A) square
B) rectangle
C) rhombus
D) trapezium
Correct Answer: B
Solution :
[b] \[AB=\sqrt{{{(9-9)}^{2}}+{{(6-0)}^{2}}}=\sqrt{{{0}^{2}}+{{6}^{2}}}=6\] |
\[BC=\sqrt{{{(-9-9)}^{2}}+{{(6-6)}^{2}}}=\sqrt{{{(-18)}^{2}}+{{0}^{2}}}=18\] |
\[CD=\sqrt{{{(-9+9)}^{2}}+{{(0-6)}^{2}}}=\sqrt{{{0}^{2}}+{{(-6)}^{2}}}=6\] |
\[DA=\sqrt{{{(9+9)}^{2}}+{{(0-0)}^{2}}}=\sqrt{{{(18)}^{2}}}=18\] |
\[\therefore \,\,\,\,\,\,\,AB=CD\] and \[BC=DA\] |
Now, \[AC=\sqrt{{{(-9-9)}^{2}}+{{(6-0)}^{2}}}=\sqrt{{{(18)}^{2}}+{{6}^{2}}}=\sqrt{360}\]and \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] |
\[\therefore \] Angle B is a right angle. |
Hence, ABCD is a rectangle. |
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