Question 12 Marks
Find the square root in decimal from: $0.00002025$
Answer
Hence, the square root of $0.00002025$ is $0.0045$
View full question & answer→Hence, the square root of $0.00002025$ is $0.0045$
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Question 22 Marks
Find the squares of the following numbers. $512$
Answer
View full question & answer→$(512)^2$ Here $a = 1, b = 2$
$\therefore$ $(5ab)^2 = (250 + ab) \times 1000 + (ab)^2$
$\therefore (512)^2= (250 + 12) \times 1000 + (12)^2$
$= 262 × 1000 + 144$
$= 262000 + 144$
$= 262144$
$\therefore$ $(5ab)^2 = (250 + ab) \times 1000 + (ab)^2$
$\therefore (512)^2= (250 + 12) \times 1000 + (12)^2$
$= 262 × 1000 + 144$
$= 262000 + 144$
$= 262144$
Question 32 Marks
Using prime factorization method, find the following numbers are perfect squares? $441$
Answer
View full question & answer→$441 = 3 \times 3 \times 7 \times 7$
$\begin{array}{c|c} 3& 441 \\ \hline 3 & 147 \\\hline 7&49 \\\hline 7&7\\\hline&1 \end{array}$
Grouping them into pairs of equal factors, $441 = 3 \times 3 \times (7 \times 7)$
There are no left out of pairs.
Hence, $441$ is a perfect square.
$\begin{array}{c|c} 3& 441 \\ \hline 3 & 147 \\\hline 7&49 \\\hline 7&7\\\hline&1 \end{array}$
Grouping them into pairs of equal factors, $441 = 3 \times 3 \times (7 \times 7)$
There are no left out of pairs.
Hence, $441$ is a perfect square.
Question 42 Marks
Find the length of a side of a square playground whose area is equal to the area of a rectangular field of diamensions $72m$ and 338m.
Answer
View full question & answer→The area of the playground $= 72 \times 338 = 24336\ m^2$
The length of one side of a square is equal to the square root of its area.
Hence, we just need to find the square root of $24336$.
Hence, the length of one side of the playground is $156$ metres.
The length of one side of a square is equal to the square root of its area.
Hence, we just need to find the square root of $24336$.
Hence, the length of one side of the playground is $156$ metres.
Question 52 Marks
Find the least number which be added to the following numbers to make tham a perfect square: $4515600$
Answer
View full question & answer→Using the long division method,
We can see that $4515600$ is $25$ more than $2125^2$.
Hence, we have to add $25$ to $4515600$ to get a perfect square.
We can see that $4515600$ is $25$ more than $2125^2$.
Hence, we have to add $25$ to $4515600$ to get a perfect square.
Question 62 Marks
The area of a square playground is $256.6404$ square metres. Find the length of one side of the playground.
Answer
View full question & answer→The length of one side of the playground is the square root of its area.
So, the length of one side of the playground is $16.02$ metres.
So, the length of one side of the playground is $16.02$ metres.
Question 72 Marks
The area of a square field is $30\frac{1}{4}\text{m}^2$ Calculate the length of the side of the square.
Answer
View full question & answer→The length of one side is equal to the square root of the area of the field.
