If $A B C D E F$ is a regular hexagon, where two adjacent sides $\mathbf{A B}$ and $\mathbf{B C}$ are…

1. $\mathrm{b}-\mathrm{a}$
2. $a+b$
3. $a-b$
4. $a+2 b$

Solution

$A B C D E F$ is a regular hexagon where, $ \begin{aligned} \mathbf{A B} & =\mathbf{a} \\ \mathbf{B C} & =\mathbf{b} \\ \mathbf{C D} & =? \end{aligned} $
$\because A B C D E F$ is regular hexagon, then $ \mathbf{A B}=\mathbf{E D}, \mathbf{B C}=\mathbf{F E} \text { and } \mathbf{C D}=\mathbf{A F} $ Now, we know that $ \mathbf{A D}=2 \mathbf{B C}=2 \mathbf{b} $ and in $\triangle A B C$ $ \begin{aligned} \mathbf{A C} & =\mathbf{A B}+\mathbf{B C} [Using triangle law]\\ & =\mathbf{a}+\mathbf{b} \end{aligned} $ Now, In $\triangle A C D$ $ \begin{array}{rlrl} & & \mathbf{A C}+\mathbf{C D} & =\mathbf{A D} \\ \Rightarrow & \mathbf{a}+\mathbf{b}+\mathbf{C D} & =\mathbf{2} \mathbf{b} \\ \Rightarrow & & \mathbf{C D} & =\mathbf{2} \mathbf{b}-\mathbf{a}-\mathbf{b} \\ \Rightarrow & & \mathbf{C D} & =\mathbf{b}-\mathbf{a} \end{array} $

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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