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 $987$.
Question 512 Marks
Find the square root of the following by long division method: $1471369$
Answer
Hence, the square root of $1471369$ is $1213$
View full question & answer→Hence, the square root of $1471369$ is $1213$
Question 522 Marks
Find the square root of: $\frac{324}{841}$
Answer
View full question & answer→We know, $\sqrt{\frac{324}{841}}=\frac{\sqrt{324}}{\sqrt{841}}$
Now, let compute the square roots of the numberator and the denominator separately. $\sqrt{324}=\sqrt{2\times2\times3\times3\times3\times3}$
$\sqrt{841}=\sqrt{29\times29}=29$
$\therefore\sqrt{\frac{324}{841}}=\frac{81}{29}$
Now, let compute the square roots of the numberator and the denominator separately. $\sqrt{324}=\sqrt{2\times2\times3\times3\times3\times3}$
$\sqrt{841}=\sqrt{29\times29}=29$
$\therefore\sqrt{\frac{324}{841}}=\frac{81}{29}$
Question 532 Marks
Find the square root of the following by long division method: $9653449$
Answer
Hence, the square root of $9653449$ is $3107$
View full question & answer→Hence, the square root of $9653449$ is $3107$
Question 542 Marks
Find the squares of the following numbers using the identity $ (a+b)^2=a^2+2 a b+b^2 : 405$
Answer
View full question & answer→$ (a+b)^2=a^2+2 a b+b^2 $
$ (405)^2=(400+5)^2 $
$ =(400)^2+2 \times 400 \times 5+(5)^2 $
$ =160000+4000+25 $
$ =164025 $
$ (405)^2=(400+5)^2 $
$ =(400)^2+2 \times 400 \times 5+(5)^2 $
$ =160000+4000+25 $
$ =164025 $
Question 552 Marks
Find the squares of the following numbers. $575$
Answer
View full question & answer→$(575)^2$
Here $n = 57$
$\therefore$ $n(n + 1) = 57(57 + 1)$
$= 57 \times 58 = 3306$
$\therefore$ $(575)^2$$ = 330625$
Here $n = 57$
$\therefore$ $n(n + 1) = 57(57 + 1)$
$= 57 \times 58 = 3306$
$\therefore$ $(575)^2$$ = 330625$
Question 562 Marks
Find the square root of: $2\frac{14}{25}$
Answer
View full question & answer→We know, $\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{65}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}$
Question 572 Marks
Find the value of: $\frac{\sqrt{441}}{\sqrt{625}}$
Answer
View full question & answer→Computing the square roots, $\sqrt{441}=\sqrt{(3\times3)\times(7\times7)}=3\times7=21$ $\sqrt{625}=\sqrt{(5\times5)\times(5\times5)=5\times5}=25$
$\therefore\frac{\sqrt{441}}{\sqrt{625}}=\frac{21}{25}$
$\therefore\frac{\sqrt{441}}{\sqrt{625}}=\frac{21}{25}$
Question 582 Marks
Find the square root of: $21\frac{2797}{3364}$
Answer
View full question & answer→We know, $\sqrt{21\frac{2797}{3364}}=\sqrt{\frac{73441}{3364}}=\frac{\sqrt{73441}}{\sqrt{3364}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{21\frac{2797}{3364}}=\frac{271}{58}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{21\frac{2797}{3364}}=\frac{271}{58}$
Question 592 Marks
Find the squares of the following numbers. $405$
Answer
View full question & answer→$(405)^2$
Here $n = 40$
$\therefore$ $n(n + 1) = 40(40 + 1)$
$= 40 \times 41 = 1640$
$\therefore$ $(405)^2= 164025$
Here $n = 40$
$\therefore$ $n(n + 1) = 40(40 + 1)$
$= 40 \times 41 = 1640$
$\therefore$ $(405)^2= 164025$
Question 602 Marks
Find the square root of: $25\frac{544}{729}$
Answer
View full question & answer→We know, $\sqrt{25\frac{544}{729}}=\sqrt{\frac{18769}{729}}=\frac{\sqrt{18769}}{\sqrt{729}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{729}=27$ $\therefore\sqrt{25\frac{544}{729}}=\frac{137}{27}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{729}=27$ $\therefore\sqrt{25\frac{544}{729}}=\frac{137}{27}$
Question 612 Marks
Find the square root in decimal from: $0.7225$
Answer
Hence, the square root of $0.7225$ is $0.85$
View full question & answer→Hence, the square root of $0.7225$ is $0.85$
Question 622 Marks
Find the square root of the following by long division method: $363609$
Answer
Hence, the square root of $363609$ is $603$
View full question & answer→Hence, the square root of $363609$ is $603$
Question 632 Marks
Find the smallest number which must be added to $2300$ so that it becomes a perfect square.
