Question

An aeroplane flies along the sides of an equilateral triangle with speed of 300 km/hr, 200 km/hr, 240 km/hr. the average speed of the plane while flying around the triangle is

Answer 1

Average speed = $\frac{Total distance covered}{Time taken}$
Let side of equilateral triangle = a km
Total distance covered = (a + a +a)km = 3a km
Time taken = sum of time taken by each side
                   = ($\frac{a}{300}$+$\frac{a}{200}$+$\frac{a}{240}$) hr
                   = $\frac{4a+6a+5a}{1200}$ hr
                   = $\frac{15a}{1200}$ hr
                   =$\frac{a}{80}$ hr
Hence, average speed = $\frac{3a}{a/80}$
                                     = 240 km/hr

Answer 2

Answer:

Step-by-step explanation:

speed =distance/time

distance =3a

total time=t1+t2+t3=a/300+a/200+a/240=15a/1200=a/80

Avg speed=total distance/total time=3a*80/a=240km/hr