Q1. The following numbers are not perfect squares. Give reason.
$$ (i) \quad 1547 \quad (ii) \quad 45743 \quad (iii) \quad 8948 \quad (iv) \quad 333333 $$Sol. We know that the numbers having 2,3,7 or 8 at unit’s place is never a perfect square.
Therefore,
(i) 1547 : Unit place is 7 so not a perfect square.
(ii) 45743 : Unit place is 3 so not a perfect square.
(iii) 8948 : Unit place is 8 so not a perfect square.
(iv) 333333 : Unit place is 3 so not a perfect square.
Q2. Show that the following numbers are not perfect squares:
$$ (i) \quad 9327 \quad (ii) \quad 4058 \\ (iii) \quad 22453 \quad (iv) \quad 743522 $$Sol. We know that the numbers having 2,3,7 or 8 at unit’s place is never a perfect square.
(i) 9327 : Unit place is 7 so Not a perfect square.
(ii) 4058 : Unit place is 8 so Not a perfect square.
(iii) 22453 : Unit place is 3 so Not a perfect square.
(iv) 743522 : Unit place is 2 so Not a perfect square.
Q3. The square of which of the following numbers would be an odd number?
$$ (i) \quad 731 \quad (ii) \quad 3456 \\ (iii) \quad 5559 \quad (iv) \quad 42008 $$Sol. We know that square of an odd number is an odd number and square of an even number is an even number.
Therefore,
(i) 731 is an odd number so its square will be odd.
(ii) 3456 is an even number so its square will be even.
(iii) 5559 is an odd number so its square will be odd.
(iv) 42008 is an even number so its square will be even.
Q4. What will be the units digit of the squares of the following
$$ (i) \quad 52 \quad (ii) \quad 977 \quad (iii) \quad 4583 $$ $$ (iv) \quad 78367 \quad (v) \quad 52698 \quad (vi) \quad 99880 $$ $$ (vii) \quad 12796 \quad (viii) \quad 55555 \quad (ix) \quad 53924 $$Sol. (i) 52 : Its last digit is 2.
$$ \text {Hence, the unit’s digit of } 52^{2} \text { is } 2^{2} \text {, which is equal to 4. } $$(ii) 977 : Its last digit is 7.
$$ \text {Hence, the unit’s digit of } 977^{2} \text { is equal to unit digit of } 7^{2} \text {, which is equal to 9. } $$(iii) 4583 : Its last digit is 3.
$$ \text {Hence, the unit’s digit of } 4583^{2} \text { is unit digit of } 3^{2} \text {, which is equal to 9. } $$(iv) 78367 : Its last digit is 7.
$$ \text {Hence, the unit’s digit of } 78367^{2} \text { is unit digit of } 7^{2} \text {, which is equal to 9. } $$(v) 52698 : Its last digit is 8.
$$ \text {Hence, the unit’s digit of } 52698^{2} \text { is unit digit of } 8^{2} \text {, which is equal to 4. } $$(vi) 99880 : Its last digit is 0.
$$ \text {Hence, the unit’s digit of } 99880^{2} \text { is 0. } $$\text {(vii) 12796 : Its last digit is 6.
$$ \text {Hence, the unit’s digit of } 12796^{2} \text { is unit digit of } 6^{2} \text {, which is equal to 6. } $$(viii) 55555 : Its last digit is 5.
$$ \text {Hence, the unit’s digit of } 55555^{2} \text { is unit digit of } 5^{2} \text {, which is equal to 5. } $$(ix) 53924 : Its last digit is 4.
$$ \text {Hence, the unit’s digit of } 53924^{2} \text { is unit digit of } 4^{2} \text {, which is equal to 6. } $$Q5. 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.
Sol. Looking at 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.
$$ \Rightarrow \quad 1 + 3 + 5 + 7+ ……. \text { upto nth term }= n^{2} $$Q6. Observe the following pattern
$$ 2^{2}-1^{2}=2+1 $$ $$ 3^{2}-2^{2}=3+2 $$ $$ 4^{2}-3^{2}=4+3 $$ $$ 5^{2}-4^{2}=5+4 $$ and find the value of
$$ (i) \quad 100^{2}-99^{2} \quad (ii) \quad 111^{2}-109^{2} \quad (iii) \quad 99^{2}-96^{2} $$Sol. Looking at the pattern, we found that difference of squares of two consecutive numbers is equat to sum of those numbers.
