Math, asked by kiratliburcu95, 6 hours ago

test paper for grade 8 chapter 6 ncert

Answers

Answered by dsselokar2306
1

Answer:

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Step-by-step explanation:

Q.1: How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

Solution: As we know, between n2 and (n+1)2, the number of non–perfect square numbers are 2n.

(i) Between 122 and 132 there are 2×12 = 24 natural numbers.

(ii) Between 252 and 262 there are 2×25 = 50 natural numbers.

(iii) Between 992 and 1002 there are 2×99 =198 natural numbers.

Q.2: Write a Pythagorean triplet whose one member is:

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution:

We know, for any natural number m, 2m, m2–1, m2+1 is a Pythagorean triplet.

(i) 2m = 6

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

Therefore, (6, 8, 10) is a Pythagorean triplet.

(ii) 2m = 14

⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

Therefore, (14, 48, 50) is not a Pythagorean triplet.

(iii) 2m = 16

⇒ m = 16/2 = 8

m2–1 = 82–1 = 64–1 = 63

m2+ 1 = 82+1 = 64+1 = 65

Therefore, (16, 63, 65) is a Pythagorean triplet.

(iv) 2m = 18

⇒ m = 18/2 = 9

m2–1 = 92–1 = 81–1 = 80

m2+1 = 92+1 = 81+1 = 82

Therefore, (18, 80, 82) is a Pythagorean triplet.

Q.3: (n+1)2-n2 = ?

Solution:

(n+1)2-n2

= (n2 + 2n + 1) – n2

= 2n + 1

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