NCERT Solutions Class 8 Maths Chapter 5 - Square and Square Roots

3.0NCERT Questions with Solutions Class 8 Maths Chapter 5 - Detailed Solutions

EXERCISE : 6.1

• What will be the unit digit of the squares of the following numbers? (i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555

Sol. The unit digit of the squares of the given numbers is shown against the number in the following table:

• The following numbers are not perfect squares. Give reason. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050

Sol. A number that ends either with $2,3,7$ or 8 cannot be a perfect square. Also, a number that ends with odd number of zero(s) cannot be a perfect square. (i) Since the given number 1057 ends with 7, so it cannot be a perfect square. (ii) Since the given number 23453 ends with 3 , so it cannot be a perfect square. (iii) Since the given number 7928 ends with 8, so it cannot be a perfect square. (iv) Since the given number 222222 ends with 2 , so it cannot be a perfect square. (v) Since the given number 64000 ends with odd number of ' 0 ', so it cannot be a perfect square. (vi) Since the given number 89722 ends with 2, so it cannot be a perfect square. (vii) Since the given number 222000 ends with odd number of ' 0 ', so it cannot be a perfect square. (viii) Since the given number 505050 ends with odd number of ' 0 ', so it cannot be a perfect square.

3. The squares of which of the following would be odd numbers? (i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Sol. (i) The given number 431 is odd, so its square must be odd. (ii) The given number 2826 is even, so its square must be even. (iii) The given number 7779 is odd, so its square must be odd. (iv) The given number 82004 is even, so its square must be even.

4. Observe the following pattern and find the missing digits:

$\begin{array}{rl}& 11^2=121\\ & 101^2=10201\\ & 1001^2=1002001\\ & 100001^2=1\dots \dots ..................\end{array}.$

Sol. The missing digits are as under: $100001^2=10000200001$ $10000001^2=100000020000001$

5. Observe the following pattern and find the missing numbers:

$\begin{array}{rl}& 11^2=121\\ & 101^2=10201\\ & 10101^2=102030201\\ & 1010101^2=\\ & \dots ...........{}^2=10203040504030201\end{array}$

Sol. $1010101^2=1020304030201$ $101010101^2=10203040504030201$

6. Using the given pattern, find the missing numbers: $1^2+2^2+2^2=3^2$ $2^2+3^2+6^2=7^2$ $3^2+4^2+12^2=13^2$ $4^2+5^2+$ _ ${}^2=21^2$ $5^2+\dots +30^2=31^2$ $6^2+7^2+{}^2={}^2$ Sol. The missing numbers are us under: $4^2+5^2+20^2=21^2$ $5^2+6^2+30^2=31^2$ $6^2+7^2+42^2=43^2$

7. Without adding, find the sum: (i) $1+3+5+7+9$ (ii) $1+3+5+7+9+11+13+15+17+19$ (iii) $1+3+5+7+9+11+13+15+17+9+21+23$

Sol. (i) $1+3+5+7+9=$ Sum of first 5 odd numbers $=5^2=25$ (ii) $1+3+5+7+9+11+13+15+17+19$ $=$ Sum of first 10 odd numbers $=10^2=100$ (iii) $1+3+5+7+9+11+13+15+17+19+21$ $+23=$ Sum of first 12 odd numbers $=12^2=$ 144

8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.

Sol. (i) $49=7^2=1+3+5+7+9+11+13$ (ii) $121=11^2=1+3+5+7+9+11+13+15+17+19+21$

9. How many numbers lie between squares of the following numbers? (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

Sol. (i) Between $12^2(=144)$ and $13^2(=169)$ there are twenty four (i.e., $2\times 12$ ) numbers. (ii) Between $25^2(=625)$ and $26^2(=676)$, there are 50 (i.e. $2\times 25$ ) numbers. (iii) Between $99^2(=9801)$ and $100^2(=$ 10000), there are '198' (i.e. $2\times 99$ ) numbers.

