# Find the Area of δAbc with Vertices A(0, -1), B(2,1) and C(0, 3). Also, Find Area of Triangle Formed by Joining the Midpoints of Its Sides. Show that the Ratio of the Areas of Two Triangles is 4:1. - Mathematics

Find the area of ΔABC with vertices A(0, -1), B(2,1) and C(0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4:1.

#### Solution

Let A (x_1=0,y_1=-1), B (x_2 =2,y_2=1) and C (x_3 = 0, y_3=3) be the given points. Then

Area (Δ ABC)=1/2[x_1 (y_2-y_3)+ x_2(y_3-y_1) +x_3(y_1-y_3)]

=1/2 [0(1-3)+2(3+1)+0(-1-1)]

=1/2xx8=4 sq . units

So, the area of the triangle 4  ΔABC is sq units.

Let D(a_1,b_1),E)a_2,b_2) and F (a_3,b_3) be the midpoints of AB, BC and AC respectively

Then

a_1 = (0+2)/2=1  b_2=(-1+1)/2=0

a_2=(2+0)/2=1  b_2=(1+3)/2=2

a_3=(0+0)/2=0   b_3 = (-1+3)/2=1

Thus, the coordinates of D,E and F are D(a_1=1,b_1=0),E(a_2=1,b_2=2) and F(a_3=0, b_3=1). Now

Area (ΔDEF)= 1/2 [a_1(b_2-b_2)+a_2(b_3-b_1)+a_3(b_1-b_2)]

=1/2 [1(2-1)+1(1-0)+0(0-2)]

=1/2[1+1+0]=1 sq. units

So, the area of the triangle ΔDEF  is 1 sq. unit.

Hence, ΔABC : ΔDEF = 4:1.

Concept: Area of a Triangle
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 23