# If area formed by (0,0 ), ( 6a,6a ) and ( 12a,0 ) is A. Then area of tringle by ( 3a,3a ), ( 6a,0 ) and ( 9a,3a ) is ?

2
by dhiraj3

Log in to add a comment

by dhiraj3

Log in to add a comment

A=1/2(0(6a)+6a(0)+12a(-6))

A=36a²

now.

Area=1/2((3a(-3a)+6a(0)+9a(3a))

Area=18a²

Ans=A/2

1/2| {0(6a-0) + 6a(0-0) + 12a(0-6a}|

= 1/2 |{-72a²}|

A = 36a²

Area of triangle formed by (3a, 3a), (6a, 0) and (9a, 3a) =

1/2 |{3a(0-3a) + 6a(3a-3a) + 9a(3a-0)}|

= 1/2 |{-9a² + 0 + 27a²}|

= 1/2 |{18a²}|

= 9a²

So, area of new triangle = A/4