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If $A B C D E F$ is a regular hexagon, where two adjacent sides $\mathbf{A B}$ and $\mathbf{B C}$ are $\mathbf{a}$ and $\mathbf{b}$ respectively. Then $\mathbf{C D}$ is
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Verified Answer
The correct answer is:
$\mathrm{b}-\mathrm{a}$
$A B C D E F$ 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 $\triangle A C D$
$$
\begin{array}{rlrl}
& & \mathbf{A C}+\mathbf{C D} & =\mathbf{A D} \\
\Rightarrow & \mathbf{a}+\mathbf{b}+\mathbf{C D} & =\mathbf{2} \mathbf{b} \\
\Rightarrow & & \mathbf{C D} & =\mathbf{2} \mathbf{b}-\mathbf{a}-\mathbf{b} \\
\Rightarrow & & \mathbf{C D} & =\mathbf{b}-\mathbf{a}
\end{array}
$$
$$
\begin{aligned}
\mathbf{A B} & =\mathbf{a} \\
\mathbf{B C} & =\mathbf{b} \\
\mathbf{C D} & =?
\end{aligned}
$$
$\because A B C D E F$ 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 $\triangle A C D$
$$
\begin{array}{rlrl}
& & \mathbf{A C}+\mathbf{C D} & =\mathbf{A D} \\
\Rightarrow & \mathbf{a}+\mathbf{b}+\mathbf{C D} & =\mathbf{2} \mathbf{b} \\
\Rightarrow & & \mathbf{C D} & =\mathbf{2} \mathbf{b}-\mathbf{a}-\mathbf{b} \\
\Rightarrow & & \mathbf{C D} & =\mathbf{b}-\mathbf{a}
\end{array}
$$
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