EXERCISE : 6.2

• Find the square of the following numbers. (i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Sol. (i) $32^2=(30+2{)}^2$

$\begin{array}{rl}& =30(30+2)+2(30+2)\\ & =30^2+30\times 2+2\times 30+2^2\\ & =900+60+60+4=1024\end{array}$

(ii) $35^2=(30+5{)}^2=30(30+5)+5(30+5)$ $=30^2+30\times 5+5\times 30+5^2$ $=900+150+150+25=1225$ (iii) $86^2=(80+6{)}^2$ $=80(80+6)+6(80+6)$ $=80^2+80\times 6+6\times 80+6^2$ $=6400+480+480+36=7396$ (iv) $93^3=(90+3{)}^2$ $=90(90+3)+3(90+3)$ $=90^2+90\times 3+3\times 90+3^2$ $=8100+270+270+9=8649$ (v) $71^2=(70+1{)}^2$ $=70(70+1)+1(70+1)$ $=70^2+70\times 1+1\times 70+1^2$ $=4900+70+70+1=5041$ (vi) $46^2=(40+6{)}^2$ $=40(40+6)+6(40+6)$ $=40^2+40\times 6+6\times 40+6^2$ $=1600+240+240+36=2116$

2. Write a Pythagorean triplet whose one member is (i) 6 (ii) 14 (iii) 16 (iv) 18

Sol. (i) $2\mathrm{m}=6\Rightarrow \mathrm{m}=3$ Put $m=3$ in $m^2-1,m^2+1$ $\therefore 2\mathrm{m}=6,{\mathrm{m}}^2-1=3^2-1=9-1=8$ and ${\mathrm{m}}^2+1=9+1=10$ Thus, 6, 8 and 10 is Pythagorean triplets. (ii) $2\mathrm{m}=14\Rightarrow \mathrm{m}=7$

Put $m=7$ in $m^2-1,m^2+1$ $\therefore 2\mathrm{m}=14,{\mathrm{m}}^2-1=7^2-1=49-1=48$ and $m^2+1=7^2+1=49+1=50$ Thus, 14, 48 and 50 is Pythagorean triplets. (iii) $2\mathrm{m}=16\Rightarrow \mathrm{m}=8$

Put $\mathrm{m}=8$ in ${\mathrm{m}}^2-1\&{\mathrm{m}}^2+1$ $\therefore 2\mathrm{m}=16,{\mathrm{m}}^2-1=8^2-1=64-1=63$ and $m^2+1=8^2+1=64+1=65$ Thus 16, 63, and 65 is Pythagorean triplets. (iv) $2\mathrm{m}=18\Rightarrow \mathrm{m}=9$

Put $m=9$ in $m^2-1$ and $m^2+1$ $\therefore 2\mathrm{m}=18,{\mathrm{m}}^2-1=9^2-1=81-1=80$ And $m^2+1=9^2+1=81+1=82$ Thus 18, 80 and 82 is Pythagorean triplets.

EXERCISE : 6.3

• What could be the possible one's digits of the square root of each of the following numbers? (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Sol. The possible one's digits of the square root of the numbers:

• Without doing any calculation, find the numbers which are surely not perfect squares: (i) 153 (ii) 257 (iii) 408 (iv) 441

Sol. We know that a number ending in 2,3,7 or 8 is never a perfect square. Therefore, (i) 153 is not a perfect square. (ii) 257 is not a perfect square. (iii) 408 is not a perfect square (iv) 441 may be a perfect square.

So, 153, 257 and 408 are surely not perfect squares.

3. Find the square roots of 100 and 169 by the method of repeated subtraction. Sol. From 100, we subtract successive odd numbers starting from 1 as under. 100-1 = 99 $99-3=96$ $96-5=91$ $91-7=84$ $84-9=75$ $75-11=64$ $64-13=51$ $51-15=36$ $36-17=19$ 19-19 = 0 and obtain 0 at $10^{\text{th }}$ step. $\therefore \sqrt{100}=10$ From 169, we subtract successive odd numbers starting from 1 as under.

