Question paper, Set 6

$p$ is multiplied by itself $2$ times.

$\therefore p \times p = p^2$ (read as 'p squared')

Answer: $p^2$

$26^2 = 26 \times 26$

$= 676$

Answer: $676$

$16^2 - 9^2 = 256 - 81$

$= 175$

Answer: $175$

$30^2 + 40^2 = 900 + 1600$

$= 2500$

Answer: $2500$

$15$: $1 \times 2 = 2$, write $25$ after it: $15^2 = 225$

$25$: $2 \times 3 = 6$, write $25$ after it: $25^2 = 625$

Answer: $15^2 = 225$, $25^2 = 625$

$a^2 = b \Rightarrow \sqrt{b} = a$

$167^2 = 27889 \Rightarrow \sqrt{27889} = 167$

Answer: $167$

From the table, $\sqrt{324} = 18$ and $\sqrt{484} = 22$

$\sqrt{324} + \sqrt{484} = 18 + 22 = 40$

Answer: $40$

Prime factorization method:

Divide by primes: $3969 \div 3 = 1323$, $1323 \div 3 = 441$, $441 \div 3 = 147$, $147 \div 3 = 49$, $49 \div 7 = 7$

$\sqrt{3969} = \sqrt{(3 \times 3) \times (3 \times 3) \times (7 \times 7)}$

$= 3 \times 3 \times 7$

$= 63$

Answer: $63$

Long division method:

Pair the digits from the right: $\overline{62}\;\overline{41}$

$7 \times 7 = 49 \le 62$: first digit $7$, and $62 - 49 = 13$

Bring down $41$: we get $1341$. New divisor: $7 \times 2 = 14$; $149 \times 9 = 1341$; $1341 - 1341 = 0$

Remainder $0$. $\therefore \sqrt{6241} = 79$

Answer: $79$

(i) Prime factorization method:

Divide by primes: $9604 \div 2 = 4802$, $4802 \div 2 = 2401$, $2401 \div 7 = 343$, $343 \div 7 = 49$, $49 \div 7 = 7$

$\sqrt{9604} = \sqrt{(2 \times 2) \times (7 \times 7) \times (7 \times 7)}$

$= 2 \times 7 \times 7$

$= 98$

(ii) Long division method:

Pair the digits from the right: $\overline{96}\;\overline{04}$

$9 \times 9 = 81 \le 96$: first digit $9$, and $96 - 81 = 15$

Bring down $04$: we get $1504$. New divisor: $9 \times 2 = 18$; $188 \times 8 = 1504$ ($189 \times 9 = 1701$ is too big); $1504 - 1504 = 0$

Remainder $0$. $\therefore \sqrt{9604} = 98$

Answer: $98$

Long division method:

Pair the digits from the right: $\overline{3}\;\overline{68}\;\overline{64}$

$1 \times 1 = 1 \le 3$: first digit $1$, and $3 - 1 = 2$

Bring down $68$: we get $268$. New divisor: $1 \times 2 = 2$; $29 \times 9 = 261$; $268 - 261 = 7$

Bring down $64$: we get $764$. New divisor: $19 \times 2 = 38$; $382 \times 2 = 764$ ($383 \times 3 = 1149$ is too big); $764 - 764 = 0$

Remainder $0$. $\therefore \sqrt{36864} = 192$

Answer: $192$

$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$

$\sqrt{12} \times \sqrt{27} = \sqrt{12 \times 27} = \sqrt{324}$

$= \sqrt{18 \times 18} = 18$

Answer: $18$

$(\sqrt{a})^2 = \sqrt{a} \times \sqrt{a} = \sqrt{a^2} = a$

$\therefore \sqrt{6} \times \sqrt{6} = \sqrt{6^2} = 6$

Answer: $6$

$\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$

$\displaystyle \frac{\sqrt{108}}{\sqrt{75}} = \sqrt{\frac{108}{75}} = \sqrt{\frac{3 \times 36}{3 \times 25}}$

$\displaystyle = \sqrt{\frac{36}{25}} = \frac{6}{5}$

Answer: $\displaystyle \frac{6}{5}$

$15 = \sqrt{15} \times \sqrt{15}$

$\displaystyle \frac{15}{\sqrt{15}} = \frac{\sqrt{15} \times \sqrt{15}}{\sqrt{15}} = \sqrt{15}$

Answer: $\sqrt{15}$

LHS $= \sqrt{289} - \sqrt{64} = 17 - 8 = 9$

RHS $= \sqrt{289 - 64} = \sqrt{225} = 15$

$\therefore$ LHS $\ne$ RHS

Answer: LHS $= 9$, RHS $= 15$; LHS $\ne$ RHS

(i) LHS $= \sqrt{20^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29$

RHS $= a + b = 20 + 21 = 41$

$29 \ne 41$, so $\sqrt{a^2 + b^2} \ne a + b$

(ii) LHS $= \sqrt{(20 + 21)^2} = \sqrt{41^2} = 41$

RHS $= 20 + 21 = 41$

LHS $=$ RHS, so $\sqrt{(a + b)^2} = a + b$

Answer: (i) $29 \ne 41$ (ii) $41 = 41$

$\displaystyle \sqrt{0.0576} = \sqrt{\frac{576}{10000}}$

$\displaystyle = \frac{\sqrt{576}}{\sqrt{10000}}$

$\displaystyle = \frac{24}{100}$

$= 0.24$

Answer: $0.24$

$\displaystyle \sqrt{0.007225} = \sqrt{\frac{7225}{1000000}}$

$\displaystyle = \frac{\sqrt{7225}}{\sqrt{1000000}}$

$85^2 = 7225$, so $\sqrt{7225} = 85$

$\displaystyle = \frac{85}{1000}$

$= 0.085$

Answer: $0.085$

$\displaystyle \sqrt{67.24} = \sqrt{\frac{6724}{100}} = \frac{\sqrt{6724}}{\sqrt{100}}$

$\sqrt{6724}$ by the long division method:

Pair the digits from the right: $\overline{67}\;\overline{24}$

$8 \times 8 = 64 \le 67$: first digit $8$, and $67 - 64 = 3$

Bring down $24$: we get $324$. New divisor: $8 \times 2 = 16$; $162 \times 2 = 324$ ($163 \times 3 = 489$ is too big); $324 - 324 = 0$

$\sqrt{6724} = 82$

$\displaystyle \therefore \sqrt{67.24} = \frac{82}{10} = 8.2$

Answer: $8.2$