question_answer

Area of quadrilateral whose vertices are \[(2,\,\,3)\] \[(3,\,\,4),\,\,(4,\,\,5)\] and \[(5,\,\,6)\] is equal to

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\[=0\].