Question

A boy walks 17 m due north, then 12 m due east and then 12 m south on a flat ground. How far away is he from the starting point?​

Answer 1

he is from starting point to , north east

Answer 2

Given,

The boy walks 17m towards north.

Then 12m towards east.

Then 12m towards south.

To find,

The distance between the initial point of journey and finishing point of journey.

Solution,

We can simply solve this mathematical problem by using the following mathematical process.

From the attached diagram,

The boy starts journey from point A, and walk towards north for 17m in order to reach point B. And, from point B he walks 12m towards east to reach point C. Then, from point C, he walks 12m towards south to reach point D.

Now, we have to calculate the distance between A and D.

Additionally in the diagram, we have dropped a perpendicular from point D on AB, which intersects AB at point X.

Now, we know that, North, East, West and South directions are perpendicularly situated to each other.

So, Angle ABC = Angle XBC = 90°

Angle BCD = 90°

We know that, Angle BXD = 90°

In the four sided polygon BCDX, the three internal angles are = 90°,90°,90°

Fourth internal angle of BCDX = 360° - (3×90°) = 90°

So, all the four internal angles of BCDX = 90°

And, two adjacent sides are equal = BC = CD = 12m

This implies BCDX is a square.

So, XD = BC = BX = CD = 12m

Now,

AB = AX + BX

17 = AX + 12

AX = 17-12 = 5m

To find out the length of AD we have to apply the Pythagoras theorem, (because, Angle AXD is 90° and ∆AXD is a right angle triangle.)

(AX)²+(XD)² = (AD)²

(AD)² = (12)²+(5)²

(AD)² = 144+25

(AD)² = 169

AD = 13

Hence, he is 13m far away from the starting point.