$\begin{array}{rl}& 169-1=168\\ & 168-3=165\\ & 165-5=160\\ & 160-7=153\\ & 153-9=144\\ & 144-11=133\\ & 133-13=120\\ & 120-15=105\\ & 105-17=88\\ & 88-19=69\\ & 69-21=48\\ & 48-23=25\\ & 25-25=0\end{array}$

and obtain 0 at $13^{\text{th }}$ step. $\therefore \sqrt{169}=13$ 4. Find the square roots of the following numbers by the Prime Factorisation Method: (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100

Sol. (i) By prime factorisation, we get

$729=\underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ $\therefore \sqrt{729}=3\times 3\times 3=27$ (ii) By prime factorisation, we get

$400=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}$ $\therefore \sqrt{400}=2\times 2\times 5=20$ (iii) By prime factorisation, we get

$1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{1764}=2\times 3\times 7=42$ (iv) By prime factorisation, we get

$4096=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}$ $\therefore \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$ (vi) By prime factorisation, we get

$9604=\underline{2\times 2\times 7\times 7\times \underline{7\times 7}}$ $\therefore \sqrt{9604}=2\times 7\times 7=98$ (vii) By prime factorisation, we get

$5929=\underline{7\times 7}\times \underline{11\times 11}$ $\therefore \sqrt{5929}=7\times 11=77$ (viii) By prime factorisation, we get

$9216=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}$ $\therefore \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3$ $=96$ (ix) By prime factorisation, we get

$529=23\times 23$ $\therefore \sqrt{529}=23$ (x) By prime factorisation, we get

$8100=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}$ $\therefore \sqrt{8100}=2\times 3\times 3\times 5=90$

• For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained. (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768 Sol. (i) By prime factorisation we get

$252=\underline{2\times 2}\times \underline{3\times 3}\times 7$ It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. $\therefore 252\times 7=1764$ is a perfect square. Thus, $1764=\underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{1764}=2\times 3\times 7=42$ (ii) By prime factorisation, we get

$180=\underline{2\times 2}\times \underline{3\times 3}\times 5$ It is clear that in order to get a perfect square, one more 5 is required. So, the given number should be multiplied by 5 to make the product a perfect square. $\therefore 180\times 5=900$ is a perfect square. Thus, $900=\underline{2\times 2}\times \underline{3\times 3}\times \underline{5\times 5}$ $\therefore \sqrt{900}=2\times 3\times 5=30$ (iii) By prime factorisation, we get

$1008=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times 7$ It is clear that in order to get a perfect square, one more 7 is required. So, the given number should be multiplied by 7 to make the product a perfect square. $\therefore 1008\times 7=7056$ is a perfect square. Thus, $7056=\underline{2\times 2}\times \underline{2\times 2}\times \underline{3\times 3}\times \underline{7\times 7}$ $\therefore \sqrt{7056}=2\times 2\times 3\times 7=84$ (iv) By prime factorisation, we get

$2028=\underline{2\times 2}\times 3\times \underline{13\times 13}$ It is clear that the given number should be multiplied by 3 to make the product a perfect square $\therefore 2028\times 3=6084$ is a perfect square Thus $6084=\underline{2\times 2}\times \underline{3\times 3}\times \underline{13\times 13}$ $\therefore \sqrt{6084}=2\times 3\times 13=78$ (v) By prime factorisation, we get $1458=2\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ It is clear that the given number should be multiplied by 2 to make the product a perfect square. $\therefore 1458\times 2=2916$ is a perfect square. Thus $2916=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times \underline{3\times 3}$ $\therefore \sqrt{2916}=2\times 3\times 3\times 3=54$ (vi) By prime factorisation, we get

$768=\underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times \underline{2\times 2}\times 3$ It is clear that the given number should be multiplied by 3 to make the product a perfect square. $\therefore 768\times 3=2304$ $\sqrt{2304}=2\times 2\times 2\times 2\times 3=48$

