Area of the Triangle

Consider a simple case: a triangle with one vertex at the origin O(0, 0), and the other two vertices on the coordinate axes, say P(a, 0) on the x-axis and Q(0, b) on the y-axis.

This forms a right-angled triangle with: Base = units and Height = units

Area of △OPQ = 12 × base × height = 12 × a × b square units

This simple case gives us intuition, but what about triangles in any position on the coordinate plane?

General Formula for Area of a Triangle

For any triangle with vertices at A(x1, y1), B(x2, y2), and C(x3, y3), we need a formula that works regardless of the triangle's position or orientation.

Let's derive this formula systematically.

Step 1: Draw Perpendiculars to the X-axis

From each vertex A, B, and C, draw perpendiculars to the x-axis, meeting it at points P, Q, and R respectively.

This creates:

AP ⊥ x-axis, with length = y1

BQ ⊥ x-axis, with length = y2

CR ⊥ x-axis, with length = y3

Example 1: Find the area of the triangle whose vertices are A(2, 3), B(-1, 4), and C(5, -2).

Solution: Using the formula with A(2, 3), B(-1, 4), C(5, -2):

x1 = , y1 = ; x2 = , y2 = ; x3 = , y3 =

Area = 12 | x1y2−y3+x2y3−y1+x3y1−y2 |

= 12 |(4 - ) + ( - ) + (3 - 4)| = 12 |2() + (-1)() + 5()|

= 12 | + + | = 12 | | = square units

Example 2: Find the area of the triangle formed by the points P(0, 4), Q(3, -1), and R(-2, 5).

Solution: With P(0, 4), Q(3, -1), R(-2, 5):

Area = 12 |0( - 5) + (5 - 4) + (4 - )| = 12 |0() + 3() + (-2)()|

= 12 | + + | = 12 |-7| = 12 × = square units

Example 3: Calculate the area of the triangle with vertices at A(6, 1), B(2, 8), and C(-3, 0).

Solution: Area = 12 |(8 - ) + (0 - 1) + ( - )| = 12 |6() + 2() + (-3)()|

= 12 | + + | = 12 || = square units

Example 4: Find the area of the triangle whose vertices are L(4, -3), M(8, 2), and N(1, 6).

Solution: Area = 12 |4( - ) + (6 - (-3)) + 1( - 2)| = 12 |4() + 8() + 1()|

= 12 | + + | = 12 || = square units

Area of a Quadrilateral Using Triangles

When we have a quadrilateral with four vertices, we can find its area by dividing it into two triangles using a diagonal, then calculating the area of each triangle separately.

Example 1: Find the area of quadrilateral PQRS with vertices P(2, 1), Q(6, 3), R(8, 7), and S(3, 5).

Solution: We'll divide the quadrilateral by drawing diagonal PR.

Area of △PQR: Area = 12 |( - ) + ( - ) + ( - )|

= 12 |2() + 6() + 8()| = 12| + + | = 12 | | = square units

Area of △PRS: Area = 12 |( - ) + ( - ) + ( - )|

= 12|2() + 8() + 3()| = 12 | + + | = 12 | | = square units

Total area of quadrilateral PQRS = + = square units

Example 2: Calculate the area of quadrilateral ABCD with vertices A(1, 4), B(5, 2), C(7, 6), and D(3, 8).

Solution: Using diagonal AC to divide into △ABC and △ACD:

Area of △ABC: Area = 12 | ( - ) + ( - ) + ( - ) |

= 12 | 1() + 5() + 7() | = 12 | + + |

= square units

Area of △ACD: Area = 12 | ( - ) + ( - ) + ( - ) |

= 12 | 1() + 7() + 3() | = 12 | + + |

= square units

Total area = + = square units

Collinearity of Points

Three points are said to be collinear, if they lie on the same straight linetwo parallel lines. When three points are collinear, they cannot form a triangle, so the area enclosed by them is .

Example 3: Show that the points A(4, 1), B(7, 4), and C(10, 7) are collinear.

Solution: We have:

Area = 12 | ( - ) + ( - ) + ( - ) |

= 12 |4() + 7() + 10()|

= 12 | + + | =

Since the area is zero, the three points are .

Example 4: Verify whether the points P(2, 5), Q(-1, 2), and R(5, 8) are collinear.

Solution: For points to be collinear: is equal to .

Area = 12 | ( - ) + ( - ) + ( - )|

= 12 |2() + (-1)() + 5()|

= 12 | + + | =

The points are collinear.

Example 5: Find the value of 'k' for which the points A(2, 3), B(k, 5), and C(-4, 1) are collinear.

Solution: For collinear points, area =

Thus, |( - 1) + k(1 - ) + ( - 5)| = 0

⇒ 2() + k() + (-4)() = 0

⇒ + + = 0

⇒ k = ⇒ k =

Example 6: Determine the value of 'm' if points P(-3, m), Q(1, 4), and R(5, 10) are collinear.

Solution:

Using the collinearity condition: ( -+× ) +-× ( -+ ) +-× ( -+ ) =

⟹ + - m + 5m + = 0

+ 4m6m = 0 ⟹ m =

When to Use Heron's Formula ?

The coordinate formula works when we know the vertices of a triangle.

But what if we only know the lengths of the three sides (a, b, and c)?

For such cases, we use Heron's Formula, named after the ancient Greek mathematician Heron of Alexandria.

Heron's Formula: Area = ss−as−bs−c

where:

a, b, c are the lengths of the three sides

s is the semi-perimeter: s = a+b+c2

Why is it called semi-perimeter? The perimeter of a triangle is a + b + c. Half of this (semi means half) is s = a+b+c2.

Example 7: Find the area of a triangle whose sides measure 8 cm, 11 cm, and 13 cm.

Solution: Given: a = 8, b = 11, c = 13

First, find the semi-perimeter: s = 8+11+132 = 322 = cm

Calculate each difference: s - a = - =

s - b = - =

s - c = - =

Apply Heron's formula: A = ss−as−bs−c = 16×8×5×316×12×5×3

= 19203840 = 8301215 ≈ cm2 (Round off to two decimal places)

Example 8: Calculate the area of a triangle with sides 6 m, 8 m, and 10 m.

Solution: s = 6+8+102 = 242 = m

s - a = - =

s - b = - =

s - c = - =

A = 12×6×4×212×6×8×2 = 5761152 = m2

Example 9: Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm using Heron's formula.

Solution: s = 5+12+132 = 302 = cm

s - a = - =

s - b = - =

s - c = - =

Area = 15×10×3×2 = 900 = cm2

Example 10: Find the area of triangle formed by points A(0, 0), B(5, 0), and C(5, 4) using Heron's formula.

Solution: First, find the distances (side lengths):

AB = 5−02+0−02 = 25 =

BC = 5−52+4−02 = 16 =

CA = 0−52+0−42 = 25+16 = 41 ≈ (upto one decimal place)

Semi-perimeter: s = 5+4+6.4032 ≈ (upto one decimal place)

Area = 7.77.7−57.7−47.7−6.4 = 7.7×2.7×3.7×1.2 ≈ square units (round off to nearest whole number)