A than goes 5 km due east, then 5 Ian due south and then 5 km south east. His distance from the starting point is

A) 15 km.

B) 10 km.

C) \( \Large 5 \left(\sqrt{2}+1\right)\ km. \)

D) \( \Large \left(5\sqrt{3}-1\right) km. \)

Correct answer:
C) \( \Large 5 \left(\sqrt{2}+1\right)\ km. \)

Description for Correct answer:


A is the starting point.

D is the finishing point.

To find AD.

From \( \Large \triangle ABC,\ AC^{2} = AB^{2} + BC^{2} \)

= \( \Large 5^{2} + 5^{2} = 50 \)

=\( \Large AC = \sqrt{50} = 5\sqrt{2} \)

= \( \Large AD = 5 \sqrt{2}+5 = 5 \left(\sqrt{2}+1\right)\ km. \)


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