Answer
View full question & answer→To find the square root of $2300$, we use the long division method,
$23000$ is $4\ (704 - 700)$ less than $48^2$.
Hence, $4$ must be added to $2300$ to get a perfect square.
$23000$ is $4\ (704 - 700)$ less than $48^2$.
Hence, $4$ must be added to $2300$ to get a perfect square.
Question 642 Marks
Which of the following triplets are pythagorean? $(14, 48, 51)$
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.
He two smallest numbers are $14$ and $48$. The sum of their squares is,
$14^2+48^2=2500$, which is not equal to $51^2=2601$
Hence, $(14, 48, 51)$ is not a Pythagorean triplet.
He two smallest numbers are $14$ and $48$. The sum of their squares is,
$14^2+48^2=2500$, which is not equal to $51^2=2601$
Hence, $(14, 48, 51)$ is not a Pythagorean triplet.
Question 652 Marks
Which of the following triplets are pythagorean? $(16, 63, 65)$
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 $16$ and $63$. The sum of their squares is,
$16^2+63^2=4225=65^2$
Hence, $(16, 63, 65)$ is a Pythagorean triplet.
The two smallest numbers are $16$ and $63$. The sum of their squares is,
$16^2+63^2=4225=65^2$
Hence, $(16, 63, 65)$ is a Pythagorean triplet.
Question 662 Marks
Find the squares of the following numbers using the identity $(a + b)^2= a^2+ 2ab + b^2: 605$
Answer
View full question & answer→$ (a+b)^2=a^2+2 a b+b^2 $
$ (605)^2=(600+5)^2 $
$ =(600)^2+2 \times 600 \times 5 \times(5)^2$
$ =360000+6000+25 $
$ =366025 $
$ (605)^2=(600+5)^2 $
$ =(600)^2+2 \times 600 \times 5 \times(5)^2$
$ =360000+6000+25 $
$ =366025 $
Question 672 Marks
Find the square root of: $23\frac{394}{729}$
Answer
View full question & answer→We know, $\sqrt{23\frac{394}{729}}=\sqrt{\frac{17161}{729}}=\frac{\sqrt{17161}}{\sqrt{729}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{729}=27$
$\therefore\sqrt{23\frac{394}{729}}=\frac{131}{27}=4\frac{23}{27}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{729}=27$
$\therefore\sqrt{23\frac{394}{729}}=\frac{131}{27}=4\frac{23}{27}$
Question 682 Marks
The area of a square field is $325m^2$. Find the approximate length of one side of the field.
