A) \[0\]
B) \[4\]
C) \[6\]
D) none of these
Correct Answer: A
Solution :
Key Idea: For cyclic quadrilateral Area\[=\sqrt{(s-a)(s-b)(s-c)(s-d)}\] \[\therefore \]Let points are \[A(2,\,\,3),\,\,B(3,\,\,4),\,\,C(4,\,\,5),\,\,D(5,\,\,6)\] \[\therefore \] \[AB=\sqrt{{{(3-2)}^{2}}+{{(4-3)}^{2}}}=\sqrt{2}\] Similarly, \[BC=\sqrt{2},\,\,CD=\sqrt{2}\] and \[DA=3\sqrt{2}\] \[\therefore \] \[a=\sqrt{2}=b=c,\,\,d=3\sqrt{2}\] Now, \[s=\frac{a+b+c+d}{2}\] \[=\frac{\sqrt{2}+\sqrt{2}+\sqrt{2}+3\sqrt{2}}{2}=\frac{6\sqrt{2}}{2}=3\sqrt{2}\] \[\therefore \]Area\[=\sqrt{(s-a)(s-b)(s-c)(s-d)}\] \[=\sqrt{(3\sqrt{2}-\sqrt{2})(3\sqrt{2}-\sqrt{2})}\] \[=0\] Alternate Method: Area of quadrilateral\[=\]Area of\[\Delta ABD+\]Area of\[\Delta BDC\] ? (i) Now, Area of\[\Delta ABD=\frac{1}{2}\left| \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 5 & 6 & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}[2(4-6)-3(3-5)+1(18-20)]\] \[=\frac{1}{2}[-4+6-2]=0\] ? (ii) Similarly, Area of\[\Delta ABC=\frac{1}{2}\left| \begin{matrix} 5 & 6 & 1 \\ 4 & 5 & 1 \\ 3 & 4 & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}[5(5-4)-6(4-3)+1(16-15)]\] \[=\frac{1}{2}[5-6+1]\] \[=0\] ? (iii) From (i), (ii), (iii) Area of quadrilateral\[=0+0=0\] Note: Notice that all the given point lies on the straight line\[y=x+1\] \[\therefore \]Points are collinear\[\Rightarrow \]Area\[=0\].You need to login to perform this action.
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