Example 2 - Chapter 5 Class 8 Squares and Square Roots

Last updated at April 16, 2024 by Teachoo

Transcript

Example 2 Write a Pythagorean triplet whose smallest member is 8. We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet.
We know 2m, 𝒎^𝟐−𝟏 and 𝒎^𝟐+𝟏 form a Pythagorean triplet.
Given,
Smallest member of the triplet = 8.
Let 2m = 8
2m = 8
m = 8/2
m = 4
Let 𝒎^𝟐−𝟏" = 8"
𝑚^2 = 8 + 1
𝑚^2 = 9
∴ m = 3
Let 𝒎^𝟐+𝟏 = 8
𝑚^2 = 8 − 1
𝑚^2 = 7
Since, 7 is not a square number,
∴ 𝑚^2+1 ≠ 8
It is not possible.
Therefore, m = 4 or m = 3
Finding Triplets for m = 3
1st number = 2m
2nd number = 𝑚^2−1
3rd number = 𝑚^2+1
∴ The required triplet is 6, 8, 10
But 8 is not a smallest member of this triplet.
So, lets try for m = 4
Finding Triplets for m = 4
1st number = 2m
2nd number = 𝑚^2−1
3rd number = 𝑚^2+1
As 8 is the smallest member of this triplet.
∴ The required triplet is 8, 15, 17

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!