Answer
View full question & answer→The length of one side of the square field will be the square root of $325$
$\therefore\sqrt{325}=\sqrt{5\times5\times13}$
$=5\times\sqrt{13}$
$=5\times3.605$
$=18.030$
Hence, the length of one side of the field is $18.030\ m$
$\therefore\sqrt{325}=\sqrt{5\times5\times13}$
$=5\times\sqrt{13}$
$=5\times3.605$
$=18.030$
Hence, the length of one side of the field is $18.030\ m$
Question 692 Marks
Find the square root in decimal form:$176.252176$
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 702 Marks
Find the squares of the following numbers using the identity $ (a-b)^2=a^2-2 a b+b^2 : 995$
Answer
View full question & answer→$ (a-b)^2=a^2-2 a b+b^2 $
$ (995)^2=(1000-5)^2 $
$ =(1000)^2-2 \times 1000 \times 5+(5)^2 $
$ =1000000-10000+25 $
$ =1000025-10000 $
$ =990025 $
$ (995)^2=(1000-5)^2 $
$ =(1000)^2-2 \times 1000 \times 5+(5)^2 $
$ =1000000-10000+25 $
$ =1000025-10000 $
$ =990025 $
Question 712 Marks
Find the squares of the following numbers. $995$
Answer
View full question & answer→$(995)^2$
Here $n = 99$
$\therefore$ $n(n + 1) = 99(99 + 1)$
$= 99 × 100 = 9900$
$\therefore$ $(995)^2= 990025$
Here $n = 99$
$\therefore$ $n(n + 1) = 99(99 + 1)$
$= 99 × 100 = 9900$
$\therefore$ $(995)^2= 990025$
Question 722 Marks
Find the square root of: $\frac{441}{961}$
Answer
View full question & answer→We know, $\sqrt{\frac{441}{961}}=\frac{\sqrt{441}}{\sqrt{961}}$
Now, let compute the square roots of the numberator and the denominator separately. $\sqrt{441}=\sqrt{(3\times3)\times(7\times7)}=3\times7=21$
$\sqrt{961}=\sqrt{31\times31}=31$
$\therefore\sqrt{\frac{441}{961}}=\frac{21}{31}$
Now, let compute the square roots of the numberator and the denominator separately. $\sqrt{441}=\sqrt{(3\times3)\times(7\times7)}=3\times7=21$
$\sqrt{961}=\sqrt{31\times31}=31$
$\therefore\sqrt{\frac{441}{961}}=\frac{21}{31}$
Question 732 Marks
Find the least number which be added to the following numbers to make tham a perfect square: $4931$
Answer
View full question & answer→Using the long division method,
We can see that $4931$ is $110$ more than $71^2$.
Hence, we have to add $110$ to $4931$ to get a perfect square.
We can see that $4931$ is $110$ more than $71^2$.
Hence, we have to add $110$ to $4931$ to get a perfect square.
Question 742 Marks
Using prime factorization method, find the following numbers are perfect squares? $225$
Answer
View full question & answer→$225 = 3 \times 3 \times 5 \times 5$
$\begin{array}{c|c} 3& 225 \\ \hline 3 & 75 \\\hline 5&25 \\\hline 5&5 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $225 = (3 \times 3) \times (5 \times 5)$
There are no left out of pairs. Hence, $225$ is a perfect square.
$\begin{array}{c|c} 3& 225 \\ \hline 3 & 75 \\\hline 5&25 \\\hline 5&5 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $225 = (3 \times 3) \times (5 \times 5)$
There are no left out of pairs. Hence, $225$ is a perfect square.