Therefore,
$$ (n+1)^{2} – n^{2} = (n+1) + n $$ $$ (i) \quad 100^{2}-99^{2} =100+99=199 $$ $$ (ii) \quad 111^{2}-109^{2} $$ $$ = 111^{2}-110^{2}+ 110^{2}-109^{2} $$ $$ = (111+110)+(110+109) $$ $$ = 221 + 219 = 440 $$$$ (iii) \quad 99^{2}-96^{2} = (99^{2}-98^{2})+(98^{2}-97^{2})+(97^{2}-96^{2}) $$ $$ =(99+98)+(98+97)+(97+96) = 585 $$Q7. Which of the following triplets are Pythagorean?
$$ (i) \quad (8,15,17) \quad \quad (ii) \quad (18,80,82) $$ $$ (iii) \quad (14,48,51) \quad \quad (iv) \quad (10,24,26) $$ $$ (v) \quad (16,63,65) \quad \quad (vi) \quad (12,35,38) $$$$ \text {Sol. } (i) 8^{2}+15^{2} = 64 + 225 =289 =17^{2} $$ Therefore, (8,15,17) triplet is Pythagorean.
$$ (ii) \quad 18^{2}+80^{2}= 324 + 6400 =6724 =82^{2} $$ Therefore (18,80,82) triplet is Pythagorean.
$$ (iii) \quad 14^{2}+48^{2}= 196 + 2304 =2500 =50^{2} $$ Therefore (14,48,51) triplet is not a Pythagorean.
$$ (iv) \quad 10^{2}+24^{2}= 100 + 576 =676 =26^{2} $$ Therefore (10,24,26) triplet is Pythagorean.
$$ (v) \quad 16^{2}+63^{2}= 256 + 3969 =4225 =65^{2} $$ Therefore (16,63,65) triplet is Pythagorean.
$$ (vi) \quad 12^{2}+35^{2}= 144 + 1225 =1369 =37^{2} $$ Therefore (12,35,38) triplet is not a Pythagorean.
$$ (i) \quad 1547 \quad (ii) \quad 45743 \quad (iii) \quad 8948 \quad (iv) \quad 333333 $$Sol. We know that the numbers having 2,3,7 or 8 at unit’s place is never a perfect square.
Therefore,
(i) 1547 : Unit place is 7 so not a perfect square.
(ii) 45743 : Unit place is 3 so not a perfect square.
(iii) 8948 : Unit place is 8 so not a perfect square.
(iv) 333333 : Unit place is 3 so not a perfect square.
Q2. Show that the following numbers are not perfect squares:
$$ (i) \quad 9327 \quad (ii) \quad 4058 \\ (iii) \quad 22453 \quad (iv) \quad 743522 $$Sol. We know that the numbers having 2,3,7 or 8 at unit’s place is never a perfect square.
(i) 9327 : Unit place is 7 so Not a perfect square.
(ii) 4058 : Unit place is 8 so Not a perfect square.
(iii) 22453 : Unit place is 3 so Not a perfect square.
(iv) 743522 : Unit place is 2 so Not a perfect square.
Q3. The square of which of the following numbers would be an odd number?
$$ (i) \quad 731 \quad (ii) \quad 3456 \\ (iii) \quad 5559 \quad (iv) \quad 42008 $$Sol. We know that square of an odd number is an odd number and square of an even number is an even number.
Therefore,
(i) 731 is an odd number so its square will be odd.
(ii) 3456 is an even number so its square will be even.
(iii) 5559 is an odd number so its square will be odd.
(iv) 42008 is an even number so its square will be even.
