Question

a travels 5 km towards north and Travels 12 km towards west. how far is A from B , now?

please answer me​

Answer 1

Correct Question:

A travels 5 km towards north and B travels 12 km towards west, how far is A from B , now?

Solution:

let F be the initial point of A and B

FA = 5km

FB = 12 km

let AB = x

[ ΔAFB is a right triangle, as the both directions are perpendicular ]

According to Pythagorean theorem,

↬ ( FA )² + ( FB )² = ( AB )²

↬(5km)² + (12km)² = ( AB )²

↬25km² + 144km² = ( AB )²

↬169km² = ( AB)²

↬$\rm \sqrt{169 \: {km}^{2} } = AB$

↬ 13km = AB

A is 13 km away from B

Answer 2

$\large{\boxed{\boxed{\sf Correct \: Question}}}$

A travels 5 km towards north and B travels 12 km towards west. Now, how far is A from B?

$\large{\boxed{\boxed{\sf Let's \: Understand \: Question \: F1^{st}}}}$

Here, we have said that A travels 5km towards North and from the same point B travels 12km towards West and have to find the distance between A and B.

$\large{\boxed{\boxed{\sf How \: to \: do \: it?}}}$

Here, after understanding Question we f1st made its fig. and then observing it's fig. carefully we apply Pythogoras Theorem in it and substituting given values in it we can obtain distance between A and B as our required answer.

Let's Do It ⚡

$\huge{\underline{\boxed{\sf AnSwer}}}$

_____________________________

Given:-

• A travels 5km towards North
• B travels 12km towards West

Find:-

• Distance between A and B

Diagram:-

Let, Initial position/starting point from which A and B starts travelling be C

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line( - 1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(-3.5,1)\put(1.2,2.5){\large\bf 5 km}\put( -1.7,.3){\large\bf 12 km}\put(0.68,1.02){\framebox(0.3,0.3)}\put(.7,4.6){\large\bf A}\put( - 3.7,.6){\large\bf B}\put(.8,.6){\large\bf C}\end{picture}$

Solution:-

Here, In ∆ABC

$\sf \mapsto H^2 = P^2 + B^2 \qquad \bigg\lgroup Pythogoras\: Theorem\bigg\rgroup$

$\sf \leadsto AB^2 = AC^2 + BC^2$

$\sf where \small{\begin{cases} \sf AC = 5km \\ \sf BC = 12km \end{cases}}$

$\pink\bigstar$ Substituting these values:-

$\sf \dashrightarrow AB^2 = AC^2 + BC^2$

$\sf \dashrightarrow AB^2 = 5^2 + 12^2$

$\sf \dashrightarrow AB^2 = 25 + 144$

$\sf \dashrightarrow AB^2 = 169$

$\sf \dashrightarrow AB = \sqrt{169}$

$\sf \dashrightarrow AB = 13km$

$\underline{\boxed{\sf \therefore A\:is\:13km\:far\:from\:B}}$