Distance & Direction Test-2

A diagram is made according to the directions given in the question. Clearly A will be South-West of C.

From the figure given, clearly the direction is OC which is South-East.

We can solve this problem using the Pythagorean theorem.
First, let 's break down Ayush 's movements:
1. He walks 100m south.
2. He turns left (which means he is now facing east) and walks 200m.
3. He turns right (which means he is now facing south again) and walks 500m.
Now, let 's determine his horizontal (east-west) and vertical (north-south) displacements.
Horizontal displacement:
Ayush only moved horizontally during step 2, where he walked 200m east. Therefore, his total horizontal displacement is 200m.
Vertical displacement:
Ayush moved vertically during steps 1 and 3. In step 1, he walked 100m south, and in step 3, he walked 500m south. Therefore, his total vertical displacement is 100m + 500m = 600m.
Now we can use the Pythagorean theorem to calculate the distance between his starting point (his house) and his final position:
Distance = √(horizontal displacement² + vertical displacement² )
Distance = √(200m² + 600m² )
Distance = √(40,000m² + 360,000m² )
Distance = √(400,000m² )
Distance = 200 √10m
So the correct answer is 3 (200 √10m).

Step 1:  Tanuj starts from point  P  and walks  40 metres south .
Step 2:  He takes a  left turn  (facing east) and walks  30 metres  to reach point  Q .
Applying the Pythagorean Theorem:
The path forms a right-angled triangle with:
One side = 40 metres (southward movement)
Another side = 30 metres (eastward movement)
Using the Pythagorean theorem to find the distance from  P  to  Q :
PQ  = √(40 ²+ 30 ²) = √(1600 + 900) = √2500 = 50 metres
Direction of Q from P:
Since Tanuj moved south first and then east, Q  is located  South-East  of  P .
Answer:
c) 50 metres, South-East

Required distance= 20 +12
= 32 m in south direction

The last position of the man is P and OEP is a Right angled triangle in which
EP = 3 km and OE = 4 km
Thus,
OP = √(3²  +4² )
OP = √25
OP = 5 km.

Let 's analyze the path taken by A and the distances involved:

1. A moves towards B 's house : A starts moving north towards B 's house which is 4 km away.
2. A covers three-fourth of the distance : He covers 4 ×3/4 = 3 km towards the north.
3. A takes a left turn : At this point, A is 1 km away from B 's house (since he has covered 3 km out of 4 km). He then takes a left turn, which would be towards the west and runs 2 km.

Now, A is 1 km south of B 's house and 2 km west of a direct line north from B ’s house. To find the distance from point P (A 's new location) to B 's house, we use the Pythagorean theorem:

Thus, A would need to cover a distance of: B: √5km   

1. A walks three-fourths of the distance towards B 's house, which is:
(3/4) ×4 km = 3 km  (north).
After walking 3 km north, A takes a left turn and runs 2 km west.

2. Now, we have a right-angled triangle, where:
- One leg (north) = 3 km,
- Another leg (west) = 2 km.

3. The distance from A ’s current position (P) to his house is the hypotenuse of this right-angled triangle:
Distance = √(3 ²+ 2 ²) = √(9 + 4) = √13 km.

Thus, the distance from A ’s current position to his own house is √13 km.

P is in South-West of Y.