1. Find the square of the following numbers. (1) 32 (ii) 35 (i) 86 (V) 71 (vi) 46 2. Write a Pythagorean triplet whose one member is. (1) 6 (ii) 14 (i) 16 (iv) 18
Answers
Step-by-step explanation with answers:
1) Square of small numbers are easy to find but for large numbers multiplication may prove to be laborious and time consuming.
The method for squaring a two digit number by using the identity
(a+b)²= a²+2ab+b²
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Solution:
1)
32 can be written as (30+2)
32² = (30+2)²= (30+2)(30+2)
we know the identity
(a+b)²= a² + b² +2ab
= 30² + 2 x 30 x 2 + 2²
= 900 + 120 + 4 = 1024
(32)² = 1024
2) (35)² = (30+5)²
= 30²+ 2 x 30 x 5 + 5²
= 900 + 300 + 25 = 1225
(35)² =1225
3) 86² = (80 + 6)²
= 80²+ 2 x 80 x 6 + 6²
= 6400 + 960 + 36 = 7396
86² = 7396
4) 93² = (90+3)²
= 90² +2 x 90 x 3 + 3 ²
= 8100 + 540 + 9 = 8649
93² = 8649
5) 71² = (70 + 1)²
= 70² +2 x 70 x 1 + 1²
= 4900 + 140 + 1 = 5040 + 1 = 5041
71² =5041
6) 46² = (40+6)²
= 40²+ 2 x 40 x 6 + 6²
= 1600 + 480 + 36 = 2080 + 36 = 2116
46² =2116
2)Pythagorean triplets:
For any natural number m > 1, we have (2m) ² + (m2 – 1)²= (m2 + 1)² . So, 2m, m² – 1 and m² + 1 forms a Pythagorean triplet.
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Solution:
i)
Let 2m = 6
m = 3
m²+ 1 = 3² + 1= 9 + 1 = 10
m²- 1 = 3 2 - 1 = 9 - 1 = 8
check:
6 ² + 8² = 36 + 64 = 100 = 10²
Hence, the triplet is 6, 8 & 10
ii)
Let 2 m = 14
m = 7
m² + 1 = 7²+ 1 = 49 + 1 = 50
m²- 1 = 7² - 1 = 49 - 1 = 48
check:
14² + 48² = 196 + 1304 = 2500 = 50²
Hence, the triplet is 14, 48, and 50
iii)
Let 2 m = 16,
m = 8
m² + 1 = 8² + 1 = 64 + 1 = 65
m²- 1 = 8 2 - 1 = 64 - 1 = 63
check:
16² + 63² = 256 + 3969 = 4225 = 65²
Hence, the triplet is 16, 63 & 65
iv)
Let 2 m = 18
m = 9
m² + 1 = 9² + 1 = 81 + 1 = 82
m² - 1 = 9² - 1 = 81 - 1 = 80
check:
18²+ 80² = 324 + 6400 = 6724 = 82²
Hence, the triplet is 18, 80 & 82
Answer:
Q1=1. 1024 ,2.1225, 3.7296, 4.5041, 5.2126
Step-by-step explanation:
question 2 answer after some time