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Question

Given a square of one unit side length. Four points are chosen on the square as shown. Then the quadrilateral EFGH so formed is a


A

Square

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B

Parallelogram

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C

Rhombus

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D

Rectangle

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Solution

The correct option is A

Square



Using Phythagoras theorem in Δ AEH we get
EH=104
Similarly we can get EH = FG = EF = GH
In ΔAEH and ΔHDG
AH = DG
AE = DH
EAH = HDG (900)
Δ EAH Δ HDG
Let AHE = x0
Then DGH = x0
DHG = 900x0 (Using sum of angles in a triangle is 1800)
EHG = 900 (using sum of angles on a straight line)
Similarly we can prove others also.
So EHGF is a square.


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