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Write a Pythagoras triplet whose one member is: (i) 6 (ii) 14 (iii) 16 (iv) 18

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NCERT Solutions for Class 8 Mathematics Chapter 6
Important NCERT Questions
Square and Square Roots Chapter 6 Exercise 6.2
NCERT Books for Session 2022-2023
CBSE Board
Questions No: 2

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  1. (i) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
    Here, 2m=6 ⇒ m=6/2=3
    Therefore,
    Second number (m²-1)=(3)²-1=9-1=8
    Third number m²+1=(3)²+1=9+1=10
    Hence, Pythagorean triplet is (6,8,10).

    (ii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
    Here, 2m=14
    ⇒ m=14/2=7
    Therefore,
    Second number (m²-1)=(7)²-1=49-1=48
    Third number m²+1=(7)²+1=49+1=50
    Hence, Pythagorean triplet is (14,48,50).

    (iii) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
    Here, 2m=16
    ⇒ m=16/2=8
    Therefore,
    Second number (m²-1)=(8)²-1=64-1=63
    Third number m²+1 =(8)²+1=64+1=65
    Hence, Pythagorean triplet is (16,63,65)..

    (iv) There are three numbers 2m,m²-1 and m²+1 in a Pythagorean Triplet.
    Here, 2m=18
    ⇒ m=18/2=9
    Therefore,
    Second number (m²-1)=(9)²-1=81-1=80
    Third number m²+1 =(9)²+1=81+1=82
    Hence, Pythagorean triplet is (18, 80, 82).

    Class 8 Maths Chapter 6 Exercise 6.2 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/

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