium PQAB = (1/2)(QA+PB) Ã— AB

QA = y_{2}

PB =y_{1}

AB = OB – OA =Â x_{1}-x_{2}

Area of trapezium PQAB = (1/2)(y_{1}+y_{2})(x_{1}-x_{2} )Â —-(2)

**Finding Area of a TrapeziumÂ PBCR**

Area of trapezium PBCR =(1/2)Â (PB + CR) Ã— BC

PB =y_{1}

CR = y_{3}

BC = OC – OB =x_{3}-x_{1}

Area of trapezium PBCR =(1/2)Â (y_{1}+ y_{3} )(x_{3}-x_{1})Â —-(3)

**Finding Area of a TrapeziumÂ QACR**

Area of trapezium QACR = (1/2)Â (QA + CR) Ã— AC

QA=y_{2}

CR = y_{3}

AC = OC – OA = x_{3}-x_{2}

Area of trapezium QACR =(1/2)(y_{2}+ y_{3} )Â (x_{3}-x_{2} )—-(4)

Substituting (2), (3) and (4) in (1) gives,

Area of âˆ†PQR = (1/2)[(y_{1}+y_{2})(x_{1}-x_{2} )+(y_{1}+ y_{3} )(x_{3}-x_{1})-(y_{2}+ y_{3} )Â (x_{3}-x_{2} )]

A = (1/2)Â [x_{1} (y_{2-}Â y_{3} )+x_{2} (y_{3}-y_{1} )+x_{3}(y_{1}-y_{2})]

**Special Case:**

If one of the vertices of the triangle is origin, then

Area of triangle with vertices are (0,0) P(a,b), and Q(c,d) is

A = (1/2)[0(b – c) + a(d – 0) + c(0 – b)]

A = Â (ac-bc)/2Â

If area of triangle with vertices P(x_{1}, y_{1}), Q(x_{2}, y_{2})Â and R(x_{3}, y_{3})Â is zero, then (1/2)Â [x_{1} (y_{2-}Â y_{3} )+x_{2} (y_{3}-y_{1} )+x_{3}(y_{1}-y_{2})] = 0 and the points P(x_{1}, y_{1}), Q(x_{2}, y_{2})Â and R(x_{3}, y_{3})are collinear.

### Area of a Triangle in Coordinate Geometry Example

**Example:** What is the area of the âˆ†ABC whose vertices are A(1,2), B(4,2) and C(3,5)?

**Solution:**

Using the formula,

A = Â (1/2)Â [x_{1} (y_{2-}Â y_{3} )+x_{2} (y_{3}-y_{1} )+x_{3}(y_{1}-y_{2})]

A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]

A = (1/2) [-3 +12]= Â 9/2Â square units.

Therefore, the area of a triangle ABC isÂ Â 9/2Â square units.

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