Question: The vertices of a quadrilateral ABCD are A (0, 0), B (4, 4), C (4, 8) and D (0, 4). Then, ABCD is a
Options:
A) square
B) rhombus
C) rectangle
D) parallelogram
Show Answer
Answer:
Correct Answer: D
Solution:
- $ AB^{2}={{(4-0)}^{2}}+{{(4-0)}^{2}}=32 $
$ BC^{2}={{(4-4)}^{2}}+{{(8-4)}^{2}} $
$ =0+16=16 $
$ CD^{2}={{(0-4)}^{2}}+{{(4-8)}^{2}} $
$ =16+16=32 $
$ AD^{2}={{(0-0)}^{2}}+{{(4-0)}^{2}} $
$ =(0+16)=16 $
$ AB=CD=\sqrt{32}=4\sqrt{2} $
$ BC=AD=\sqrt{16}=4 $
$ AC^{2}={{(4-0)}^{2}}+{{(8-0)}^{2}}=16+64=80 $
$ BD^{2}={{(0-4)}^{2}}+{{(4-4)}^{2}}=16+0=16 $
$ \therefore $ Diagonal AC $ \ne $ Diagonal BD
So, ABCD is a parallelogram.