Question

Sachin walks 12 km towards east and turn to his right and walk 8 km. Then he again took right turn and walk 6 km. How far is sachin from the starting point

Answer 1

Answer:

10

Step-by-step explanation:

$= \sqrt{8 ^{2} + 6 ^{2} }$

$= \sqrt{64 + 36}$

$= \sqrt{100}$

$= 10$

Answer 2

Given,

Sachin walks 12 km towards the east and turn to his right and walk 8 km. Then he again took the right turn and walk 6 km.

To find,

The distance from the starting point.

Solution,

We can easily solve this problem by following the given steps.

According to the question,

Sachin walks 12 km towards the east and turn to his right and walk 8 km. Then he again took the right turn and walk 6 km.

Now, let's take his starting point to be A, the point when he travelled 12 km to be B, the point when he had travelled 8 km to be C and the last point after 6 km to be D.

So,

AB = 12 km

AB = 12 kmBC = 8 km

AB = 12 kmBC = 8 kmCD = 6 km

Now, draw a perpendicular from D to AB. Let's take this point to be E.

AE = AB-CD

AE = (12-6) km

AE = 6 km

BC = DE = 8 km

So,

∆ADE is a right-angled triangle.

Using the Pythagoras theorem,

AD² = AE²+DE²

AD² = (6)²+(8)²

AD² = 36+64

AD² = 100

AD =√100

AD = 10 km

Hence, Sachin is 10 km away from his starting point.