Question

Q4-Find the square root of 47 correct up to 3 decimal places.
Q5-Write a Pythagorean triplets having one of the number as 6. full explanation ​

Answer 1

Answer:

Ncert solutions

Grade 8

Mathematics

Science

Chapters in NCERT Solutions - Mathematics , Class 8

Exercises in Squares and Square Roots

Question 2

Q2) Write a Pythagoras triplet whose one member is:

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution

Transcript

Solution:

(i) There are three numbers 2m,\ m^2-1\ and\ m^2+12m, m

2

−1 and m

2

+1 in a Pythagorean Triplet.

Here, 2m = 6 \Rightarrow\ m=\frac{6}{2}=3⇒ m=

2

6

=3

Therefore, Second number \left(m^2-1\right)=\left(3\right)^2-1=9-1=8(m

2

−1)=(3)

2

−1=9−1=8

Third number m^2+1=\left(3\right)^2+1=9+1=10m

2

+1=(3)

2

+1=9+1=10

Hence, Pythagorean triplet is (6, 8, 10).

(ii) There are three numbers

2m,\ m^2-1\ and\ m^2+12m, m

2

−1 and m

2

+1 in a Pythagorean Triplet.

Here, 2m = 14 \Rightarrow m=\frac{14}{2}=7⇒m=

2

14

=7

Therefore, Second number m^2-1=\left(7\right)^2-1=49-1=48m

2

−1=(7)

2

−1=49−1=48

Third number m^2+1=\left(7\right)^2+1=49+1=50m

2

+1=(7)

2

+1=49+1=50

Hence, Pythagorean triplet is (14, 48, 50).

(iii) There are three numbers 2m,m^2-1\ and\ m^2+12m,m

2

−1 and m

2

+1 in a Pythagorean Triplet.

Here, 2m = 16 \Rightarrow m=\frac{16}{2}=8⇒m=

2

16

=8

Therefore, Second number \left(m^2-1\right)=\left(8\right)^2+1=64+1=65(m

2

−1)=(8)

2

+1=64+1=65

Hence, Pythagorean triplet is (16, 63, 65).

(iv) There are three numbers 2m,\ m^2-1\ and\ m^2+12m, m

2

−1 and m

2

+1 in a Pythagorean Triplet.

Here, 2m = 18 \Rightarrow m=\frac{18}{2}=9⇒m=

2

18

=9

Therefore, Second number \left(m^2-1\right)=\left(9\right)^2-1=81-1=80(m

2

−1)=(9)

2

−1=81−1=80

Third number m^2+1=\left(9\right)^2+1=81+1=81m

2

+1=(9)

2

+1=81+1=81

Hence, Pythagorean triplet is (18, 80, 82).