A) \[\frac{5}{4}{{\operatorname{n}}^{2}}\]
B) \[\frac{5}{2}{{\operatorname{n}}^{2}}\]
C) \[\frac{25{{a}^{2}}}{4}\]
D) \[5{{a}^{2}}\]
Correct Answer: B
Solution :
Let y-coordinate of C =b \[\therefore \]C = (2a,b) \[AB=\sqrt{4{{a}^{2}}+{{a}^{2}}}=\sqrt{5}a\] Now,\[AC=BC\Rightarrow b=\sqrt{4{{a}^{2}}+{{(b-a)}^{2}}}\] \[\Rightarrow \]\[{{b}^{2}}=4{{a}^{2}}+{{b}^{2}}+{{a}^{2}}-2ab\] \[\Rightarrow \]\[2ab=5{{a}^{2}}\Rightarrow b=\frac{5a}{2}\] \[\therefore \]\[C=\left( 2a,\frac{5a}{2} \right)\]Hence area of the triangle \[=\frac{1}{2}\left| \begin{matrix} 2a & 0 & 1 \\ 0 & a & 1 \\ 2a & \frac{5a}{2} & 1 \\ \end{matrix} \right|=\frac{1}{2}\left| \begin{matrix} 2a & 0 & 1 \\ 0 & a & 1 \\ 0 & \frac{5a}{2} & 0 \\ \end{matrix} \right|\] \[=\frac{1}{2}\times 2a\left( -\frac{5a}{2} \right)=-\frac{5{{a}^{2}}}{1}\] Since area is always +ve, hence area \[=\frac{5{{a}^{2}}}{2}\]sq. unitYou need to login to perform this action.
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