Question 752 Marks
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2: 99$
Answer
View full question & answer→$(a-b)^2=a^2-2 a b+b^2$
$ (99)^2=(100-1)^2 $
$ =(100)^2-2 \times 100 \times 1+(1)^2 $
$ =10000-200+1 $
$ =10001-200 $
$ =9801$
$ (99)^2=(100-1)^2 $
$ =(100)^2-2 \times 100 \times 1+(1)^2 $
$ =10000-200+1 $
$ =10001-200 $
$ =9801$
Question 762 Marks
Find the value of $\sqrt{103.0225}$ and hence find the value of: $\sqrt{10302.25}$
Answer
View full question & answer→The value of $103.0225$ is,
Hence, the square root of $103.0225$ is $10.15$
$\sqrt{10302.25}=\sqrt{103.0225\times100}$
$=\sqrt{103.0225}\times{100}=10.15\times10=101.5$
Hence, the square root of $103.0225$ is $10.15$
$\sqrt{10302.25}=\sqrt{103.0225\times100}$
$=\sqrt{103.0225}\times{100}=10.15\times10=101.5$
Question 772 Marks
Find the square root of: $21\frac{51}{169}$
Answer
View full question & answer→We know, $\sqrt{21\frac{51}{169}}=\sqrt{\frac{3600}{169}}=\frac{\sqrt{3600}}{\sqrt{169}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{3600}=\sqrt{60\times60}=60$
$\sqrt{169}=\sqrt{13\times13}=13$
$\therefore\sqrt{21\frac{51}{169}}=\frac{60}{13}=4\frac{8}{13}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{3600}=\sqrt{60\times60}=60$
$\sqrt{169}=\sqrt{13\times13}=13$
$\therefore\sqrt{21\frac{51}{169}}=\frac{60}{13}=4\frac{8}{13}$
Question 782 Marks
Find the square root of the following by long division method: $6407522209$
Answer
Hence, the square root of $6407522209$ is $80047$
View full question & answer→Hence, the square root of $6407522209$ is $80047$
Question 792 Marks
Find the least number of three digits which is perfect square.
Answer
View full question & answer→Let us make a list of the squares starting from $1$.
$ 1^2=1 $
$ 2^2=4 $
$ 3^2=9 $
$ 4^2=16 $
$ 5^2=25 $
$ 6^2=36 $
$ 7^2=49 $
$ 8^2=64 $
$ 9^2=81 $
$ 10^2=100 $
The square of $10$ has three digits.
Hence, the least three-digit perfect square is $100$
$ 1^2=1 $
$ 2^2=4 $
$ 3^2=9 $
$ 4^2=16 $
$ 5^2=25 $
$ 6^2=36 $
$ 7^2=49 $
$ 8^2=64 $
$ 9^2=81 $
$ 10^2=100 $
The square of $10$ has three digits.
Hence, the least three-digit perfect square is $100$
Question 802 Marks
Using prime factorization method, find the following numbers are perfect squares? $11025$
Answer
View full question & answer→$11025 = 3 \times 3 \times 5 \times 5 \times 7 \times 7$
$\begin{array}{c|c} 3& 11025 \\ \hline 3 & 3675 \\\hline 5&1225 \\\hline 5&245 \\\hline 7&49 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
$11025 = (3 \times 2) \times (5 \times 5) \times (7 \times 7)$
There are no left out of pairs.
Hence, $11025$ is a perfect square.
$\begin{array}{c|c} 3& 11025 \\ \hline 3 & 3675 \\\hline 5&1225 \\\hline 5&245 \\\hline 7&49 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors,
$11025 = (3 \times 2) \times (5 \times 5) \times (7 \times 7)$
There are no left out of pairs.
Hence, $11025$ is a perfect square.
Question 812 Marks
Find the least number which be added to the following numbers to make tham a perfect square: $37460$
Answer
View full question & answer→Using the long division method,
We can see that $37460$ is $176$ more than $194^2$.
Hence, we have to add $176$ to $37460$ to get a perfect square.
We can see that $37460$ is $176$ more than $194^2$.
Hence, we have to add $176$ to $37460$ to get a perfect square.
Question 822 Marks
What is the fraction which when multiplied by itself gives $0.00053361$?
Answer
View full question & answer→We have to find the square root of the given number.
Hence, the fraction which multiplied by itself, gives $0.00053361$ is $0.0231$.
Hence, the fraction which multiplied by itself, gives $0.00053361$ is $0.0231$.
Question 832 Marks
Find the squares of the following numbers: $127$
Answer
View full question & answer→$(127)^2=(120+7)^2$
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(120)^2+2 \times 120 \times 7+(7)^2 $
$ =14400+1680+49 $
$ =16129$
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(120)^2+2 \times 120 \times 7+(7)^2 $
$ =14400+1680+49 $
$ =16129$
Question 842 Marks
Find the least number which must be subtracted from the following numbers to make tham a perfact square: $2361$
Answer
View full question & answer→Using the long division method,
We can see that $2361$ is $57$ more than $47^2$.
