Math, asked by Pratyush6803, 15 days ago

Q: What least monumber should be added to 3150 to
get a ferfect square ? Also find the square root
of the new
poéflect square obtained.​

Answers

Answered by hr905655
0

Answer:

I think right answer is 2

Answered by rudranshsharma2011
0

Answer:

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get the remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square.

\therefore402-2=400∴402−2=400

Hence, the square root of 400 is 20.

(ii) 1989

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get the remainder 53. Therefore 53must be subtracted from 1989 to get a perfect square.

\therefore1989-53=1936∴1989−53=1936

Hence, the square root of 1936 is 44.

(iii) 3250

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get the remainder 1. Therefore 1 must be subtracted from 3250 to get a perfect square.

\therefore3250-1=3249∴3250−1=3249

Hence, the square root of 3249 is 57.

(iv) 825

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 41. Therefore 41 must be subtracted from 825 to get a perfect square.

\therefore825-41=784∴825−41=784

Hence, the square root of 784 is 28.

(v) 4000

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get the remainder 31. Therefore 31 must be subtracted from 4000 to get a perfect square.

\therefore4000-31=3969∴4000−31=3969

Hence, the square root of 3969 is 63.

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