# Selina Solutions Class 9 Mathematics Solutions for Exercise 13(A) in Chapter 13 - Chapter 13- Pythagoras Theorem

Question 2 Exercise 13(A)

A man goes 40m due north then 50 m due west. Find his distance from the starting point.

Solution:

Here, we need to measure the distance AB as shown in the figure below,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to

the sum of the squares on the remaining two sides.

Therefore, in this case,

Therefore the required distance is 64.03 m.

Video transcript
"hello students welcome to lido homework i am tutor aaron bhatti from leader learning now let us solve this interesting question a man goes 40 meter dew north then 50 meter dew best find the distance from the starting point as this is related with directions first we will know how to write the directions so i will draw it here always the top will be our north and below it will be south and in the right side we have our east direction and the left is the west direction so for this question let us draw a reference diagram so they have told the person goes 40 meter from north so let this point be a and it will be the starting point from a he is moving 40 meters north so up 40 meters and let me name this point as b so the distance between a and b is 40 meter and next from the point b he is moving 50 meters west so from b he is moving west by 50 meters let me name this point as c and this is 50 meters the question is we have to find the distance from the starting point that is we have to measure the distance from a to c so we have write the things what we have done so let a be the starting point and a b is equal to 40 meter and c is equal to sorry bc is equal to 50 meter we have to find the distance between a and c now so here we could see the directions where the person has formed is forming a right angled triangle so we could take this as a right angled triangle so in a right angled triangle abc by using the pythagoras theorem we know that the square of the hypotenuse side will be equal to the sum of the square of the other two sides that is ac the whole square is equal to a b the whole square plus bc the whole square here we know what is a b and b c so we'll get ac the whole square is equal to a b is 40 the whole square plus bc is 50 the whole square now on squaring we'll get 40 as plus 50 the whole square will be now on adding them we will get ac the whole square is equal to six thousand hundred as we need ac that is equal to root of six one double zero on taking the square root we will get ac is equal to sixty four point zero three therefore we have got our ac as sixty four point zero three therefore the required distances sixty four point meter if you have any questions please post your questions in the comment box and subscribe the channel for regular updates thank you "
Related Questions
Exercises

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!