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. |
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