Hence, we just need to calculate the value of $\sqrt{30\frac{1}{4}}$
Calculate the value of $\sqrt{30\frac{1}{4}}$
We have, $\sqrt{30\frac{1}{4}}=\frac{\sqrt{121}}{\sqrt{14}}$
Now, calculating the square root of the numerator and the denominator, $\sqrt{121}=\sqrt{11\times11}=11$ $\sqrt{4}=2$
Therefore, the length of the side of the square $\sqrt{30\frac{1}{4}}=\frac{11}{2}=5\frac{1}{2}\text{m}$
Hence, we just need to calculate the value of $\sqrt{30\frac{1}{4}}$
Calculate the value of $\sqrt{30\frac{1}{4}}$
We have, $\sqrt{30\frac{1}{4}}=\frac{\sqrt{121}}{\sqrt{14}}$
Now, calculating the square root of the numerator and the denominator, $\sqrt{121}=\sqrt{11\times11}=11$ $\sqrt{4}=2$
Therefore, the length of the side of the square $\sqrt{30\frac{1}{4}}=\frac{11}{2}=5\frac{1}{2}\text{m}$
Question 82 Marks
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2: 498$
Answer
View full question & answer→$ (a-b)^2=a^2-2 a b+b^2$
$ (498)^2=(500-5)^2$
$ =(500)^2-2 \times 500 \times 2+(2)^2 $
$ =250000-2000+4 $
$ =250004-2000 $
$ =248004 $
$ (498)^2=(500-5)^2$
$ =(500)^2-2 \times 500 \times 2+(2)^2 $
$ =250000-2000+4 $
$ =250004-2000 $
$ =248004 $
Question 92 Marks
Find the square root of: $75\frac{46}{49}$
Answer
View full question & answer→We know, $\sqrt{75\frac{46}{49}}=\sqrt{\frac{3721}{49}}=\frac{\sqrt{3721}}{\sqrt{49}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{49}=7$ $\therefore\sqrt{75\frac{46}{49}}=\frac{61}{7}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{49}=7$ $\therefore\sqrt{75\frac{46}{49}}=\frac{61}{7}$
Question 102 Marks
Find the square root of the following by long division method:$4008004$
Answer
Hence, the square root of $4008004$ is $2002$
View full question & answer→Hence, the square root of $4008004$ is $2002$
Question 112 Marks
Find the squares of the following numbers. $95$
Answer
View full question & answer→$(95)^2$
Here $n = 9$
$\therefore$ $n(n + 1)= 9(9 + 1)$
$= 9 × 10 = 90$
$\therefore$ $(95)^2= 9025$
Here $n = 9$
$\therefore$ $n(n + 1)= 9(9 + 1)$
$= 9 × 10 = 90$
$\therefore$ $(95)^2= 9025$
Question 122 Marks
Find the square root of: $3\frac{942}{2209}$
Answer
View full question & answer→We know, $\sqrt{3\frac{942}{2209}}=\sqrt{\frac{7569}{2209}}=\frac{\sqrt{7569}}{\sqrt{2209}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{3\frac{942}{2209}}=\frac{87}{47}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{3\frac{942}{2209}}=\frac{87}{47}$
Question 132 Marks
Find the square root of: $10\frac{151}{225}$
Answer
View full question & answer→We know, $\sqrt{10\frac{151}{225}}=\sqrt{\frac{2401}{225}}=\frac{\sqrt{2401}}{\sqrt{225}}$
Now, let us compute the square roots of the numerator and the denominator separately. $\sqrt{2401}=\sqrt{7\times7\times7\times7}=7\times7=49$
$\sqrt{225}=\sqrt{3\times3\times5\times5}=3\times5=15$
$\therefore\sqrt{10\frac{151}{225}}=\frac{49}{15}=3\frac{4}{15}$
Now, let us compute the square roots of the numerator and the denominator separately. $\sqrt{2401}=\sqrt{7\times7\times7\times7}=7\times7=49$
$\sqrt{225}=\sqrt{3\times3\times5\times5}=3\times5=15$
$\therefore\sqrt{10\frac{151}{225}}=\frac{49}{15}=3\frac{4}{15}$
Question 142 Marks
Which of the following triplets are pythagorean? $(18, 80, 82)$
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 $18$ and $80$. The sum of their squares is,
$18^2+80^2=6724=82^2$
Hence, $(18, 80, 82)$ is a Pythagorean triplet.
The two smallest numbers are $18$ and $80$. The sum of their squares is,
$18^2+80^2=6724=82^2$
Hence, $(18, 80, 82)$ is a Pythagorean triplet.
