Question 1012 Marks
Find the square root of the following by long division method:3226694416
Answer
Hence, the square root of 3226694416 is 56804
View full question & answer→Hence, the square root of 3226694416 is 56804Questions · Page 3 of 3
2 Mark Question
Question 1022 Marks
Find the squares of the following numbers.
745
745
Answer
View full question & answer→$(745)^2$
Here $n =74$
$\therefore n(n+1)=74(74+1)$
$=74 \times 75=5550$
$\therefore(745)^2=555025$
Here $n =74$
$\therefore n(n+1)=74(74+1)$
$=74 \times 75=5550$
$\therefore(745)^2=555025$
Question 1032 Marks
Using prime factorization method, find the following numbers are perfect squares?
189
189
Answer
View full question & answer→189 = 3 × 3 × 3 × 7
$\begin{array}{c|c} 3& 189 \\ \hline 3 & 63 \\\hline 3&21 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
189 = (3 × 3) × 3 × 7
The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.
$\begin{array}{c|c} 3& 189 \\ \hline 3 & 63 \\\hline 3&21 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
189 = (3 × 3) × 3 × 7
The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.
Question 1042 Marks
Find the square root of the following correct to three places of decimal:
$2\frac{1}{2}$
$2\frac{1}{2}$
Answer
View full question & answer→We can find the square root up to four decimal places by expanding $2\frac{1}{2}$ into decimal form up to eight digits to the right of the decimal point as shown below,
$2\frac{1}{2}=2.50000000$
But, this is the same with the value 2.5 in problem (ix). Hence, the square root of $2\frac{1}{2}$ is 1.581
$2\frac{1}{2}=2.50000000$
But, this is the same with the value 2.5 in problem (ix). Hence, the square root of $2\frac{1}{2}$ is 1.581
Question 1052 Marks
Find the square root of:
$38\frac{11}{25}$
$38\frac{11}{25}$
Answer
View full question & answer→We know,
$\sqrt{38\frac{11}{25}}=\sqrt{\frac{961}{25}}=\frac{\sqrt{961}}{\sqrt{25}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{961}=31$
$\sqrt{25}=5$
$\therefore\sqrt{38\frac{11}{25}}=\frac{31}{5}$
$\sqrt{38\frac{11}{25}}=\sqrt{\frac{961}{25}}=\frac{\sqrt{961}}{\sqrt{25}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{961}=31$
$\sqrt{25}=5$
$\therefore\sqrt{38\frac{11}{25}}=\frac{31}{5}$
Question 1062 Marks
Find the squares of the following numbers:
$862$
$862$
Answer
View full question & answer→$(862)^2=(800+62)^2$
$\left\{(a+b)^2=a^2+2 a b+b^2\right\}$
$=(800)^2+2 \times 800 \times 62+(62)^2$
$=640000+99200+3844$
$=743044$
$\left\{(a+b)^2=a^2+2 a b+b^2\right\}$
$=(800)^2+2 \times 800 \times 62+(62)^2$
$=640000+99200+3844$
$=743044$
Question 1072 Marks
Find the square root in decimal form:0.00059049
Answer
Hence, the square root of 0.00059049 is 0.0243
View full question & answer→Hence, the square root of 0.00059049 is 0.0243
Question 1082 Marks
Find the least number which must be subtracted from the following numbers to make tham a perfact square:
$4401624$
$4401624$
Answer
View full question & answer→Using the long division method,
We can see that 4401624 is 20 more than $2098^2$. Hence, 20 must be subtracted from 4401624 to get a perfect square.
We can see that 4401624 is 20 more than $2098^2$. Hence, 20 must be subtracted from 4401624 to get a perfect square.
Question 1092 Marks
Find the square root of the following by long division method:
291600
291600
Answer
Hence, the square root of 291600 is 540
View full question & answer→Hence, the square root of 291600 is 540Question 1102 Marks
Find the squares of the following numbers using the identity $(a+b)^2=a^2+2 a b+b^2$ :
1001
1001
Answer
View full question & answer→$(a+b)^2=a^2+2 a b+b^2$
$(1001)^2=(1000+1)^2$
$=(1000)^2+2 \times 1000 \times 1 \times(1)$
$=1000000+2000+1$
$=1002001$
$(1001)^2=(1000+1)^2$
$=(1000)^2+2 \times 1000 \times 1 \times(1)$
$=1000000+2000+1$
$=1002001$
Question 1112 Marks
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2$ :
$599$
$599$
Answer
View full question & answer→$(a-b)^2=a^2-2 a b+b^2$
$(599)^2=(600-1)^2$
$=(600)^2-2 \times 600 \times 1+(1)^2$
$=360000-1200+1$
$=360001-1200$
$=358801$
$(599)^2=(600-1)^2$
$=(600)^2-2 \times 600 \times 1+(1)^2$
$=360000-1200+1$
$=360001-1200$
$=358801$
Question 1122 Marks
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2$ :
$999$
$999$
Answer
View full question & answer→$(a-b)^2=a^2-2 a b+b^2$
$(999)^2=(1000-1)^2$
$=(1000)^2-2 \times 1000 \times 1+(1)^2$
$=1000000-2000+1$
$=10000001-2000$
$=998001$
$(999)^2=(1000-1)^2$
$=(1000)^2-2 \times 1000 \times 1+(1)^2$
$=1000000-2000+1$
$=10000001-2000$
$=998001$
Question 1132 Marks
Find the least number which must be subtracted from the following numbers to make tham a perfact square:
$194491$
$194491$
Answer
View full question & answer→Using the long division method,
We can see that 194491 is 10 more than $441^2$. Hence, 10 must be subtracted from 194491 to get a perfact square.
