Question

mohan goes 8m to west and then 6m due north. How far is he away from the starting point??​

Answer 1

Answer:

use the Pythagorean theorem to solve it ; answer = 10m

Step-by-step explanation:

$AB^{2} +BC^{2}= AC^{2}$

where CD is the hypotenuse in right angled triangle ABC ;

angle ABC= 90°;

C is the starting point and A is the ending point

BC=8m

AB= 6m

        AB^{2} +BC^{2}= AC^{2}

⇒8^2 + 6^2 = AC^2

⇒$AC^{2} = \sqrt{ 64+36}$

           =$\sqrt{100} = 10$

Answer 2

According to Pythagoreans theorem

= AB² + BC² = AC²

= 6m² + 8m² = AC²

= (6m×6m) + (8m×8m) = AC²

= 36m + 64m = AC²

= AC = √100m

= 10m

Hence, Mohan was 10m far from the starting point.