Question

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 triangle, in square units, is :

• A. $\frac{5}{4}{a}^2$
• B. $\frac{5}{2}{a}^2$
• C. $\frac{25a^2}{4}$
• D. $5a^2$

Answer 1

Answer: (B)

Let $y$-coordinate of $C=b$
$AB=\sqrt{4a^2+a^2}=\sqrt{5}a$
Now, $AC=BC\Rightarrow b=a^2\sqrt{4a^2+{\left(b-a\right)}^2}$
$b^2=4a^2+b^2+a^2-2ab$
$\Rightarrow 2ab=5a^2\Rightarrow b=\frac{5a}{2}$
$\therefore C=\left(2a,\frac{5a}{2}\right)$
Hence area of the triangle
$=\frac{1}{2}\left|\begin{array}{ccc}2a& 0& 1\\ 0& a& 1\\ 2a& \frac{5a}{2}& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}2a& 0& 1\\ 0& a& 1\\ 0& \frac{5a}{2}& 0\end{array}\right|$
$=\frac{1}{2}\times 2a\left(-\frac{5a}{2}\right)=-\frac{5a^2}{2}$
Since area is always +ve, hence area
$=\frac{5a^2}{2}$ sq.unit