Exercise 3.6
Page-3.48Question 1:
Find the square root of:
(i) 441961
(ii) 324841
(iii) 42929
(iv) 21425
(v) 2137196
(vi) 2326121
(vii) 25544729
(viii) 754649
(ix) 39422209
(x) 33343025
(xi) 2127973364
(xii) 381125
(xiii) 23394729
(xiv) 2151169
(xv) 10151225
Answer 1:
(i) We know:
√441961=√441√961
Now, let us compute the square roots of the numerator and the denominator separately.
√441=√(3×3)×(7×7)=3×7=21√961=√31×31=31∴√441961=2131
(ii)We know:
√324841=√324√841
Now, let us compute the square roots of the numerator and the denominator separately.
√324=√2×2×3×3×3×3=2×3×3=18√841=√29×29=29∴√324841=1829
(iii) By looking at the book's answer key, the fraction should be √42949, not √42929.
We know:
√42949=√22549=√225√49√225=15√49=7∴√42949=157
(iv) We know:
√21425=√6425=√64√25=85
(v) We know:
√2137196=√529196=√529√196
Now, let us compute the square roots of the numerator and the denominator separately.
√529=√23×23=23√196=√2×2×7×7=2×7=14∴√2137196=2314
(vi) We know:
√2326121=√2809121=√2809√121
Now, let us compute the square roots of the numerator and the denominator separately.
√121=11∴√2326121=5311
(vii) We know:
√25544729=√18769729=√18769√729
Now, let us compute the square roots of the numerator and the denominator separately.
√729=27∴√25544729=13727
(viii) We know:
√754649=√372149=√3721√49
Now, let us compute the square roots of the numerator and the denominator separately.
√49=7∴√754649=617
(ix) We know:
√39422209=√75692209=√7569√2209
Now, let us compute the square roots of the numerator and the denominator separately.
∴√39422209=8747
(x) We know:
√33343025=√94093025=√9409√3025
Now, let us compute the square roots of the numerator and the denominator separately.
∴√33343025=9755
(xi) We know:
√2127973364=√734413364=√73441√3364
Now, let us compute the square roots of the numerator and the denominator separately.
∴√2127973364=27158
(xii) We know:
√381125=√96125=√961√25
Now, let us compute the square roots of the numerator and the denominator separately.
√961=31√25=5∴√381125=315
(xiii) We know:
√23394729=√17161729=√17161√729
Now, let us compute the square roots of the numerator and the denominator separately.
√729=27∴√23394729=13127=42327
(xiv) We know:
√2151169=√3600169=√3600169
Now, let us compute the square roots of the numerator and the denominator separately.
√3600=√60×60=60√169=√13×13=13∴√2151169=6013=4813
(xv) We know:
√10151225=√2401225=√2401√225
Now let us compute the square roots of the numerator and the denominator separately.
√2401=√7×7×7×7=7×7=49√225=√3×3×5×5=3×5=15∴√10151225=4915=3415
Question 2:
Find the value of:
(i) √80√405
(ii) √441√625
(iii) √1587√1728
(iv) √72×√338
(v) √45×√20
Answer 2:
(i) We have:
(ii) Computing the square roots:
∴
(iii) We have:
(by dividing both numbers by 3)
Computing the square roots of the numerator and the denominator:
∴
(iv) We have:
=
(v) We have:
= 30
Question 3:
The area of a square field is 80244729 square metres. Find the length of each side of the field.
Answer 3:
The length of one side is the square root of the area of the field. Hence, we need to calculate the value of
We have
Now, to calculate the square root of the numerator and the denominator:
We know that:
Therefore, length of one side of the field = 24227 = 82627 m
Question 4:
The area of a square field is 3014m2. Calculate the length of the side of the square.
Answer 4:
The length of one side is equal to the square root of the area of the field. Hence, we just need to calculate the value of .
We have:
Now, calculating the square root of the numerator and the denominator:
Therefore, the length of the side of the square = √3014= 112 = 512 m
Question 5:
Find the length of a side of a square playground whose area is equal to the area of a rectangular field of diamensions 72 m and 338 m.
Answer 5:
The area of the playground = 72 × 338 = 24336 m2
The length of one side of a square is equal to the square root of its area. Hence, we just need to find the square root of 24336.
Hence, the length of one side of the playground is 156 metres.
No comments:
Post a Comment