第3章平方根和平方根–练习3.4 |套装1
问题11.一个正方形场的面积是5184 m 2 。长度为宽度的两倍的矩形场的周长等于正方形场的周长。找到矩形区域的面积。
解决方案：
Let the side of square field as x
x2 = 5184 m2
x = √5184m
x = 2 × 2 ×2 × 9
= 72 m
Perimeter of square = 4x
= 4(72)
= 288 m
Perimeter of rectangle = 2 (l + b) = perimeter of the square field
= 288 m
l = 2b
2 (2b + b) = 288
2(3b) = 288
6b = 288
b = 288/6 (Transposing 6)
b = 48m
l = 2 × 48
= 96m
Area of rectangle = l × b
Area of rectangle = 96 × 48 m2
= 4608 m2
问题12：找到最小二乘数，该数可以被每个数整除：
(i)6、9、15和20
解决方案：
L.C.M of 6, 9, 15, 20 is 180
Prime factorization of 180 = 22 × 32 × 5 (Pairing of 2 and 3)
5 is left out
Multiplying the number with 5
180 × 5 = 22 × 32 × 52
= 900
Therefore, 900 is the least square number divisible by 6, 9, 15 and 20
(ii)8、12、15和20
解决方案：
L.C.M of 8, 2, 15, 20 is 360
Prime factorization of 360 = 22 × 32 ×2 × 5
2 and 5 are left out
Multiplying the number with 2 × 5 = 10
360 × 10 = 22 × 32 × 52 × 22
Therefore, 3600 is the least square number divisible by 8, 12, 15 and 20
问题13.通过反复相减的方法求出121和169的平方根。
解决方案：
In repeated subtraction method, odd numbers are subtracted one by one from the previous result and number of times subtraction is carried out is the square root.
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
11 times subtraction operation is carried out
Therefore, √121 = 11
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
13 times subtraction operation is carried out
Therefore, √169 = 13
问题14：写下以下数字的素因式分解，从而找到它们的平方根。
(i)7744
解决方案：
Prime factorization of 7744 is
7744 = 22 × 22 × 22 × 112
Therefore, the square root of 7744 is
√7744 = 2 × 2 × 2 × 11
= 88
(ii)9604
解决方案：
Prime factorization of 9604 is
9604 = 22 × 72 × 72
Therefore, the square root of 9604 is
√9604 = 2 × 7 × 7
= 98
(iii)5929
解决方案：
Prime factorization of 5929 is
5929 = 112 × 72
Therefore, the square root of 5929 is
√5929 = 11 × 7
= 77
(iv)7056
解决方案：
Prime factorization of 7056 is
7056 = 22 × 22 × 72 × 32
Therefore, the square root of 7056 is
√7056 = 2 × 2 × 7 × 3
= 84
问题15：学校VIII类的学生为PM的国家救济基金捐赠了2401卢比。每个学生捐赠的卢比与班上学生的数量一样多，找到班上学生的数量。
解决方案：
Let the number of students be x
Each student denoted x rupees
Total amount collected is x × x rupees = 2401
x2 = 2401
x = √2401
x = 49
Therefore, there are 49 students in the class.
问题16. PT老师希望在一个字段中最多安排6000名学生，以使行数等于列数。如果排列后遗漏了71，则查找行数。
解决方案：
Let the number of rows be x
Number of columns = x
Total number of students in the arrangement = x2
71 students are left out
Total students x2 + 71 = 6000
x2 = 5929
x = √5929
x = 77
Therefore, total number of rows are 77.