Q4. What will be the units digit of the squares of the following
$$ (i) \quad 52 \quad (ii) \quad 977 \quad (iii) \quad 4583 $$ $$ (iv) \quad 78367 \quad (v) \quad 52698 \quad (vi) \quad 99880 $$ $$ (vii) \quad 12796 \quad (viii) \quad 55555 \quad (ix) \quad 53924 $$Sol. (i) 52 : Its last digit is 2.
$$ \text {Hence, the unit’s digit of } 52^{2} \text { is } 2^{2} \text {, which is equal to 4. } $$(ii) 977 : Its last digit is 7.
$$ \text {Hence, the unit’s digit of } 977^{2} \text { is equal to unit digit of } 7^{2} \text {, which is equal to 9. } $$(iii) 4583 : Its last digit is 3.
$$ \text {Hence, the unit’s digit of } 4583^{2} \text { is unit digit of } 3^{2} \text {, which is equal to 9. } $$(iv) 78367 : Its last digit is 7.
$$ \text {Hence, the unit’s digit of } 78367^{2} \text { is unit digit of } 7^{2} \text {, which is equal to 9. } $$(v) 52698 : Its last digit is 8.
$$ \text {Hence, the unit’s digit of } 52698^{2} \text { is unit digit of } 8^{2} \text {, which is equal to 4. } $$(vi) 99880 : Its last digit is 0.
$$ \text {Hence, the unit’s digit of } 99880^{2} \text { is 0. } $$\text {(vii) 12796 : Its last digit is 6.
$$ \text {Hence, the unit’s digit of } 12796^{2} \text { is unit digit of } 6^{2} \text {, which is equal to 6. } $$(viii) 55555 : Its last digit is 5.
$$ \text {Hence, the unit’s digit of } 55555^{2} \text { is unit digit of } 5^{2} \text {, which is equal to 5. } $$(ix) 53924 : Its last digit is 4.
$$ \text {Hence, the unit’s digit of } 53924^{2} \text { is unit digit of } 4^{2} \text {, which is equal to 6. } $$Q5. 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.
Sol. Looking at 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.
$$ \Rightarrow \quad 1 + 3 + 5 + 7+ ……. \text { upto nth term }= n^{2} $$Q6. Observe the following pattern
$$ 2^{2}-1^{2}=2+1 $$ $$ 3^{2}-2^{2}=3+2 $$ $$ 4^{2}-3^{2}=4+3 $$ $$ 5^{2}-4^{2}=5+4 $$ and find the value of
$$ (i) \quad 100^{2}-99^{2} \quad (ii) \quad 111^{2}-109^{2} \quad (iii) \quad 99^{2}-96^{2} $$Sol. Looking at the pattern, we found that difference of squares of two consecutive numbers is equat to sum of those numbers.
Therefore,
$$ (n+1)^{2} – n^{2} = (n+1) + n $$ $$ (i) \quad 100^{2}-99^{2} =100+99=199 $$ $$ (ii) \quad 111^{2}-109^{2} $$ $$ = 111^{2}-110^{2}+ 110^{2}-109^{2} $$ $$ = (111+110)+(110+109) $$ $$ = 221 + 219 = 440 $$$$ (iii) \quad 99^{2}-96^{2} = (99^{2}-98^{2})+(98^{2}-97^{2})+(97^{2}-96^{2}) $$ $$ =(99+98)+(98+97)+(97+96) = 585 $$Q7. Which of the following triplets are Pythagorean?
$$ (i) \quad (8,15,17) \quad \quad (ii) \quad (18,80,82) $$ $$ (iii) \quad (14,48,51) \quad \quad (iv) \quad (10,24,26) $$ $$ (v) \quad (16,63,65) \quad \quad (vi) \quad (12,35,38) $$$$ \text {Sol. } (i) 8^{2}+15^{2} = 64 + 225 =289 =17^{2} $$ Therefore, (8,15,17) triplet is Pythagorean.
$$ (ii) \quad 18^{2}+80^{2}= 324 + 6400 =6724 =82^{2} $$ Therefore (18,80,82) triplet is Pythagorean.
$$ (iii) \quad 14^{2}+48^{2}= 196 + 2304 =2500 =50^{2} $$ Therefore (14,48,51) triplet is not a Pythagorean.
$$ (iv) \quad 10^{2}+24^{2}= 100 + 576 =676 =26^{2} $$ Therefore (10,24,26) triplet is Pythagorean.
$$ (v) \quad 16^{2}+63^{2}= 256 + 3969 =4225 =65^{2} $$ Therefore (16,63,65) triplet is Pythagorean.
$$ (vi) \quad 12^{2}+35^{2}= 144 + 1225 =1369 =37^{2} $$ Therefore (12,35,38) triplet is not a Pythagorean.