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 $

AcademicMathematicsNCERTClass 10

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.

raja
Updated on 10-Oct-2022 13:28:28

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