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
• B. Parallelogram
• C. Rhombus
• D. Rectangle

Answer 1

The correct option is A

Square

Using Phythagoras theorem in $\Delta$ AEH we get
$EH=\frac{\sqrt{10}}{4}$
Similarly we can get EH = FG = EF = GH
In $\Delta$AEH and $\Delta$HDG
AH = DG
AE = DH
$\angle$ EAH = $\angle$ HDG $\left({90}^0\right)$
$\Delta$ EAH $\cong \Delta$ HDG
Let $\angle$ AHE = $x^0$
Then $\angle$ DGH = $x^0$
$\angle$ DHG = ${90}^0-x^0$ (Using sum of angles in a triangle is ${180}^0$)
$\angle$EHG = ${90}^0$ (using sum of angles on a straight line)
Similarly we can prove others also.
So EHGF is a square.