
Examine the value of the following square roots to the nearest whole number
$(i)\sqrt {440} $
$(ii)\sqrt {800} $
$(iii)\sqrt {1020} $
Answer
464.1k+ views
Hint: First we have to define what the terms we need to solve the problem are. Since we need to find the nearest square roots to the number of integer values; given as above, Square means multiplying a number itself twice like two squares is ${2^2} = 4$and the square root means its inverse like the common terms will cancels once like, $\sqrt 4 = 2$, hence by using the square and square root concept we will solve the above question,
Complete answer:
(i) $(i)\sqrt {440} $ now let us consider a number $20$and the square of this number will gets ${20^2} = 400$(which is close to the required number not too close enough) now again consider a number ${21^2} = 441$which is too close to the number $\sqrt {440} $also if we solve this root of $\sqrt {440} = 20.97$which is more close to the number 21. And hence $(i)\sqrt {440} $the nearest whole number is $21$
(ii) By the same way we solve the $(ii)\sqrt {800} $now let us consider a number $28$and the square of this number will gets ${28^2} = 784$(which is close to the required number too close enough) now again consider a number ${29^2} = 841$which is not too close to the number $\sqrt {800} $also if we solve this root of \[\sqrt {800} = 28.24\]which is more close to the number 28. And hence $(ii)\sqrt {800} $the nearest whole number is $28$
(iii) Since approaching the same way $(iii)\sqrt {1020} $ now let us consider a number $31$and the square of this number will gets ${31^2} = 961$(which is close to the required number not too close enough) now again consider a number ${32^2} = 1024$which is too close to the number $\sqrt {1020} $also if we solve this root of $\sqrt {1020} = 31.93$which is more close to the number $31$. And hence $(iii)\sqrt {1020} $the nearest whole number is $31$
Note: We can also able to check the nearest point with the square values of $\sqrt {440} $ lies in-between $400 < 440 < 441$(20 and 21 square comparing) since if we see that, the nearest number of the root is 441 and hence nearest number is $21$(mostly occurs in decimals and they are asking whole numbers)
Complete answer:
(i) $(i)\sqrt {440} $ now let us consider a number $20$and the square of this number will gets ${20^2} = 400$(which is close to the required number not too close enough) now again consider a number ${21^2} = 441$which is too close to the number $\sqrt {440} $also if we solve this root of $\sqrt {440} = 20.97$which is more close to the number 21. And hence $(i)\sqrt {440} $the nearest whole number is $21$
(ii) By the same way we solve the $(ii)\sqrt {800} $now let us consider a number $28$and the square of this number will gets ${28^2} = 784$(which is close to the required number too close enough) now again consider a number ${29^2} = 841$which is not too close to the number $\sqrt {800} $also if we solve this root of \[\sqrt {800} = 28.24\]which is more close to the number 28. And hence $(ii)\sqrt {800} $the nearest whole number is $28$
(iii) Since approaching the same way $(iii)\sqrt {1020} $ now let us consider a number $31$and the square of this number will gets ${31^2} = 961$(which is close to the required number not too close enough) now again consider a number ${32^2} = 1024$which is too close to the number $\sqrt {1020} $also if we solve this root of $\sqrt {1020} = 31.93$which is more close to the number $31$. And hence $(iii)\sqrt {1020} $the nearest whole number is $31$
Note: We can also able to check the nearest point with the square values of $\sqrt {440} $ lies in-between $400 < 440 < 441$(20 and 21 square comparing) since if we see that, the nearest number of the root is 441 and hence nearest number is $21$(mostly occurs in decimals and they are asking whole numbers)
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