• For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained. (i) 252 (ii) 2925 (iii) 396 (iv) 2645 (v) 2800 (vi) 1620

Sol. (i) By prime factorisation, we get

$252=\underline{2\times 2}\times \underline{3\times 3}\times 7$ Since the prime factor 7 cannot be paired. $\therefore$ The given number should be divided by $\therefore \frac{252}{7}=\frac{2\times 2\times 3\times 3\times 7}{7}=\underline{2\times 2}\times \underline{3\times 3}$ $=36$ is a perfect square and, $\sqrt{36}=\sqrt{\underline{2\times 2\times 3\times 3}}$ $=2\times 3=6$ (ii) By prime factorisation, we get

$2925=\underline{3\times 3}\times \underline{5\times 5}\times 13$ Since the prime factor 13 cannot be paired. $\therefore$ The given number should be divided by 13 . $\therefore \frac{2925}{13}=\frac{3\times 3\times 5\times 5\times 13}{13}$ $=\underline{3\times 3}\times \underline{5\times 5}=225$ is a perfect square and, $\sqrt{225}=\sqrt{\underline{3\times 3}\times \underline{5\times 5}}$ $=3\times 5=15$ (iii) By prime factorisation, we get

$396=\underline{2\times 2}\times \underline{3\times 3}\times 11$ Since the prime factor 11 cannot be paired so 396 should be divided by 11 . $\frac{396}{11}=36$ is a perfect square. and $\sqrt{36}=\sqrt{2\times 2\times \underline{3\times 3}}=2\times 3=6$ (iv) By prime factorisation, we get

$2645=5\times 23\times 23$ Since the prime factor 5 cannot be paired, so 2645 should be divided by 5 . $\frac{2645}{5}=529$ is a perfect square and $\sqrt{529}=\sqrt{23\times 23}=23$ (v) By prime factorisation, we get

$2800=\underline{2\times 2}\times \underline{2\times 2}\times \underline{5\times 5}\times 7$ Since the prime factor 7 cannot be paired, so 2800 should be divided by 7 . $=\frac{2800}{7}$ $=400$ is a perfect square and $\sqrt{400}=20$ (vi) By prime factorisation, we get

$1620=\underline{2\times 2}\times \underline{3\times 3}\times \underline{3\times 3}\times 5$ Since the prime factor 5 cannot be paired, so 1620 should be divided by 5 . $\frac{1620}{5}$ $=324$ is a perfect square. and $\sqrt{324}=18$

• The students of class VIII of a school donated ₹2401 in all, for Prime Minister's National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. Sol. Let x be the number of students in the class. Therefore, total money denoted by students $=₹(x\times x)=₹x^2$. But it is given as $₹2401$. $\therefore x^2=2401$ $\Rightarrow \mathrm{x}=\sqrt{2401}$

$=\sqrt{\underline{7\times 7}\times \underline{7\times 7}}$ $=7\times 7=49$ $\Rightarrow \mathrm{x}=\sqrt{\underline{3\times 3}\times \underline{3\times 3}\times \underline{5\times 5}}$ $\Rightarrow \mathrm{x}=3\times 3\times 5$ $\Rightarrow \mathrm{x}=45$ number of rows as well as number of plants.

• Find the smallest square number that is divisible by each of the numbers 4,9 and 10. Sol. The smallest number divisible by each one of the numbers 4,9 and 10 is their LCM., which is $(2\times 2\times 9\times 5)$, i.e., 180 . Now, $180=\underline{2\times 2}\times \underline{3\times 3}\times 5$ To make it a perfect square, it must be multiplied by 5 . $\therefore$ Required number $=180\times 5=900$
• Find the smallest square number that is divisible by each of the numbers 8,15 and 20. Sol. LCM of 8, 15 and 20

$=2\times 2\times 2\times 3\times 5=120$ To make 120 a perfect square, we will multiply this number by $2\times 3\times 5$. So the number will be $2\times 2\times 2\times 2\times 3\times 3\times 5\times 5=3600$ Hence, 3600 is the smallest square number, which is divisible by 8,15 and 20.

