Question

sagrika walks 8mtr west and 15 mtr north how far is she away her initial position

Answer 1

Answer :

Sagarika walks 8 m West and 15 m North; we can draw a right-angled triangle whose two sides be 8 m and 15 m.

Thus, the required distance be

= √(15² + 8²) m

= √(225 + 64) m

= √289 m

= 17 m

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Answer 2

Step-by-step explanation:

$\sqrt{(8^{2}+15^{2}) }$

$\sqrt{64+225}$

$\sqrt{289}$ =17m

By:

D.Kamal (HitmanBeastman50)