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 is

• A. $5a^2$
• B. $\frac{5a^2}{2}$
• C. $\frac{25a^2}{2}$
• D. None of these

Answer 1

The correct option is B

$\frac{5a^2}{2}$

Explanation for the correct option:

Step 1: Solve for the third vertex of the triangle

Given vertices of the base of the triangle be $A=(2a,0)$ and $B=(0,a)$

Let the third vertex on the line $x=2a$ be $C=\left(2a,t\right)$

Given that the triangle is Isosceles $\Rightarrow AC=BC$

Applying distance formula i.e. distance between two points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$ is given by $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\begin{array}{crclc}\therefore & AC& =& \sqrt{{\left(2a-2a\right)}^2+{\left(t-0\right)}^2}& \\ & & =& \sqrt{t^2}& \\ & & =& t& \\ & BC& =& \sqrt{{\left(2a-0\right)}^2+{\left(t-a\right)}^2}& \\ & & =& \sqrt{4a^2+t^2-2at+a^2}& \\ & & =& \sqrt{t^2-2at+5a^2}& \\ \Rightarrow & t& =& \sqrt{t^2-2at+5a^2}& \left[\because AC=BC\right]\end{array}$

Squaring on both sides, we get,

$\begin{array}{crcl}& t^2& =& t^2-2at+5a^2\\ \Rightarrow & 2at& =& 5a^2\\ \Rightarrow & t& =& \frac{5a^2}{2a}\\ \Rightarrow & t& =& \frac{5a}{2}\end{array}$

Therefore the coordinates of $C=\left(2a,\frac{5a}{2}\right)$

Step 2: Solve for area of the triangle

We know that the area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ is $\frac{1}{2}\left|\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ 1& 1& 1\end{array}\right|$

Therefore area of the $△ABC=\frac{1}{2}\left|\begin{array}{ccc}2a& 0& 2a\\ 0& a& \frac{5a}{2}\\ 1& 1& 1\end{array}\right|$

$\begin{array}{cl}=& \frac{1}{2}\left|2a(a-\frac{5a}{2})-0+2a(0-a)\right|\\ =& \frac{1}{2}\left|2a\left(\frac{-3a}{2}\right)-2a^2\right|\\ =& \frac{1}{2}\left|\frac{-6a^2-4a^2}{2}\right|\\ =& \frac{10a^2}{4}\\ =& \frac{5a^2}{2}units\end{array}$

Hence, the correct answer is option(B) i.e. $\frac{5a^2}{2}$.