EXERCISE : 6.4

• Find the square root of each of the following numbers by division method : (i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369 (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900 Sol. (i) By long division, we have

• $\therefore \sqrt{2304}=48$ (ii) By long division, we have

• $\therefore \sqrt{4489}=67$ (iii) By long division, we have

• $\therefore \sqrt{3481}=59$ (iv) 529 By long division, we have

• $\therefore \sqrt{529}=23$ (v) 3249 By long division, we have

• $\therefore \sqrt{3249}=57$ (vi) 1369 By long division, we have

• $\therefore \sqrt{1369}=37$ (vii) 5776

• $\therefore \sqrt{5776}=76$ (viii) 7921 By long division, we have

• $\therefore \sqrt{7921}=89$ (ix) 576

• By long division, we have $\therefore \sqrt{576}=24$ (x) 1024 By long division, we have

• $\therefore \sqrt{1024}=32$ (xi) 3136 By long division, we have

• $\therefore \sqrt{3136}=56$ (xii) 900

• Find the number of digits in the square root of each of the following numbers (without any calculation) (i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625 Sol. We know that if a perfect square is of $n$ digits, then its square root will have $\frac{n}{2}$ digits, if n is even $\frac{(\mathrm{n}+1)}{2}$ digits, if n is odd. (i) Given number is 64 . It is a 2 -digit number, i.e., even number of digits. $\therefore$ The number of digits in $\sqrt{64}$ is $\frac{2}{2}$, i.e., 1 . (ii) Given number is 144. It is a 3-digit number, i.e., odd number of digits. $\therefore \sqrt{144}$ contains $\frac{3+1}{2}$, i.e., 2-digits. (iii) Given number is 4489. It is a 4-digit number, i.e. even number of digits. $\therefore$ The number of digits in $\sqrt{4489}$ is $\frac{4}{2}$, i.e., 2 digits. (iv) Given number is 27225. It is a 5 -digit number, i.e., odd number of digits $\therefore$ The number of digits in $\sqrt{27225}$ is $\frac{5+1}{2}$, i.e., $\frac{6}{2}=3$ digits. (v) Given number has 6-digits i.e., even number of digits. $\therefore \sqrt{390625}$ contains $\frac{6}{2}$, i.e., 3-digits.
• Find the square root of the following decimal numbers. (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36 Sol. (i) 2.56 Here, the number of decimal places is already even. So, mark off periods and proceed as under

• $\therefore \sqrt{2.56}=1.6$ (ii) 7.29 Here, the number of decimal places is already even. So, mark off periods and proceed as under

• $\therefore \sqrt{7.29}=2.7$ (iii) 51.84 Here, the number of decimal places are already even. So, mark off periods and proceed as under

• $\therefore \sqrt{51.84}=7.2$ (iv) 42.25 Here, the number of decimal places is already even. So, mark off periods and proceed as under

• $\therefore \sqrt{42.25}=6.5$ (v) 31.36 Here, the number of decimal places is already even. So, mark off periods and proceed as under

• $\therefore \sqrt{31.36}=5.6$
• Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 402 (ii) 1989 (iii) 3250 (iv) 825 (v) 4000 Sol. (i) Let us try to find the square root of 402. This shows that $(20{)}^2$ is less than 402 by 2 . So, in order to get a perfect square, 2 must be subtracted from the given number.

• $\therefore$ Required perfect square number $=402-2=400$ Also, $\sqrt{400}=20$ (ii) Let us try to find the square root of 1989. This shows that (44) ${}^2$ is less than 1989 by 53. So, in order to get a perfect square, 53 must be subtracted from the given number.

• $\therefore$ Required perfect square number = 1989-53 = 1936 Also, $\sqrt{1936}=44$ (iii) Let us try to find the square root of 3250 . This shows that (57) ${}^2$ is less than 3250 by 1 . So, in order to get a perfect square, 1 must be subtracted from the given number.

