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
5a2sq. units
5a22sq. units
None of these
25a22sq. units
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Solution
The correct option is B5a22sq. 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 ⇒√(2a−2a)2+(t−0)2=√(2a−0)2+(t−a)2 ⇒t2=4a2+t2+a2−2at t=5a2 ∴Coordinates of third vertex is(2a,5a2) The area ofΔABC is ±12∣∣
∣
∣
∣∣2a5a212a010a1∣∣
∣
∣
∣∣ =±12|−2a2−5a2+2a2|=5a22sq. units