Question

A submarine travels 10 km due North from its base and then turns and travels due West for 8.8 km.
How far away is the submarine from its base?

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Distance=13.32\:km\:\:North-West}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a submarine travels 10 km due North from its base and then turns and travels due West for 8.8 km.

• We have to find the distance from submarine to its base.

$\underline \bold{Given : } \\ \implies AB = 10 \: km \\ \\ \implies BC = 8.8 \:km \\ \\ \underline \bold{to \: find : } \\ \implies AC = ?$

• According to given question :

$\bold{ \angle B = 90 \degree}\\ \\ \bold{Using \: Phythagoras \: theoram : } \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \\ \implies {AC}^{2} = ( {10})^{2} + {(8.8)}^{2} \\ \\ \implies {AC}^{2} = 100 + 77.44 \\ \\ \implies {AC}^{2} = 177.44 \\ \\ \implies AC= \sqrt{177.44} \\ \\ \bold{\implies AC = 13.32 \: km} \\ \\ \bold{\therefore Submarine \: is \: 13.32 \: km \:\: North -West\: away \: from \: base}$

Answer 2

Answer_____

• 13.32 km

Step by step explanation-

$\bold{using \: phythagoras \: theoram : } \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies {ac}^{2} = ( {10})^{2} + {(8.8)}^{2} \\ \\ \implies {h}^{2} = 100 + 77.44 \\ \\ \implies {h}^{2} = 177.44 \\ \\ \implies h = \sqrt{177.44} \\ \\ \bold{\implies h = 13.32 \: km}$