question_answer

                If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and the equation of one of the sides is \[x=2a,\] then the area of the square units, is:     JEE Main Online Paper ( Held On 23  April 2013 )

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. unit