Exam

JEE Main

Year

2013

Date

23 April

Mode

Online

Subject

Mathematics

Chapter

Properties of Triangles

JEE Main 2013 · 23 April · Online · Mathematics · Properties of Triangles

If the extremities of the base of an isosceles triangle are the...

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 :

Step-by-step solution

\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} \Rightarrowb=\sqrt{4 a^2+(b-a)^2}\\ & \Rightarrow b²=4 a²+b²+a²-2 a b \\ & \Rightarrow 2 a b=5 a² \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} \times2 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 }