TS EAMCET · Maths · Properties of Triangles
\(A B C D E F\) is a regular hexagon whose centre is O. Then, \(\mathbf{A B}+\mathbf{A C}+\mathbf{A D}+\mathbf{A E}+\mathbf{A F}\) is equal to
- A \(2 \mathrm{AO}\)
- B 3 AO
- C \(5 \mathrm{AO}\)
- D \(6 \mathrm{AO}\)
Answer & Solution
Correct Answer
(D) \(6 \mathrm{AO}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \because \quad {AB}={ED}, {AF}={CD} \\ & \therefore \quad {AB}+{AC}+{AD}+{AE}+{AF} \\ & =A E+E D+A C+C D+A D \\ & =A D+A D+A D \\ & =3 {AD} \\ & \end{aligned} \) \(\because O\) is the centre of regular hexagon.…
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