Question 152 Marks
Find the squares of the following numbers using the identity $ (a+b)^2=a^2+2 a b+b^2 : 510$
Answer
View full question & answer→$ (a+b)^2=a^2+2 a b+b^2 $
$ (510)^2=(500+10)^2 $
$ =(500)^2+2 \times 500 \times 10 \times(10)^2 $
$ =250000+10000+100 $
$ =260100 $
$ (510)^2=(500+10)^2 $
$ =(500)^2+2 \times 500 \times 10 \times(10)^2 $
$ =250000+10000+100 $
$ =260100 $
Question 162 Marks
Find the square root of the following by long division method:$82264900$
Answer
Hence, the square root of $82264900$ is $9070$
View full question & answer→Hence, the square root of $82264900$ is $9070$
Question 172 Marks
Find the squares of the following numbers: $451$
Answer
View full question & answer→$ (451)^2=(400+51)^2 $
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(400)^2+2 \times 400 \times 51+(51)^2 $
$ =160000+4080+2601 $
$ =203401 $
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(400)^2+2 \times 400 \times 51+(51)^2 $
$ =160000+4080+2601 $
$ =203401 $
Question 182 Marks
Which of the following triplets are pythagorean? $(8, 15, 17)$
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 $8$ and $15$. The sum of their squares is,
$8^2+15^2=289=17^2$
Hence, $(8, 15, 17)$ is a Pythagorean triplet.
The two smallest numbers are $8$ and $15$. The sum of their squares is,
$8^2+15^2=289=17^2$
Hence, $(8, 15, 17)$ is a Pythagorean triplet.
Question 192 Marks
Using square root table, find the square root: $25725$
Answer
View full question & answer→Using the table to find $\sqrt{3}$ and $\sqrt{7}$
$\sqrt{25725}=\sqrt{3\times5\times5\times7\times7\times7}$
$=\sqrt{3}\times5\times7\times\sqrt{7}$
$=1.732\times5\times7\times2.646$
$=160.41$
$\sqrt{25725}=\sqrt{3\times5\times5\times7\times7\times7}$
$=\sqrt{3}\times5\times7\times\sqrt{7}$
$=1.732\times5\times7\times2.646$
$=160.41$
Question 202 Marks
Find the square root of: $2\frac{137}{196}$
Answer
View full question & answer→We know, $\sqrt{2\frac{137}{196}}=\sqrt{\frac{529}{196}}=\frac{\sqrt{529}}{\sqrt{196}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{529}=\sqrt{23\times23}=23$
$\sqrt{196}=\sqrt{2\times2\times7\times7}=2\times7=14$
$\therefore\sqrt{2\frac{137}{196}}=\frac{23}{14}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{529}=\sqrt{23\times23}=23$
$\sqrt{196}=\sqrt{2\times2\times7\times7}=2\times7=14$
$\therefore\sqrt{2\frac{137}{196}}=\frac{23}{14}$
Question 212 Marks
What is that fraction which when multiplied by itself gives $227.798649$?
Answer
View full question & answer→We have to find the square root of the given number.
Hence, the fraction, which when multiplied by itself, gives $227.798649$ is $15.093$.
Hence, the fraction, which when multiplied by itself, gives $227.798649$ is $15.093$.
