The area of a triangle when its base and corresponding height (altitude) are given:

Area of a triangle =$1/2$ × base × altitude


Let ABC be any triangle whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3). Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezia . Now, from Fig

area of ∆ ABC = area of trapezium ABQP + area of trapezium APRC – area of trapezium BQRC.

Since, area of a trapezium = ${1/2}$(sum of parallel sides)(perpendicular distance )

Therefore,
Area of ∆ ABC = ${1/2}$(BQ + AP) QP +${1/2}$ (AP + CR) PR -${1/2}$(BQ + CR) QR

Area of ∆ ABC =$1/2(y_2+y_1) (x_1-x_2) +1/2 (y_1+y_3)(x_3-x_1) -1/2 (y_2 + y_3)(x_3-x_2) $

Area of ∆ ABC =${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$

Thus, the area of ∆ ABC is the numerical value of the expression, = ${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$

Example: Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5).

Solution : The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and C (–3, –5), by using the formula above, is given by,

${1/2}{[x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 – y_2)]}$

i.e ${1/2}{(1(6 - (-5)) + (-4) ((-5) + 1) + (-3) ((-1) – 6))}$

=$24$

=So, the area of the triangle is 24 square units.

HomeWork.

1)Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4).

2)Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4).

3)Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

4) If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.