Question

The vertices of a triangle abc are a(-5,-1), b(3,-5),c(5,2).Show that the area of the triangle abc is four times the area of the triangle formed by joining the midpointd of the sides of the triangle abc

Answer 1

Given info : The vertices of a triangle ABC are A(-5,-1), B(3,-5), C(5,2).

To show : the area of the triangle ABC is 4 times the area of the triangle formed by joining the midpoint of the side of the triangle ABC.

proof : Area of triangle ABC = 1/2 |[-5(-5 -2) + 3(2 + 1) + 5(-1 + 5)]|

= 1/2 |35 + 9 + 20|

= 32 sq unit

now let D , E and F are the midpoint of AB , BC and CA respectively.

from midpoint section formula,

D = [(-5 + 3)/2, (-1 - 5)/2 ] = (-1, -3)

E = [(3 + 5)/2, (-5 + 2)/2 ] = (4, -1.5)

F = [(5 - 5)/2, (2 - 1)/2 ] = (0, 0.5)

area of triangle formed by joining the points D, E and F = 1/2 [-1(-1.5 - 0.5) + 4(0.5 + 3) + 0 ]

= 1/2 [ 2 + 14]

= 8 sq unit

Therefore it is clear that,

ar (∆ABC) = 4 × ar (∆DEF)

Hence area of triangle ABC is 4 times the area of triangle formed by joining the midpoint of the sides of the triangle ABC.