Q.

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

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a

5a2  sq.  units

b

5a22   sq.  units

c

25a22   sq.  units

d

none of these

answer is B.

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Detailed Solution

Given vertices are A2a,0 and B0,aLet the coordinates of the third vertex C are 2a, t. Now, AC=BC. Hence,t=4a2+a−t2 or t=5a2So, the coordinates of the third vertex C are 2a,  5a/2.Therefore, area of the triangle is 122a5a/212a010a1=5a22sq. units
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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