A) 14 sq. units
B) 2tsq. Units
C) 5 sq. units
D) 4 sq. units
Correct Answer: D
Solution :
Let \[A(t,\,t-2),\,B(t+2,t+2)\] and \[C(t+3,t)\] be the vertices of the given triangle. Then, Area of \[\Delta \,ABC=\frac{1}{2}|\{t(t+2-t)\] \[+(t+2)(t-t+2)+(t+3)(t-2-t-2)\}|\] \[\Rightarrow \] Area of \[\Delta ABC\] \[=\frac{1}{2}|\{2t+2t+4-4t-12\}|=|-4|=4\]sq. unitsYou need to login to perform this action.
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