### Examples to find perfect squares near to a given number

**Example 5: If the area of a square is 90 cm ^{2}, what is its side-length rounded to the nearest integer?**

**Solution:**

Since, area A = a^{2}, we have a^{2} = 90.

But 81 < 90 < 100 and 81 is the nearer to 90 than 100. Hence, the nearest integer to √90 is √81 = 9.

**Example 6: A square piece of land has area 112m ^{2}. What is the closest integer which approximates the perimeter of the land?**

**Solution:**

If a is the side length of a square, its perimeter is 4a. We know that a^{2} = 112 m^{2}.

Hence, (4a)^{2} = 16a^{2} = 16 x 112 = 1792

But 42^{2} = 1764 < 1792 < 1849 = 43^{2} .

!764 is nearer to 1792 than 1849. Therefore, the integer approximation for √1792 is 42. The approximate value of perimeter is 42cm.

**Perfect squares near to a given number – Exercise 1.2.6**

**Find the nearest integer to the square root of the following numbers:**

**i) 232**

**Solution:**

We have, 15^{2 }= 225 and 16^{2} = 256.

We know, 225 < 232 < 256

15^{2} < 232 < 16^{2}

∴ Square root of 232 is nearest to 15

**ii) 600**

**Solution:**

We have, 24^{2} = 576 and 25^{2} = 625.

We know, 576 < 600 < 625

24^{2} < 600 < 25^{2}

∴ Square root of 600 s nearest to 24.

**iii) 728**

**Solution:**

We have, 26^{2} = 676 and 27^{2} = 729

676 < 728 < 729

26^{2} < 728 < 27^{2}

∴ Square root of 728 is nearest to 27.

**iv) 824**

**Solution:**

We have, 28^{2} = 784 and 29^{2} = 841

784 < 824 < 841

28^{2} < 824 < 29^{2}

∴Square root of 824 is nearest to 29

**v) 1729**

**Solution:**

We have, 41^{2} = 1681 and 42^{2} = 1764

1681 < 1729 < 1764

41^{2} < 1729 < 42^{2}

∴Square root of 1729 is nearest to 42

**A piece of land is in the shape of a square and its area is 1000m**^{2}. This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose.

**Solution:**

Area of the land = 1000m^{2}

Area = a2 = 1000m^{2}

Perimeter = 4a

Squaring , (Perimeter)^{2} = ( 4a)^{2} = 16a^{2} = 16 x 1000

(perimeter)^{2} = 16,000

We have, square of the perimeter i.e., 16000. now we have to find the square root of 16000. As 16000 does not have a perfect square root, let us find the square root nearest to square root of 16000.

We know, 126^{2 } = 15876 and 127^{2} = 16129.

Then, 15876 < 16000 < 16129

126^{2} < 16000< 127^{2}

∴ Nearest number is 126 But this is not enough to cover the hard.

∴Length of barbed wire required = 127 m.

**A student was asked to find √961. He read it wrongly and found √691 to the nearest integer. How much small was his number forms the correct answer?**

**Solution:**

We know, √ 961 = 31

It is given that student read it wrongly and found the result for √691. So, we have to find out the difference between √961 and √691.

√676 < √691 < √729

26^{2} < √691 < 27^{2}

26 is nearest number to √ 691

∴ difference = 31 – 26 = 5

**Squares, Square roots, cube and cube roots**

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