Q62 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

If $O$ is any point $O A+O B+O C+O D=x O E$, then find $x$, given that $A B C D$ is quadrilateral, $B$ is the point of intersection of the line joining the mid-points of opposite sides.

  • Correct answer
    A
    4
  • B
    3
  • C
    5
  • D
    9

Solution

Let $P, Q, R$ and $S$ are the mid-point of sides of a quadrilateral $A B C D$ respectively. $$ \begin{aligned} & P=\text { mid-point of } \mathbf{A B}=\frac{\mathbf{a}+\mathbf{b}}{2} \\ & Q=\text { mid-point of } \mathbf{B C}=\frac{\mathbf{b}+\mathbf{c}}{2} \\ & R=\text { mid-point of } \mathbf{C D}=\frac{\mathbf{c}+\mathbf{d}}{2} \\ & S=\text { mid-point of } \mathbf{A D}=\frac{\mathbf{a}+\mathbf{d}}{2} \\ & \underline{\mathbf{a}+\mathbf{b}}+\frac{\mathbf{c}+\mathbf{d}}{2} \end{aligned} $$ Mid-point of $P R=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4}$ Mid-point of $S Q=\frac{\frac{\mathbf{a}+\mathbf{b}}{2}+\frac{\mathbf{c}+\mathbf{d}}{2}}{2}$ $$ =\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} $$ $\therefore E=$ mid-point of $P R=$ mid-point of $S Q$ $$ \begin{aligned} \therefore \quad \mathbf{O E}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}}{4} & \\ 4 \mathbf{O E}=\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d} & \\ \therefore \mathbf{O A}+\mathbf{O B}+\mathbf{O C}+\mathbf{O D} & =4 \mathbf{O E} \\ x \mathbf{O E} & =4 \mathbf{O E} \\ x & =4 \end{aligned} $$ Hence, option (1) is correct.

Hence, the correct answer is option: 'A'.

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