We can see that 194491 is 10 more than $441^2$. Hence, 10 must be subtracted from 194491 to get a perfact square.
Question 1142 Marks
Write the possible unit's digits of the square root of the following numbers. these numbers are odd square roots?
998001
998001
Answer
View full question & answer→The unit digit of the number 998001 is 1 . So, the possible unit digits are 1 or 9 . Note that 998001 is equal to $\left(3^3 \times 37\right)^2$. Hence, the square root is an odd number.
Question 1152 Marks
Which of the following triplets are pythagorean?
$(14,35,38)$
$(14,35,38)$
Answer
View full question & answer→A triplet ( $a, b, c$ ) is called Pythagorean if the sum of the squares of the two smallest numbers is equal to the square of the biggest number.
The two smallest numbers are 12 and 35 . The sum of their squares is,
$12^2+35^2=1369$, which is not equal to $38^2=1444$
Hence, $(12,35,38)$ is not a Pythagorean triplet.
The two smallest numbers are 12 and 35 . The sum of their squares is,
$12^2+35^2=1369$, which is not equal to $38^2=1444$
Hence, $(12,35,38)$ is not a Pythagorean triplet.
Question 1162 Marks
Find the squares of the following numbers:
$503$
$503$
Answer
View full question & answer→$(503)^2=(500+3)^2$
$\left\{(a+b)^2=a^2+2 a b+b^2\right\}$
$=(500)^2+2 \times 500 \times 3+(3)^2$
$=250000+3000+9$
$=253009$
$\left\{(a+b)^2=a^2+2 a b+b^2\right\}$
$=(500)^2+2 \times 500 \times 3+(3)^2$
$=250000+3000+9$
$=253009$
Question 1172 Marks
Using prime factorization method, find the following numbers are perfect squares?
2048
2048
Answer
View full question & answer→2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
$\begin{array}{c|c} 2& 2048 \\ \hline 2 & 1024 \\\hline 2&512 \\\hline 2&256 \\\hline 2&128 \\\hline 2&64 \\\hline 2&32 \\\hline2&16 \\\hline2&8 \\\hline2&4 \\\hline2&2 \\\hline&1 \end{array}$
Grouping them into pairs of equal factors,
2048 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × 2
The last factor, 2 cannot be paired. Hence, 2048 is not a perfect square.
$\begin{array}{c|c} 2& 2048 \\ \hline 2 & 1024 \\\hline 2&512 \\\hline 2&256 \\\hline 2&128 \\\hline 2&64 \\\hline 2&32 \\\hline2&16 \\\hline2&8 \\\hline2&4 \\\hline2&2 \\\hline&1 \end{array}$
Grouping them into pairs of equal factors,
2048 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × 2
The last factor, 2 cannot be paired. Hence, 2048 is not a perfect square.
Question 1182 Marks
Using prime factorization method, find the following numbers are perfect squares?
2916
2916
Answer
View full question & answer→2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
$\begin{array}{c|c} 2& 2916 \\ \hline 2 & 1458 \\\hline 3&729 \\\hline 3&243 \\\hline 3&81 \\\hline 3&27 \\\hline 3&9 \\\hline 3&3 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
2916 = (2 × 2) × (3 × 3) × (3 × 3) × (3 × 3)
There are no left out of pairs. Hence, 2916 is a perfect square.
$\begin{array}{c|c} 2& 2916 \\ \hline 2 & 1458 \\\hline 3&729 \\\hline 3&243 \\\hline 3&81 \\\hline 3&27 \\\hline 3&9 \\\hline 3&3 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
2916 = (2 × 2) × (3 × 3) × (3 × 3) × (3 × 3)
There are no left out of pairs. Hence, 2916 is a perfect square.