$O A B C D$ is a pentagon in which the sides $O A$ and $C B$ are parallel and the sides $O D$ and $A B$ are…

$O A B C D$ is a pentagon in which the sides $O A$ and $C B$ are parallel and the sides $O D$ and $A B$ are parallel. Also, it is given that $\frac{O A}{C B}=2$, $\frac{O D}{A B}=\frac{1}{3} \cdot$ If $O A=\mathbf{a}, \mathbf{O D}=\mathbf{d}$, then $\mathbf{A D}+\mathbf{O C}+\mathbf{D C}=$
  1. $d-a$
  2. $\frac{1}{2} a+3 d$
  3. $\frac{1}{2} a+2 d$
  4. $6 d$

Solution

Given, $\mathbf{O A}=\mathbf{a}, \mathbf{O D}=\mathbf{d}$
Also, $\mathbf{O A}=2 \mathrm{CB} \Rightarrow \mathbf{C B}=\frac{\mathbf{O A}}{2}=\frac{\mathbf{a}}{2}$ and $3 \mathrm{OD}=\mathbf{A B} \Rightarrow \mathbf{A B}=3 \mathbf{d}$ Now, $\mathbf{A D}+\mathbf{O C}+\mathbf{D C}$ $=(\mathbf{A O}+\mathbf{O D})+(\mathbf{O A}+\mathbf{A B}+\mathbf{B C})$ $+(\mathbf{D O}+\mathbf{O A}+\mathbf{A B}+\mathbf{B C})$ $\begin{aligned} & =(-\mathbf{a}+\mathbf{d})+\left(\mathbf{a}+3 \mathbf{d}-\frac{\mathbf{a}}{2}\right)+\left(-\mathbf{d}+\mathbf{a}+3 \mathbf{d}-\frac{\mathbf{a}}{2}\right) \\ & =6 \mathbf{d}\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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