Question 222 Marks
Simplify: $\frac{\sqrt{59.29}\ -\sqrt{5.29}}{\sqrt{59.29}\ +\sqrt{5.29}}$
Answer
View full question & answer→We have, $\sqrt{59.29}=\sqrt{\frac{5929}{100}}=\frac{\sqrt{7\times7\times11\times11}}{10}=\frac{7\times11}{10}=7.7$ $\sqrt{59.29}=\sqrt{\frac{5929}{100}}=\frac{\sqrt{529}}{\sqrt{100}}=\frac{23}{10}=2.3$
$\frac{\sqrt{59.29}\ -\sqrt{5.29}}{\sqrt{59.29}\ +\sqrt{5.29}}=\frac{7.7-2.3}{7.7+2.3}=\frac{5.4}{10}=0.54$
$\frac{\sqrt{59.29}\ -\sqrt{5.29}}{\sqrt{59.29}\ +\sqrt{5.29}}=\frac{7.7-2.3}{7.7+2.3}=\frac{5.4}{10}=0.54$
Question 232 Marks
Find the value of: $\frac{\sqrt{1587}}{\sqrt{1728}}$
Answer
View full question & answer→We have, $\frac{\sqrt{1587}}{\sqrt{1728}}=\sqrt{\frac{529}{576}}$ (by dividing both numbers by $3$)
Computing the square roots of the numerator and the denominator,
$\sqrt{529}=\sqrt{23\times23}=23$ $\sqrt{576}=\sqrt{24\times24}=24$
$\therefore\frac{\sqrt{1587}}{\sqrt{1728}}=\frac{23}{24}$
Computing the square roots of the numerator and the denominator,
$\sqrt{529}=\sqrt{23\times23}=23$ $\sqrt{576}=\sqrt{24\times24}=24$
$\therefore\frac{\sqrt{1587}}{\sqrt{1728}}=\frac{23}{24}$
Question 242 Marks
Find the square root of the following by long division method: $62504836$
Answer
Hence, the square root of $6250486$ is $7906$
View full question & answer→Hence, the square root of $6250486$ is $7906$
Question 252 Marks
Find the square root in decimal from:
$0.813604$
$0.813604$
Answer
Hence, the square root of $0.813604$ is $0.902$
View full question & answer→Hence, the square root of $0.813604$ is $0.902$
Question 262 Marks
Find the least number which be added to the following numbers to make tham a perfect square: $506900$
Answer
View full question & answer→Using the long division method,
We can see that $506900$ is $44$ more than $712^2$.
Hence, we have to add $44$ to $506900$ to get a perfect square.
We can see that $506900$ is $44$ more than $712^2$.
Hence, we have to add $44$ to $506900$ to get a perfect square.
Question 272 Marks
Find the least number which be added to the following numbers to make tham a perfect square: $5607$
Answer
View full question & answer→Using the long division method,
We can see that $5607$ is $18$ more than $75^2$.
Hence, we have to add $18$ to $5607$ to get a perfect square.
We can see that $5607$ is $18$ more than $75^2$.
Hence, we have to add $18$ to $5607$ to get a perfect square.
Question 282 Marks
Find the least number which must be subtracted from the following numbers to make tham a perfact square: $26535$
Answer
View full question & answer→Using the long division method,
We can see that $26535$ is $291$ more than $162^2$.
Hence, $291$ must be subtracted from $26535$ to get a perfect square.
We can see that $26535$ is $291$ more than $162^2$.
Hence, $291$ must be subtracted from $26535$ to get a perfect square.
Question 292 Marks
Find the squares of the following numbers using the identity $ (a+b)^2=a^2+2 a b+b^2: 209$
Answer
View full question & answer→$ (a+b)^2=a^2+2 a b+b^2$
$ (209)^2=(200+9)^2 $
$=(200)^2+2 \times 200 \times 9 \times(9)^2 $
$=40000+3600+81 $
$ =43681 $
$ (209)^2=(200+9)^2 $
$=(200)^2+2 \times 200 \times 9 \times(9)^2 $
$=40000+3600+81 $
$ =43681 $
Question 302 Marks
Using square root table, find the square root, $11.11$
Answer
View full question & answer→We have, $\sqrt{11}=3.317$ and $\sqrt{12}=3.464$
Their difference is $0.1474$
Thus, for the difference of $1\ (12 - 11),$ the difference in the value of the square roots is $0.1474$ For the difference of $0.11$,
the difference in the values of the square roots is, $0.11 \times 0.1474 = 0.0162$
$\therefore\sqrt{11.11}=3.3166+0.0162=3.328\approx3.333$
Their difference is $0.1474$
Thus, for the difference of $1\ (12 - 11),$ the difference in the value of the square roots is $0.1474$ For the difference of $0.11$,
the difference in the values of the square roots is, $0.11 \times 0.1474 = 0.0162$
$\therefore\sqrt{11.11}=3.3166+0.0162=3.328\approx3.333$
Question 312 Marks
Write the possible unit's digits of the square root of the following numbers. these numbers are odd square roots? $9801$
Answer
View full question & answer→The unit digit of the number $9801$ is $1$.
