A) 4 sq unit
B) \[4\sqrt{2}\,\]sq unit
C) 8 sq unit
D) \[\left( -\frac{1}{5},-\frac{22}{5} \right)\]
Correct Answer: C
Solution :
Given that coordinate of \[B(-3,6)\]and C (1, 2). \[\therefore \] \[BC=\sqrt{{{4}^{2}}+{{(-4)}^{2}}}\] \[=4\sqrt{2}\] Since, \[AB=AC\]\[\Rightarrow \]\[\angle B=\angle C={{45}^{o}}\] In \[\Delta ABC,\] \[\cos ={{45}^{o}}=\frac{AB}{BC}\] \[\Rightarrow \] \[AB=4\sqrt{2}.\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[AB=AC=4\] \[\therefore \] Area of triangle \[=\frac{1}{2}\times BA\times AC\] \[=\frac{1}{2}\times 4\times 4\] \[=8\,\text{q}\,\text{unit}\]You need to login to perform this action.
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