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  • If the extremities of the base of an isosceles triangle are points <img alt="\left ( 2a,0 \right )" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cinline%20%5Cleft%20%28%202a%2C0%20%5C

If the extremities of the base of an isosceles triangle are points \left ( 2a,0 \right ) and \left ( 0,a \right ) and the equation of one of the sides is x=2a, then the area of the triangle, in square units, is :

Option: 1

\frac{5}{4}a^{2}


Option: 2

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


Option: 3

\frac{25a^{2}}{4}


Option: 4

5a^{2}


Answers (1)

best_answer

Let y-coordinate of C is b.

therefore, C = (2a,b)

\\\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)

Hence, the area of the triangle

\\=\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}

Since area can not be -ve, so

\text{the area of the triangle }=\frac{5 a^{2}}{2}

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Nehul

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