A Square and A CubeClass 8 Mathematics NCERT Solutions

Given: Four squared numbers: $64^2, 108^2, 292^2, 36^2$.

To Find: Which of these squares has the last digit 4.

Property: The units digit of a square number is determined by the units digit of the original number.

Both $108^2$ and $292^2$ have 4 as their last digit.

Final Answer: The numbers among the given options that have 4 as their last digit are $108^2$ and $292^2$.

Given: The number 1323.

To Find: The smallest number to multiply 1323 by to make it a perfect cube.

Solution: First, we find the prime factorization of 1323. The sum of digits of 1323 is $1+3+2+3=9$, so it is divisible by 3. $1323 = 3 \times 441$ We know that $441 = 21^2 = (3 \times 7)^2 = 3^2 \times 7^2$. So, the prime factorization of 1323 is: $1323 = 3^1 \times (3^2 \times 7^2) = 3^3 \times 7^2$ For a number to be a perfect cube, the exponents of all its prime factors must be a multiple of 3. In the factorization of 1323, the exponent of 3 is 3 (which is a multiple of 3). The exponent of 7 is 2. To make this exponent a multiple of 3, we need to multiply by $7^{3-2} = 7^1 = 7$. Thus, we must multiply 1323 by 7.

The resulting number would be $1323 \times 7 = (3^3 \times 7^2) \times 7 = 3^3 \times 7^3 = (3 \times 7)^3 = 21^3 = 9261$.

Final Answer: The number to be multiplied by 1323 to make it a perfect cube is 7.

Given: $125^2 = 15625$.

To Find: The value of $126^2$.

Formula: We can use the identity $(n+1)^2 = n^2 + 2n + 1$. The difference between two consecutive squares is $(n+1)^2 - n^2 = 2n+1$.

Solution: Let $n = 125$. We need to find $(n+1)^2 = 126^2$. Using the formula for the difference of consecutive squares: $126^2 - 125^2 = 2 \times 125 + 1$ $126^2 - 125^2 = 250 + 1$ $126^2 - 125^2 = 251$ Now, we can find $126^2$ by adding 251 to $125^2$. $126^2 = 125^2 + 251$ $126^2 = 15625 + 251$ $126^2 = 15876$ This corresponds to option (iv).

Final Answer: The correct option is (iv) $15625+251$.

Given: The area of a square is $441 \text{ m}^2$.

To Find: The length of the side of the square.

Formula: Area of a square = $(\text{side})^2$.

Solution: Let the length of the side be $s$. Then, $s^2 = 441$ To find the side length, we need to find the square root of 441. $s = \sqrt{441}$ We know that $20^2 = 400$ and the number 441 ends in 1. So, its square root must end in 1 or 9. Let's try 21. $21 \times 21 = 441$ So, the square root of 441 is 21. The length of the side is 21 meters.

Final Answer: The length of the side of the square is $21 \text{ m}$.

Given: Numbers 4, 9, and 10.

To Find: The smallest perfect square that is divisible by 4, 9, and 10.

Solution: First, we find the Least Common Multiple (LCM) of 4, 9, and 10. The smallest number divisible by 4, 9, and 10 is their LCM. Prime factorization of the numbers: $4 = 2^2$ $9 = 3^2$ $10 = 2 \times 5$

To find the LCM, we take the highest power of each prime factor present in the numbers. LCM(4, 9, 10) = $2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180$.

Now, we need to find the smallest square number which is a multiple of 180. Let's look at the prime factorization of 180: $180 = 2^2 \times 3^2 \times 5^1$. For a number to be a perfect square, the exponents of all its prime factors must be even. In the factorization of 180, the exponent of 5 is 1, which is odd. To make it a perfect square, we need to multiply it by a number that makes the exponent of 5 even. The smallest such number is 5.

Smallest square number = $180 \times 5 = 900$. Let's check the prime factorization of 900: $900 = 2^2 \times 3^2 \times 5^2 = (2 \times 3 \times 5)^2 = 30^2$. Since all exponents are even, 900 is a perfect square. It is also the smallest perfect square divisible by 4, 9, and 10.

Final Answer: The smallest square number divisible by 4, 9, and 10 is 900.

Given: The number 9408.

To Find: The smallest number to multiply 9408 by to make it a perfect square, and the square root of the resulting number.

Solution: First, we find the prime factorization of 9408. $9408 = 2 \times 4704$ $= 2 \times 2 \times 2352$ $= 2^2 \times 2 \times 1176$ $= 2^3 \times 2 \times 588$ $= 2^4 \times 2 \times 294$ $= 2^5 \times 2 \times 147$ $= 2^6 \times 147$ $= 2^6 \times 3 \times 49$ $= 2^6 \times 3^1 \times 7^2$ For a number to be a perfect square, all the exponents in its prime factorization must be even. Here, the exponent of 2 is 6 (even) and the exponent of 7 is 2 (even). However, the exponent of 3 is 1 (odd). To make the exponent of 3 even, we must multiply by $3^1$, which is 3. So, the smallest number to multiply by is 3.

The new number (product) is $9408 \times 3 = 28224$. The prime factorization of the product is $(2^6 \times 3^1 \times 7^2) \times 3 = 2^6 \times 3^2 \times 7^2$.

Now, we find the square root of this new number. $\sqrt{28224} = \sqrt{2^6 \times 3^2 \times 7^2}$ $= \sqrt{(2^3 \times 3^1 \times 7^1)^2}$ $= 2^3 \times 3^1 \times 7^1$ $= 8 \times 3 \times 7$ $= 24 \times 7$ $= 168$

Final Answer: The smallest number to be multiplied is 3. The square root of the product is 168.

Given: A pattern of equations involving sums of squares.

To Find: The missing numbers in the last two equations.

Observation: The pattern can be described by the relation: $a^2 + (a+1)^2 + (a(a+1))^2 = (a(a+1)+1)^2$. Or more generally, for two numbers $n_1$ and $n_2$, the pattern is $n_1^2 + n_2^2 + (n_1 n_2)^2 = (n_1 n_2 + 1)^2$.

Solution: Using this pattern for the remaining equations:

Final Answer: $4^2+5^2+20^2=(21)^2$ $9^2+10^2+(90)^2=(91)^2$

Given: A picture of a large square made up of smaller, tiny squares. The large square is a grid.

To Find: The total number of tiny squares and its prime factorization.

Solution: By observing the picture, we can count the number of tiny squares along one side of the large square. There are 12 tiny squares along the length and 12 tiny squares along the width. Therefore, the total number of tiny squares is the area of the large square in terms of tiny squares. Total number of squares = $12 \times 12 = 144$.

Now, we find the prime factorization of 144. $144 = 2 \times 72$ $= 2 \times 2 \times 36$ $= 2 \times 2 \times 2 \times 18$ $= 2 \times 2 \times 2 \times 2 \times 9$ $= 2 \times 2 \times 2 \times 2 \times 3 \times 3$ $= 2^4 \times 3^2$

Final Answer: There are 144 tiny squares. The prime factorisation of 144 is $2^4 \times 3^2$.