So, the possible unit digits are $1$ or $9$ (Table $3.4$).
Note that $9801$ is equal to $99^2$.
Hence, the square root is an odd number.
So, the possible unit digits are $1$ or $9$ (Table $3.4$).
Note that $9801$ is equal to $99^2$.
Hence, the square root is an odd number.
Question 322 Marks
Find the square root of the following by long division method: $1745041$
Answer
Hence, the square root of $1745041$ is $1321$
View full question & answer→Hence, the square root of $1745041$ is $1321$
Question 332 Marks
Using prime factorization method, find the following numbers are perfect squares? $3549$
Answer
View full question & answer→$3549 = 3 \times 7 \times 13 \times 13$
$\begin{array}{c|c} 3& 3549 \\ \hline 7 & 1183 \\\hline 13&169 \\\hline 13&13 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $3549 = (13 \times 13) \times 3 \times 7$ The last factors, $3$ and $7$ cannot be paired.
Hence, $3549$ is not a perfect square.
Hence, the perfect squares are $225, 441, 2916$ and $11025$.
$\begin{array}{c|c} 3& 3549 \\ \hline 7 & 1183 \\\hline 13&169 \\\hline 13&13 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $3549 = (13 \times 13) \times 3 \times 7$ The last factors, $3$ and $7$ cannot be paired.
Hence, $3549$ is not a perfect square.
Hence, the perfect squares are $225, 441, 2916$ and $11025$.
Question 342 Marks
Find the square root in decimal from:$225.6004$
Answer
Hence, the square root of $225.6004$ is $15.25$
View full question & answer→Hence, the square root of $225.6004$ is $15.25$
Question 352 Marks
Find the square root of the following by long division method: $12544$
Answer
Hence, the square root of $12544$ is $112$
View full question & answer→Hence, the square root of $12544$ is $112$
Question 362 Marks
Find the squares of the following numbers. $425$
Answer
View full question & answer→$(425)^2$
Here $n = 42$
$\therefore$ $n(n + 1) = 42(42 + 1)$
$= 42 \times 43 = 1806$
$\therefore$ $(425)^2= 180625$
Here $n = 42$
$\therefore$ $n(n + 1) = 42(42 + 1)$
$= 42 \times 43 = 1806$
$\therefore$ $(425)^2= 180625$
Question 372 Marks
Find the square root in decimal form:$9998.0001$
Answer
Hence, the square root of $9998.001$ is $99.99$
View full question & answer→Hence, the square root of $9998.001$ is $99.99$
Question 382 Marks
Observe the following pattern,
$ 1+3=2^2 $
$ 1+3+5=3^2 $
$ 1+3+5+7=4^2 $
and write the value of $1 + 3 + 5 + 7 + 9 + ...... $ upto n terms.$ 1+3+5=3^2 $
$ 1+3+5+7=4^2 $
Answer
View full question & answer→From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the $n^{th}$ positive number. Putting that into formula, $41 + 3 + 5 + 7 + ....... n = n^2$, where the left hand side consists of n terms.
Question 392 Marks
Find the square root in decimal form: $236.144689$
Answer
Hence, the square root of $236.144689$ is $15.367$.
View full question & answer→Hence, the square root of $236.144689$ is $15.367$.
Question 402 Marks
Find the square root of the following by long division method:$152547201$
Answer
Hence, the square root of $152547201$ is $12351$
View full question & answer→Hence, the square root of $152547201$ is $12351$
Question 412 Marks
Find the least number which must be subtracted from the following numbers to make tham a perfact square: $16160$
Answer
View full question & answer→Using the long division method,
We can see that $16160$ is $31$ more than $127^2$.
