JEE Mains · Maths · STD 11 - 9. straight line
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 \(5{a^2}sq\). units
- B \(\frac{5}{2}{a^2}sq.\)units
- C \(\frac{{25{a^2}}}{2}sq.\)units
- D None of these
Answer & Solution
Correct Answer
(B) \(\frac{5}{2}{a^2}sq.\)units
Step-by-step Solution
Detailed explanation
(b) Let the co-ordinates of the third vertex be \((2a,\,\,t)\). \(AC = BC \Rightarrow t = \sqrt {4{a^2} + {{(a - t)}^2}} \Rightarrow \)\(t = \frac{{5a}}{2}\) So the coordinates of third vertex \(C\) are \(\left( {2a,\frac{{5a}}{2}} \right)\) Therefore area of the triangle…
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