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 ?
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2
by using co-ordinate geometry formula of area of tri.,
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
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
Answered by
0
Area of triangle formed by (0,0), (6a,6a) and (12a, 0) =
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
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
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