(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32. (ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198. In general, numbers between n² and (n+1)² are 2n. Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers. For more NCERT SRead more
(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32.
(ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198.
In general, numbers between n² and (n+1)² are 2n.
Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168. So, the smallest number to multiply is 3 and the square root of the result iRead more
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168.
So, the smallest number to multiply is 3 and the square root of the result is 168.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM: 4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180. Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.Read more
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM:
4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180.
Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.
Answer: 900.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula: Area = side × side → side = √Area This method helps in reverse-calculating side lengths from square areas. For moRead more
If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula:
Area = side × side → side = √Area
This method helps in reverse-calculating side lengths from square areas.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32. (ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198. In general, numbers between n² and (n+1)² are 2n. Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers. For more NCERT SRead more
(i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32.
(ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198.
In general, numbers between n² and (n+1)² are 2n.
Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168. So, the smallest number to multiply is 3 and the square root of the result iRead more
Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168.
So, the smallest number to multiply is 3 and the square root of the result is 168.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the smallest square number that is divisible by each of the following numbers: 4, 9 and 10.
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM: 4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180. Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.Read more
To find the smallest square divisible by 4, 9 and 10, we first compute their LCM:
4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180.
Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.
Answer: 900.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessFind the length of the side of a square whose area is 441 m².
If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula: Area = side × side → side = √Area This method helps in reverse-calculating side lengths from square areas. For moRead more
If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula:
Area = side × side → side = √Area
This method helps in reverse-calculating side lengths from square areas.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See lessGiven 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 26² (iii) 15625 + 251 (iv) 15625 + 253 (v) 15625 + 51²
To find 126² using 125², apply the identity: (a+1)² = a² + 2a + 1 Here, a = 125, so 126² = 125² + 2×125 + 1 = 15625 + 250 + 1 = 15625 + 251 Therefore, the correct option is (iv) 15625 + 251. This method allows quick calculations using known square values and avoids complete re-multiplication. Read more
To find 126² using 125², apply the identity:
(a+1)² = a² + 2a + 1
Here, a = 125, so
126² = 125² + 2×125 + 1 = 15625 + 250 + 1 = 15625 + 251
Therefore, the correct option is (iv) 15625 + 251. This method allows quick calculations using known square values and avoids complete re-multiplication.
For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/
See less