Hence, $57$ must be subtracted from $2361$ to get perfact square.
We can see that $2361$ is $57$ more than $47^2$.
Hence, $57$ must be subtracted from $2361$ to get perfact square.
Question 852 Marks
Find the square root of the following by long division method:$20421361$
Answer
Hence, the square root of $20421361$ is $4519$.
View full question & answer→Hence, the square root of $20421361$ is $4519$.
Question 862 Marks
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2 : 395$
Answer
View full question & answer→$(a-b)^2=a^2-2 a b+b^2 $
$ (395)^2=(400-5)^2 $
$ =(400)^2-2 \times 400 \times 5+(5)^2 $
$ =160000-4000+25 $
$ =160025-4000 $
$ =156025 $
$ (395)^2=(400-5)^2 $
$ =(400)^2-2 \times 400 \times 5+(5)^2 $
$ =160000-4000+25 $
$ =160025-4000 $
$ =156025 $
Question 872 Marks
Find the square root of the following by long division method: $20657025$
Answer
Hence, the square root of $20657025$ is $4545$
View full question & answer→Hence, the square root of $20657025$ is $4545$
Question 882 Marks
Find the square root of the following by long division method: $390625$
Answer
Hence, the square root of $390625$ is $625$.
View full question & answer→Hence, the square root of $390625$ is $625$.
Question 892 Marks
Find the squares of the following numbers: $265$
Answer
View full question & answer→$ (265)^2=(200+65)^2 $
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(200)^2+2 \times 200 \times 65+(65)^2 $
$ =40000+26000+4225 $
$ =70225$
$ \left\{(a+b)^2=a^2+2 a b+b^2\right\} $
$ =(200)^2+2 \times 200 \times 65+(65)^2 $
$ =40000+26000+4225 $
$ =70225$
Question 902 Marks
Find the square root in decimal from: $84.8241$
Answer
Hence, the square root of $84.8241$ is $9.21$
View full question & answer→Hence, the square root of $84.8241$ is $9.21$
Question 912 Marks
Find the square root in decimal from:$150.0625$
Answer
Hence, the square root of $150.0625$ is $12.25$.
View full question & answer→Hence, the square root of $150.0625$ is $12.25$.
Question 922 Marks
Find the square root of: $23\frac{26}{121}$
Answer
View full question & answer→We know, $\sqrt{23\frac{26}{121}}=\sqrt{\frac{2809}{121}}=\frac{\sqrt{2809}}{\sqrt{121}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{121}=11$
$\therefore\sqrt{23\frac{26}{121}}=\frac{53}{11}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\sqrt{121}=11$
$\therefore\sqrt{23\frac{26}{121}}=\frac{53}{11}$
Question 932 Marks
Find the square root of the following by long division method:
$286225$
$286225$
Answer
Hence, the square root of $286225$ is $535$
View full question & answer→Hence, the square root of $286225$ is $535$
Question 942 Marks
Find the greatest number of two digits which is a perfect square.
Answer
View full question & answer→We know that $10^2$ is equal to $100$ and $9^2$ is equal to $81$.
Since $10$ and $9$ are consecutive numbers, there is no perfect square between $100$ and $81$.
Since $100$ is the first perfect square that has more than two digits, $81$ is the greatest two-digit perfect square.
Since $10$ and $9$ are consecutive numbers, there is no perfect square between $100$ and $81$.
Since $100$ is the first perfect square that has more than two digits, $81$ is the greatest two-digit perfect square.
Question 952 Marks
Find the square root of the following by long division method: $97344$
Answer
Hence, the square root of $97344$ is $312$.
View full question & answer→Hence, the square root of $97344$ is $312$.
Question 962 Marks
Write the possible unit's digits of the square root of the following numbers. these numbers are odd square roots? $657666025$
Answer
View full question & answer→The unit digit of the number $657666025$ is $5$.
