Question

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

Answer 1

Solution:

The area of a triangle = the absolute value of Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) divided by 2

(A, B + C)
(B, C + A)
(C, A + B)

[A(C + A) - (A + B)] + [B(A + B) - (B + C)] + [C(B + C) - C + A)]/ 2

A(C + A - A - B) = A(C - B)
B(A + B - B - C) = B(A - C)
C(B + C - C - A) = C(B - A)

[A(C - B) + B(A - C) + C(B - A)]/2

[AC - AB + BA - BC + CB - AC)]/2

0/2 = 0

Therefore, the correct answer is 0 which is choice B

Hope this will be helpful to you.