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If the extremities of the base of an isosceles triangle are the points $(2 a, 0)$ and $(0, a)$ and the equation of one of the sides is $x=2 a$, then the area of the triangle, in square units, is :
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Verified Answer
The correct answer is:
$\frac{5}{2} a^2$
$\frac{5}{2} a^2$
$$
\text { Let } y \text {-coordinate of } \mathrm{C}=b
$$
$$
\begin{aligned}
& \mathrm{AB}=\sqrt{4 a^2+a^2}=\sqrt{5} a \\
& \text { Now, } \mathrm{AC}=\mathrm{BC} \Rightarrow b=\sqrt{4 a^2+(b-a)^2} \\
& \Rightarrow b^2=4 a^2+b^2+a^2-2 a b \\
& \Rightarrow 2 a b=5 a^2 \Rightarrow b=\frac{5 a}{2} \\
& \therefore \mathrm{C}=\left(2 a, \frac{5 a}{2}\right)
\end{aligned}
$$
Hence area of the triangle
$$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
2 a & \frac{5 a}{2} & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
0 & \frac{5 a}{2} & 0
\end{array}\right| \\
& =\frac{1}{2} \times 2 a\left(-\frac{5 a}{2}\right)=-\frac{5 a^2}{2}
\end{aligned}
$$
Since area is always $+\mathrm{ve}$, hence area
$$
=\frac{5 a^2}{2} \text { sq. unit }
$$
\text { Let } y \text {-coordinate of } \mathrm{C}=b
$$
$$
\begin{aligned}
& \mathrm{AB}=\sqrt{4 a^2+a^2}=\sqrt{5} a \\
& \text { Now, } \mathrm{AC}=\mathrm{BC} \Rightarrow b=\sqrt{4 a^2+(b-a)^2} \\
& \Rightarrow b^2=4 a^2+b^2+a^2-2 a b \\
& \Rightarrow 2 a b=5 a^2 \Rightarrow b=\frac{5 a}{2} \\
& \therefore \mathrm{C}=\left(2 a, \frac{5 a}{2}\right)
\end{aligned}
$$
Hence area of the triangle
$$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
2 a & \frac{5 a}{2} & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
0 & \frac{5 a}{2} & 0
\end{array}\right| \\
& =\frac{1}{2} \times 2 a\left(-\frac{5 a}{2}\right)=-\frac{5 a^2}{2}
\end{aligned}
$$
Since area is always $+\mathrm{ve}$, hence area
$$
=\frac{5 a^2}{2} \text { sq. unit }
$$
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