Question

A man travels A to B, B o C, C to D and then finally D to E. What will be shortest route that man could have taken?

Answer 1

Step-by-step explanation:

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Answer 2

Given:

( Note: refer the image given below for complete question)

A man travels A to B, B o C, C to D and then finally D to E. What will be shortest route that man could have taken?

Solution:

Refer the image given below,

The shortest route to reach D to E is $AC+CE$,

Take triangle ABC and find AC,

$\[\begin{array}{l} \Rightarrow A{C^2} = {4^2} + {3^2}\\ \Rightarrow A{C^2} = 16 + 9\\ \Rightarrow AC = \sqrt {25} \\ \Rightarrow AC = 5\,{\rm{m}}\end{array}\]$

Take triangle CDE and find CE,

$\[\begin{array}{l} \Rightarrow C{E^2} = {0.9^2} + {4^2}\\ \Rightarrow C{E^2} = 0.81 + 16\\ \Rightarrow CE = \sqrt {16.81} \\ \Rightarrow CE = 4.1\,{\rm{m}}\end{array}\]$

Therefore the shortest distance is,

$\[\begin{array}{l} = 5 + 4.1\\ = 9.1\,{\rm{m}}\end{array}\]$

Hence, shortest route that man could have taken is A to C then C to D of length 9.1 meter.