A) 7
B) 3
C) 6
D) 8
Correct Answer: C
Solution :
[c] Since ABCD is a square. |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,AB=AD\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,A{{B}^{2}}=A{{D}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,{{(1-5)}^{2}}+{{(5-p)}^{2}}={{(6-5)}^{2}}+{{(2-p)}^{2}}\] |
\[\Rightarrow \,\,\,16+25-10p+{{p}^{2}}=1+4-4p+{{p}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6p=36\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=6\] |
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