Question

An aeroplane leaves an Airport and flies due north at 300 km/h. At the same time, another

aeroplane leaves the same Airport and flies due west at 400km/h.

Distance travelled by the second aeroplane in 1.5 hours​

Answer 1

Solution:-

Formula Used

• Distance = Speed × Time

The Distance travelled by aeroplane A toward North = 300 × 1.5 = 15×30 = 450 km

and The distance travelled by the aeroplane B in West = 400 ×15/10 = 40×15 = 600 km

Now, we have to calculate the distance between Aeroplane

A & Aeroplane B .

Using Pythagoras theorem we get

• h² = p² + b²

Where,

• h denotes hypotenuse
• p denotes perpendicular
• b denotes base

Substitute the value we get

→ AC² = AO² + BO²

→ AC² = (450)² + (600)²

→ AC² = 202500 + 360000

→ AC² = 562500

→ AC = √562500

→ AC = 750 Km

• Therefore, The distance between aeroplane A & Aeroplane B is 750 Km

For More Information Refer to Attachment !!

Answer 2

Answer:

Given :-

An aeroplane leaves an Airport and flies due north at 300 km/h. At the same time, another

aeroplane leaves the same Airport and flies due west at 400km/h.

To Find :-

Distance travelled by the second aeroplane in 1.5 hours

Solution :-

As we know that

$\huge \bf \: D = S \times T$

Distance travelled by aeroplane A

$\tt \: D = 300 \times 1.5$

$\tt \: D = 450 \: km$

Distance travelled by aeroplane B

$\tt \: D = 400 \times 1.5$

$\tt \: D = 600 \: km$

Now,

Let's calculate distance travelled by using Pythagoras theorem

$\huge \bf {h}^{2} = {p}^{2} + {b}^{2}$

Here,

H = Hypotenuse

P = Perpendicular

B = Base

Putting values

H² = (450)²+(600)²

H² = 202,500 + 360,000

H² = 562,500

H = √562500

H = 750 km

Therefore,

Distance between A and B is 750 km