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The area of a triangle with vertices $ (a, b+c),(b, c+a) $ and $ (c, a+b) $ is
(A) $ (a+b+c)^{2} $
(B) 0
(C) $ a+b+c $
(D) $ a b c $
Given:
The vertices of a triangle are \( (a, b+c),(b, c+a) \) and \( (c, a+b) \).
To do:
We have to find the area of the triangle.
Solution:
Let the vertices of the triangle be
$A(a, b+c), B (b, c+a)$ and $C (c, a+b)$
We know that,
Area of a triangle $C =\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
Therefore,
Area of triangle ABC $=\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]$
$=\frac{1}{2}[a(c-b)+b(a-c)+c(b-a)]$
$=\frac{1}{2}(a c-a b+a b-b c+b c-a c)$
$=\frac{1}{2}(0)$
$=0$
Hence, the area of the given triangle is. 0.
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