The perimeter of a rectangle is 64 units. Can the length x of the rectangle can be 20 units when its width y is 11 units?

Mathematics

2 answers:

GrogVix [38]3 years ago

7 0

The answer is B. No, the rectangle cannot have x=20 and y=11 because x + y does not equal 32.

The reasoning is, 20 + 11 = 31. In order to find the perimeter around the whole object, you will need to multiply 31 by 2 and get 62.

But, x is ONLY 20 and y is ONLY 11. It is not saying x(2) and y(2), so your answer has to be B.

vekshin13 years ago

5 0

B. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32

Perimeter = 2(l + w)
Perimeter = 64

Assuming: P = 64; w = 11 ; l = ?
64 = 2(l + 11)
64/2 = l + 11
32 - 11 = l
length = 21.

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Find the square root of 201.64 by division method

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Steps involved in finding square root using division method:

Group the digits of the number in pairs. You will start pairing from the decimal point and move away from the decimal point. E.g. for 201.64, first pair would be 2, second would be 01, and third would be 64 (see the attachment for visualizing)
Now start with the first pair, and find the number whose square is equal to or just less than the first pair. So here, since first pair is 2, that number whose square is just less than or equal to 2 is 1. Now divide 2 by 1 , and also put 1 in the quotient.
After this step, you will have 1 in quotient, and 1 in divisor with remainder of 2 - 1 = 1. Now bring down the next pair to join with the remainder. The resulting number now is 101
Now, double the number that is there in the quotient (so 1 becomes 2) and use it to determine the new divisor for 101. The new divisor will be such that when the divisor is multiplied with its unit digit then the product is just less than or equal to 101 (i.e. 2x * x <= 101). So for x = 4, 24 *4 = 96, which is less than 101. Now append this 4 in the quotient. The quotient now become 14. And subtract 96 from 101 to find the remainder. The remainder now is 5
We have now reached the decimal point in the dividend, so add a decimal point in the quotient
Now bring down the next pair to join with the remainder. The resulting number now is 564
Now, double the number that is there in the quotient (so 14 becomes 28) and use it to determine the new divisor for 564. The new divisor will be such that when the divisor is multiplied with its unit digit then the product is just less than or equal to 564 (i.e. 28x * x <= 564). So for x = 2, 282 *2 = 564. Now append this 2 in the quotient. The quotient now become 14.2. And subtract 564 from 564 to find the remainder. The remainder now is 0. Hence, we have completed the problem, and we have the square root with us
The square root is the quotient obtained: 14.2

Please refer to the attached image to see how the calculations are done