Let ABCD be a square and E be a point outside ABCD such that E, A, C are collinear in that order. Suppose EB = ED = \sqrt130 and the areas of triangle EAB and square ABCD are equal. Then the area of square ABCD is

KVPY SA Exam 05-Nov-2017 Question Paper

Question : 64 of 80

Marks: +1, -0

Let ABCD be a square and E be a point outside ABCD such that E, A, C are collinear in that order. Suppose

EB = ED = √130

and the areas of triangle EAB and square ABCD are equal. Then the area of square ABCD is

Solution:

given : ar ∆EAB = ar square ABCD

EB = ED = √130

Let side of square = x

BM =x/√2=AM

\ar∆EAB = \ar∆EMB -\ar∆ABM

=1/2EM× BM-1/2AM .BM

=1/2(√{130-x^2/2})x/√2-1/2(x/√2)^2

Area of square

= x^2

Using (i)

1/2x/√2√{130-x^2/2}-1/2x^2/2=x^2

x/{2√2}√{130-x^2/2}-{5x^2}/4

solve

x =√ 10

∴

x^2=10

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