find the value of square root of 21 + square root of 11 + square root of 5 into 3 into 8 - 4 / 4 ^ - 1
Answers
Answer:
2. Find the number of digits in the square root of each of the. following numbers (without any calculation).
(i) 64(ii) 144(iii) 4489 (iv) 27225(v) 390625
Sol: If ‘n’ stands for number of digits in the given number, then
3. Find the square root of the following decimal numbers.
(i) 2.56(ii) 7.29(iii) 51.84 (iv) 42.25(v) 31.36
Here, number of decimal places, are already even.
∴ We mark off the periods and find the square root.
Here, number of decimal places are already even.
Therefore, we mark off the periods and find the square root.
4. Find the least nti,nllc’r” which must be subtracted from each of the following numbers. so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402(ii) 1989(iii) 3250 (iv) 825(v) 4000
Sol: (i) On proceeding to find the square root of 402, we have
Since, we get a remainder 2
The required least number to be subtracted from 402 is 2.
(ii) Since, we get a remainder of 53
The least number to be subtracted from the given number - 53 84
(iii) Since, we get a remainder 1.
The smallest number to be subtracted from the given number = 1
(iv) Since, we get a remainder 41.
The required smallest number tcI be subtracted from the given number = 41
(v) Since, we get a remainder 31,
The required smallest number to he subtracted from om the given number = 31
5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect ware so obtained.
(i) 525(ii) 1750(iii) 252 (iv) 1825(v) 6412
Sol: (i) Since, we get a remainder 41.
and next square number is 23.
∴ The required number to be added = 232 – 525
= 529 – 525 = 4
Now, 524 + 4 = 529, and = 23.
(ii) Since, we get a remainder 69.
and next square number is 422.
∴ The required number to be added = 422 – 1750
= 1764 – 1750 = 14
Now, 1750 + 14 = 1764, and
(ii) Since, we get a remainder 27.
(iv) Since, we get a remainder, 61.
Next square number = 43
The required number to be added = (43)2 – 1825
= 1849 – 1825 = 24
Now, 1825 + 24 = 1849, and
(v) Since, we get a remainder 12.
6. Find the length of the side of a square whose area is 441 m2.
Sol: Let the side of the square = x metre
Area = side × side
7. In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC.
(b) If AC = 13 cm, BC = 5 cm, find AB. Solution:
Sol:
REMEMBER
I. In a right triangle, the longest side is called the hypotenuse.
II. (Hypotenuse)2 = [Sum of the squares of other two sides]
(a) ∠B = 90°
Hypotenuse = AC
AC2 = AB2 + BC2
= 82 + 62
= 64 + 36 = 100
AC = 10
Thus, AC = 10 cm
(b) Here B = 90°
∴ Hypotenuse = AC
AC2 = AB2 + BC2
or 132 = AB2 = 52
= 169 – 25 = 144
or AB = 12
Thus, AB = 12 cm
8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more or this.
Sol: Since, the number of plants in a row and the number of columns are the same.
∴ Their product must be a square number.
The gardener has 1000 plants.
1000 is not a perfect square, and (31)2 < 1000
(There is a remainder of 39).
Obviously the next square number = 32
Number of plants required to be added = (32)2 – 1000
= 1024 – 1000 = 24
9. There are 500 children in a school. For a P T drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children this arrangement.
Sol: Since, the number of rows and the number of columns are same.
Total number (i.e. their product) must be a square number, we have
Since, we get a remainder of 16
500 > (22)2 or 500 – 16 = (22)2
Thus, the required number of children to be left out = 16