Hence, $31$ must be subtracted from $16160$ to get a perfect square.
We can see that $16160$ is $31$ more than $127^2$.
Hence, $31$ must be subtracted from $16160$ to get a perfect square.
Question 422 Marks
Find the square root of the following by long division method:$3915380329$
Answer
Hence, the square root of $3915380329$ is $625763$
View full question & answer→Hence, the square root of $3915380329$ is $625763$
Question 432 Marks
Find the squares of the following numbers. $205$
Answer
View full question & answer→$(205)^2$
Here $n = 20$
$\therefore$ $n(n + 1) = 20(20 + 1)$
$= 20 \times 21 = 420$
$\therefore$ $(205)^2 = 42025$
Here $n = 20$
$\therefore$ $n(n + 1) = 20(20 + 1)$
$= 20 \times 21 = 420$
$\therefore$ $(205)^2 = 42025$
Question 442 Marks
Find the square root in decimal form:$0.00038809$
Answer
Hence, the square root of $0.00038809$ is $0.0197$
View full question & answer→Hence, the square root of $0.00038809$ is $0.0197$
Question 452 Marks
Which of the following triplets are pythagorean? $(10, 24, 26)$
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 $10$ and $24$. The sum of their squares is,
$10^2+24^2=676=26^2$
Hence, $(10, 24, 26)$ is a Pythagorean triplet.
The two smallest numbers are $10$ and $24$. The sum of their squares is,
$10^2+24^2=676=26^2$
Hence, $(10, 24, 26)$ is a Pythagorean triplet.
Question 462 Marks
Find the value of $\sqrt{103.0225}$ and hence find the value of: $\sqrt{1.030225}$
Answer
View full question & answer→The value of $103.0225$ is,
Hence, the square root of $103.0225$ is $10.15$
$\sqrt{1.030225}=\sqrt{\frac{103.0225}{100}}$
$=\frac{\sqrt{103.0225}}{\sqrt{100}}=\frac{{10.15}}{10}=1.015$
Hence, the square root of $103.0225$ is $10.15$
$\sqrt{1.030225}=\sqrt{\frac{103.0225}{100}}$
$=\frac{\sqrt{103.0225}}{\sqrt{100}}=\frac{{10.15}}{10}=1.015$
Question 472 Marks
Find the square root of the following by long division method: $120409$
Answer
Hence, the square root of $120409$ is $347$
View full question & answer→Hence, the square root of $120409$ is $347$
Question 482 Marks
Using square root table, find the square root: $1312$
Answer
View full question & answer→Using the table to find $\sqrt{2}$ and $\sqrt{41}$
$\sqrt{1312}=\sqrt{2\times2\times2\times2\times2\times41}$
$=2\times2\sqrt{2}\times\sqrt{41}$
$=2\times2\times1.414\times6.4031$
$=36.222$
$\sqrt{1312}=\sqrt{2\times2\times2\times2\times2\times41}$
$=2\times2\sqrt{2}\times\sqrt{41}$
$=2\times2\times1.414\times6.4031$
$=36.222$
Question 492 Marks
Using prime factorization method, find the following numbers are perfect squares? $343$
Answer
View full question & answer→$343 = 7 \times 7 \times 7$
$\begin{array}{c|c} 7& 343 \\ \hline 7 & 49 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $343 = (7 \times 7) \times 7$ The last factor, $7$ cannot be paired.
Hence, $343$ is not a perfect square.
$\begin{array}{c|c} 7& 343 \\ \hline 7 & 49 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $343 = (7 \times 7) \times 7$ The last factor, $7$ cannot be paired.
Hence, $343$ is not a perfect square.
Question 502 Marks
Find the square root of the following by long division method: $974169$
Answer
Hence, the square root of $974169$ is $987$.
View full question & answer→Hence, the square root of $974169$ is $987$.