A) Rhombus
B) Parallelogram
C) Square
D) Rectangle
Correct Answer: B
Solution :
Let the points be \[A(6,8),\,\,B(3,\,7),\,\,C(-2,-2)\]And \[D(1,-1)\] Now, \[AB=\sqrt{{{(3-6)}^{2}}+{{(7-8)}^{2}}}=\sqrt{10}\] \[BC=\sqrt{{{(-2-3)}^{2}}+{{(-2-7)}^{2}}}=\sqrt{106}\] \[CD=\sqrt{{{(1+2)}^{2}}+{{(-1+2)}^{2}}}=\sqrt{10}\] \[DA=\sqrt{{{(6-1)}^{2}}+{{(8+1)}^{2}}}=\sqrt{106}\] Also\[AC=\sqrt{{{(8)}^{2}}+{{(10)}^{2}}}=\sqrt{64+100}=\sqrt{164}\] \[BD=\sqrt{{{(2)}^{2}}+{{(8)}^{2}}}=\sqrt{4+64}=\sqrt{68}\] Since, \[AB=DC\]and \[BC=DA\]and\[AC\ne BD.\]. \[\therefore \] It is a parallelogram.You need to login to perform this action.
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