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Question

If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and equation of one of the sides is x=2a, then the area of triangle is
  1. 5a2 sq. units
  2. 5a22 sq. units
  3. None of these
  4. 25a22 sq. units

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Solution

The correct option is B 5a22 sq. units
Given: extremities of the base of an isosceles triangle are the points (2a, 0)
and (0, a)
Equation of one of the sides is x=2a
Let A(2a,0),B(0,a)
Coordinates of the third side is C(2a,t)
As ABC is isosceles triangle
AC=BC
(2a2a)2+(t0)2=(2a0)2+(ta)2
t2=4a2+t2+a22at
t=5a2
Coordinates of third vertex is(2a,5a2)
The area of ΔABC is
±12∣ ∣ ∣ ∣2a5a212a010a1∣ ∣ ∣ ∣
=±12|2a25a2+2a2|=5a22 sq. units

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