So, the only possible unit digit is $5$.
Note that $657666025$ is equal to $(5 \times 23 \times 223)^2$.
Hence, the square root is an odd number.
So, the only possible unit digit is $5$.
Note that $657666025$ is equal to $(5 \times 23 \times 223)^2$.
Hence, the square root is an odd number.
Question 972 Marks
Find the square root of: $4\frac{29}{49}$
Answer
View full question & answer→We know, $\sqrt{4\frac{29}{49}}=\sqrt{\frac{225}{49}}=\frac{\sqrt{225}}{\sqrt{49}}$
$\sqrt{225}=15$ $\sqrt{49}=7$
$\therefore\sqrt{4\frac{29}{49}}=\frac{15}{7}$
$\sqrt{225}=15$ $\sqrt{49}=7$
$\therefore\sqrt{4\frac{29}{49}}=\frac{15}{7}$
Question 982 Marks
Write the possible unit's digits of the square root of the following numbers. these numbers are odd square roots? $99856$
Answer
View full question & answer→The unit digit of the number $99856$ is $6$.
So, the possible unit digits are $4$ or $6$ (Table $3.4$).
Since its last digit is $6$ (an even number), it cannot have an odd number as its square root.
So, the possible unit digits are $4$ or $6$ (Table $3.4$).
Since its last digit is $6$ (an even number), it cannot have an odd number as its square root.
Question 992 Marks
Find the square root of: $3\frac{334}{3025}$
Answer
View full question & answer→We know, $\sqrt{3\frac{334}{3025}}=\sqrt{\frac{9409}{3025}}=\frac{\sqrt{9409}}{\sqrt{3025}}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{3\frac{334}{3035}}=\frac{97}{55}$
Now, let us compute the square roots of the numerator and the denominator separately.
$\therefore\sqrt{3\frac{334}{3035}}=\frac{97}{55}$
Question 1002 Marks
Find the squares of the following numbers using the identity $ (a-b)^2=a^2-2 a b+b^2: 495 $
Answer
View full question & answer→$ (a-b)^2=a^2-2 a b+b^2 $
$ (495)^2=(500-5)^2 $
$ =(500)^2-2 \times 500 \times 5+(5)^2 $
$ =250000-5000+25 $
$ =250025-5000 $
$ =245025 $
$ (495)^2=(500-5)^2 $
$ =(500)^2-2 \times 500 \times 5+(5)^2 $
$ =250000-5000+25 $
$ =250025-5000 $
$ =245025 $
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 $56804$
Question 1022 Marks
Find the squares of the following numbers. $745$
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? $189$
Answer
View full question & answer→$189 = 3 \times 3 \times 3 \times 7$
$\begin{array}{c|c} 3& 189 \\ \hline 3 & 63 \\\hline 3&21 \\\hline 7&7 \\\hline &1 \end{array}$
Grouping them into pairs of equal factors, $189 = (3 \times 3) \times 3 \times 7$
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, $189 = (3 \times 3) \times 3 \times 7$
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: $2\frac{1}{2}$
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 $1.581$.
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: $38\frac{11}{25}$
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. $\sqrt{961}=31$ $\sqrt{25}=5$ $\therefore\sqrt{38\frac{11}{25}}=\frac{31}{5}$
Question 1062 Marks
Find the squares of the following numbers: $862$
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: $4401624$
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 $540$
Question 1102 Marks
Find the squares of the following numbers using the identity $(a+b)^2=a^2+2 a b+b^2: 1001$
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 $(a-b)^2=a^2-2 a b+b^2: 599$
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 $(a-b)^2=a^2-2 a b+b^2: 999$
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: $194491$
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$? 998001$
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 $(3^3× 37)^2$.
Hence, the square root is an odd number.
Hence, the square root is an odd number.
Question 1152 Marks
Which of the following triplets are pythagorean$? (14, 35, 38)$
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: $503$
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$? 2048$
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.