The area of a triangle with


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) abc


(b) Let the vertices of a triangle are, A ≡ (x1, y1) ≡ (a, b + c)

B ≡ (x2, y2) ≡ (b,c + a) and C = (x3, y3) ≡ (c, a + b)

$\because$ Area of $\triangle A B C=\Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore \quad \Delta=\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]$



$=\frac{1}{2}(a c-a b+a b-b c+b c-a c)=\frac{1}{2}(0)=0$

Hence, the required area of triangle is. 0.

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