question_answer

The vertices of a quadrilateral ABCD are A (0, 0), B (4, 4), C (4, 8) and D (0, 4). Then, ABCD is a

A) square              

B) rhombus

C) rectangle                       

D) parallelogram

Correct Answer: D

Solution :

\[A{{B}^{2}}={{(4-0)}^{2}}+{{(4-0)}^{2}}=32\]

\[B{{C}^{2}}={{(4-4)}^{2}}+{{(8-4)}^{2}}\]

\[=0+16=16\]

\[C{{D}^{2}}={{(0-4)}^{2}}+{{(4-8)}^{2}}\]

\[=16+16=32\]

\[A{{D}^{2}}={{(0-0)}^{2}}+{{(4-0)}^{2}}\]

\[=(0+16)=16\]

\[AB=CD=\sqrt{32}=4\sqrt{2}\]

\[BC=AD=\sqrt{16}=4\]

\[A{{C}^{2}}={{(4-0)}^{2}}+{{(8-0)}^{2}}=16+64=80\]

\[B{{D}^{2}}={{(0-4)}^{2}}+{{(4-4)}^{2}}=16+0=16\]

\[\therefore \] Diagonal AC \[\ne \]Diagonal BD

So, ABCD is a parallelogram.