In Geometry, a triangle is a 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 the area of a triangle is and different methods used to find the area of a triangle in 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 =
Where a, b, c are the side lengths of the triangle and s is the semi perimeter
The value of s is found using the formula
s =
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 in the coordinate plane is known.
Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x1, y1), (x2, y2), (x3, y3), respectively.
From the figure, the area of a triangle PQR, lines such as
Now, three different trapeziums are formed such as PQAB, PBCR and QACR in the coordinate plane.
Now, calculate the area of all the trapeziums.
Therefore, the area of ∆PQR is calculated as, Area of ∆PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR] —-(1)
Finding Area of a Trapezium PQAB
We know that the formula to find the area of a trapezium is
Since Area of a trapezium = (1/2) (sum of the parallel sides)×(distance between them)
Area of trapezium PQAB = (1/2)(QA + PB) × AB
QA = y2
PB = y1
AB = OB – OA = x1 – x2
Area of trapezium PQAB = (1/2)(y1 + y2)(x1 – x2 ) —-(2)
Finding Area of a Trapezium PBCR
Area of trapezium PBCR =(1/2) (PB + CR) × BC
PB = y1
CR = y3
BC = OC – OB =x3 – x1
Area of trapezium PBCR =(1/2) (y1 + y3 )(x3 – x1) —-(3)
Finding Area of a Trapezium QACR
Area of trapezium QACR = (1/2) (QA + CR) × AC
QA = y2
CR = y3
AC = OC – OA = x3 – x2
Area of trapezium QACR =(1/2)(y2 + y3 ) (x3 – x2 )—-(4)
Substituting (2), (3) and (4) in (1),
Area of ∆PQR = (1/2)[(y1 + y2)(x1 – x2 ) + (y1 + y3 )(x3 – x1) – (y2 + y3 ) (x3 – x2 )]
A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]
Special Case:
If one of the vertices of the triangle is the origin, then the area of the triangle can be calculated using the below formula.
Area of a triangle with vertices are (0,0), P(a, b), and Q(c, d) is
A = (1/2)[0(b – d) + a(d – 0) + c(0 – b)]
A = (ad – bc)/2
If area of triangle with vertices P(x1, y1), Q(x2, y2) and R(x3, y3) is zero, then (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)] = 0 and the points P(x1, y1), Q(x2, y2) and R(x3, y3)are collinear.
Area of a Triangle in Coordinate Geometry Example
Example: What is the area of the ∆ABC whose vertices are A(1, 2), B(4, 2) and C(3, 5)?
Solution:
Using the formula,
A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]
A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]
A = (1/2) [-3 + 12]= 9/2 square units.
Therefore, the area of a triangle ABC is 9/2 square units.
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