Question

3. A boy first goes 5 m due north and then 12 m due east. Find the distance between
the initial and the final positions of the boy​

Answer 1

✬ Distance = 13 m ✬

Step-by-step explanation:

Given:

• A boy first goes 5 m due north and then 12 m due east.

To Find:

• What is the distance between the initial and final position of boy ?

Solution: Let the north direction be AC and east direction be AB.

➟ AC = North = 5 m

➟ AB = East = 12 m

➟ BC = Distance from initial to final

In right angled ∆ABC, We have

• AC = Perpendicular
• AB = Base
• BC = Hypotenuse

Applying Pythagoras Theorem,

★ H² = B² + P² ★

$\implies{\rm }$ BC² = AB² + AC²

$\implies{\rm }$ BC² = 12² + 5²

$\implies{\rm }$ BC² = 144 + 25

$\implies{\rm }$ BC² = 169

$\implies{\rm }$ BC = √169

$\implies{\rm }$ BC = 13

Hence, required distance is 13 m.

Answer 2

$\large\bf{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• A boy first goes ‘5m’ due to North and ‘12m’ due to east .

To Find :-

• The distance between the initial and the final position of the boy .

CALCULATION :-

☞ See the attachment picture .

________________________________

☞ As shown in the picture,

• AB (due to North) = 5m
• BC (due to East) = 12m

▼ For measurement of AC , we use the Pythagoras theorem .

▼ According to the picture, the Pythagoras theorem is,

$\checkmark\:\rm{\purple{\boxed{(AC)^2\:=\:(AB)^2\:+\:(BC)^2\:}}}$

$\rm{\implies\:(AC)^2\:=\:(5)^2\:+\:(12)^2\:}$

$\rm{\implies\:(AC)^2\:=\:25\:+\:144\:}$

$\rm{\implies\:(AC)^2\:=\:169\:}$

$\rm{\implies\:AC\:=\:\sqrt{169}\:}$

$\rm{\implies\:AC\:=\:±13\:}$

[Note:- Distance never negative.]

$\rm{\boxed{\blue{\implies\:AC\:=\:13m\:}}}$

✪ Therefore, the distance between the initial and the final position of the boy is ‘13m’ .