Araft was sent away from point a downstream. a patrol-boat left a 5 hour and 20 min later and caught up with the raft 20 km away from point a. what was the speed of the raft if the speed of the patrol-boat was 12 km/hour?

The speed of the raft  if the speed of the patrol-boat was 12 km/hour, and it caught the raft in 5 hours 20 min later is 2.857 km/hours.

The speed of a body is the rate at which it covers the total distance is in the time taken. The speed of the body is given as,

Here, (d) is the distance travelled by the body and (t) is time taken by the body to cover that distance.

A raft was sent away from point A downstream. A patrol-boat left A 5 hour and 20 min later and caught up with the raft 20 km away from point A.

Total time the patrol boat took in hours is,

The speed of the patrol boat is 12 km/hour, and it caught up with the raft 20 km away from point A. Thus, the time taken with this speed to cover 20 km distance is,

Thus, the total time is,

Now as the raft travel 20 km in 7 hours. Therefore, find the speed of raft by putting the values in above formula as,

Hence, the speed of the raft  if the speed of the patrol-boat was 12 km/hour, and it caught the raft in 5 hours 20 min later is 2.857 km/hours.

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option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.