• $\therefore$ Required perfect square number $=3250-1=3249$ Also, $\sqrt{3249}=57$ (iv) Let us try to find the square root of 825 . This shows that $(28{)}^2$ is less than 825 by 41 . So, in order to get a perfect square, 41 must be subtracted from the given number.

• $\therefore$ Required perfect square number = $825-41=784$ Also, $\sqrt{784}=28$ (v) Let us try to find the square root of 4000.This shows that $(63{)}^2$ is less than 4000 by 31 . So in order to get a perfect square, 31 must be subtracted from the given number.

• $\therefore$ Required perfect square number. $=4000-31=3969$ Also, $\sqrt{3969}=63$
• Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412 Sol. (i) We try to find out the square root of 525.

• We observe that $(22{)}^2<525<(23{)}^2$ The required number to be added $=(23{)}^2-525$ = 529-525 = 4 $\therefore$ Required perfect square number $=525+4=529$ Clearly, $\sqrt{529}=23$ (ii) We try to find out the square root of 1750

• We observe that $(41{)}^2<1750<(42{)}^2$. The required number to be added $=(42{)}^2-1750$ = 1764-1750 $=14$ $\therefore$ Required perfect square number $=(1750+14)=1764$ Clearly, $\sqrt{1764}=42$ (iii) We try to find the square root of 252 .

• we observe that $(15{)}^2<252<(16{)}^2$ The required number to be added $=(16{)}^2-252$ $=256-252=4$ $\therefore$ Required perfect square number $=252+4=256$ Clearly, $\sqrt{256}=16$ (iv) We try to find out the square root of 1825

• we observe that $(42{)}^2<1825<(43{)}^2$ The required number to be added. $=(43{)}^2-1825$ $=1849-1825=24$ $\therefore$ Required perfect square number. $=1825+24=1849$ Clearly, $\sqrt{1849}=43$ (v) We try to find out the square root of 6412

• we observe that $(80.7{)}^2<6412<(81{)}^2$ The required number to be added $=(81{)}^2-6412$ $=6561-6412=149$ $\therefore$ Required perfect square number. $=6412+149=6561$ Clearly, $\sqrt{6561}=81$
• Find the length of the side of a square whose area is $441{\mathrm{m}}^2$. Sol. Area $=(\text{ side }{)}^2$ $=441{\mathrm{m}}^2$

• $\therefore$ Side $=\sqrt{441}\mathrm{m}=21\mathrm{m}$
• In a right triangle $\mathrm{ABC},\angle \mathrm{B}=90^{\circ }$ (a) If $AB=6\mathrm{cm},\mathrm{BC}=8\mathrm{cm}$, find AC . (b) If $\mathrm{AC}=13\mathrm{cm},\mathrm{BC}=5\mathrm{cm}$, find AB . Sol. (a) In right angled $△\mathrm{ABC}$, using Pythagoras theorem, we have

• $$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^2=6^2+8^2 \\ =36+64=100 \\ \therefore AC=\sqrt{100}\mathrm{cm}=10\mathrm{cm} \end{gathered}$$ (b) In right angled $△\mathrm{ABC}$, using Pythagoras theorem, we have

• $$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow 13^2={\mathrm{AB}}^2+5^2 \\ {\mathrm{AB}}^2=169-25=144 \\ \therefore AB=\sqrt{144}\mathrm{cm}=12\mathrm{cm} \end{gathered}$$
• A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this. Sol. Let us find out the square root of 1000 . Since number of rows and the number of columns of plants are same. So number of plants should be perfect square. We observe that

• $(31{)}^2<1000<(32{)}^2$ The required plants to be added. $(32{)}^2-1000$ $\therefore 1024-1000=24$ plants
• There are 500 children in a school. For a P.T. practice they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement? Sol. Let us find the square root of 500 .

• This shows that $(22{)}^2=484$ is less than 500 by 16 . $\therefore 16$ students have to go out for others to do the P.T. practice as per condition.