Question

Please solve this i will mark brainly​

Answer 1

$\huge{\mathfrak{\red{\underline{\underline{........Question........}}}}}$

A man travels 7 km due north then goes 3 km due east and then 3 km due south. How far is he from his starting point.

$\huge{\mathfrak{\red{\underline{\underline{........Solution........}}}}}$

Note : Diagram of this question attached in attachment file.

Given:

=> AB = 7 km

=> BC = 3 km

=> CD = 3 km

To find:

=> As per diagram AD is the shortest path between initial and final position. So, we have to find AD.

Formula used:

=> Pythagoras theorem

So, As we know AB = 7 km and AE = AB - BE

=> AE = 7 km - 3km

=> AE = 4 km

And,

=> ED = 3 km

Now, ΔADE is a right angled triangle. So, by using Pythagoras theorem we will find AD,

=> AD² = AE² + ED²

=> AD² = 4² + 3²

=> AD² = 16 + 9

=> AD² = 25

=> AD = 5 km

Hence, he is 5 km far from his starting point.