A car travelled 70 km to the East of Patna and then110 km to the West

1.	A car travelled 70 km to the East of Patna and then110 km to the West from there. How far from Patnawas the car finally ?
Answer» Answer:- $\red{\bigstar}$ Distance between Patna and final point of car $\large\leadsto\boxed{\rm\purple{130.28 \: km}}$ • GIVEN:- Car TRAVELLED 70 km to East of Patna. Again travelled 110 km to the WEST from there. • To Find:- Total distance between the car and Patna. • Solution:- Let the starting point be A. After travelling 70 km let that point be B Final point after travelling 110 km be C Diagram:- $\setlength{\unitlength}{1cm}\begin{picture}\thicklines\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.1,2.5){\large\bf 110}\put(2.8,.3){\large\bf 70}\put(.7,4.8){\large\bf C}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf A}\end{pictue}$ Hence, The distance formed is a right angle Traingle Therefore, Using Pythagoras theorem:- (CA)² = (CB)² + (BA)² → (CA)² = (110)² + (70)² → (CA)² = 12100 + 4900 → (CA)² = 17000 → CA = √17000 → CA = 130.38 Therefore, the distance between the final point and the starting point is 130.38 km.

A car travelled 70 km to the East of Patna and then110 km to the West from there. How far from Patnawas the car finally ?

Answer»

Answer:-

$\red{\bigstar}$ Distance between Patna and final point of car $\large\leadsto\boxed{\rm\purple{130.28 \: km}}$

• GIVEN:-

Car TRAVELLED 70 km to East of Patna.

Again travelled 110 km to the WEST from there.

• To Find:-

Total distance between the car and Patna.

• Solution:-

Let the starting point be A.

After travelling 70 km let that point be B

Final point after travelling 110 km be C

Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}\thicklines\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.1,2.5){\large\bf 110}\put(2.8,.3){\large\bf 70}\put(.7,4.8){\large\bf C}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf A}\end{pictue}$

Hence,

The distance formed is a right angle Traingle

Therefore,

Using Pythagoras theorem:-

(CA)² = (CB)² + (BA)²

→ (CA)² = (110)² + (70)²

→ (CA)² = 12100 + 4900

→ (CA)² = 17000

→ CA = √17000

→ CA = 130.38

Therefore, the distance between the final point and the starting point is 130.38 km.