Question paper, Set 8

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

$\therefore t \times t = t^2$ (read as 't squared')

Answer: $t^2$

$22^2 = 22 \times 22$

$= 484$

Answer: $484$

$27^2 - 14^2 = 729 - 196$

$= 533$

Answer: $533$

$50^2 = 50 \times 50 = 2500$

$\therefore$ the number is $50$

Answer: $50$

The number before $5$ is $7$: $7 \times 8 = 56$

Write $25$ after it.

$\therefore 75^2 = 5625$

Answer: $5625$

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

$183^2 = 33489 \Rightarrow \sqrt{33489} = 183$

Answer: $183$

From the table, $\sqrt{784} = 28$ and $\sqrt{169} = 13$

$\sqrt{784} - \sqrt{169} = 28 - 13 = 15$

Answer: $15$

Prime factorization method:

Divide by primes: $2916 \div 2 = 1458$, $1458 \div 2 = 729$, $729 \div 3 = 243$, $243 \div 3 = 81$, $81 \div 3 = 27$, $27 \div 3 = 9$, $9 \div 3 = 3$

$\sqrt{2916} = \sqrt{(2 \times 2) \times (3 \times 3) \times (3 \times 3) \times (3 \times 3)}$

$= 2 \times 3 \times 3 \times 3$

$= 54$

Answer: $54$

Long division method:

Pair the digits from the right: $\overline{37}\;\overline{21}$

$6 \times 6 = 36 \le 37$: first digit $6$, and $37 - 36 = 1$

Bring down $21$: we get $121$. New divisor: $6 \times 2 = 12$; $121 \times 1 = 121$ ($122 \times 2 = 244$ is too big); $121 - 121 = 0$

Remainder $0$. $\therefore \sqrt{3721} = 61$

Answer: $61$

(i) Prime factorization method:

Divide by primes: $1521 \div 3 = 507$, $507 \div 3 = 169$, $169 \div 13 = 13$

$\sqrt{1521} = \sqrt{(3 \times 3) \times (13 \times 13)}$

$= 3 \times 13$

$= 39$

(ii) Long division method:

Pair the digits from the right: $\overline{15}\;\overline{21}$

$3 \times 3 = 9 \le 15$: first digit $3$, and $15 - 9 = 6$

Bring down $21$: we get $621$. New divisor: $3 \times 2 = 6$; $69 \times 9 = 621$; $621 - 621 = 0$

Remainder $0$. $\therefore \sqrt{1521} = 39$

Answer: $39$

Long division method:

Pair the digits from the right: $\overline{18}\;\overline{57}\;\overline{61}$

$4 \times 4 = 16 \le 18$: first digit $4$, and $18 - 16 = 2$

Bring down $57$: we get $257$. New divisor: $4 \times 2 = 8$; $83 \times 3 = 249$ ($84 \times 4 = 336$ is too big); $257 - 249 = 8$

Bring down $61$: we get $861$. New divisor: $43 \times 2 = 86$; $861 \times 1 = 861$ ($862 \times 2 = 1724$ is too big); $861 - 861 = 0$

Remainder $0$. $\therefore \sqrt{185761} = 431$

Answer: $431$

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

$\sqrt{18} \times \sqrt{50} = \sqrt{18 \times 50} = \sqrt{900}$

$= \sqrt{30 \times 30} = 30$

Answer: $30$

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

$\therefore \sqrt{10} \times \sqrt{10} = \sqrt{10^2} = 10$

Answer: $10$

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

$\displaystyle \frac{\sqrt{44}}{\sqrt{99}} = \sqrt{\frac{44}{99}} = \sqrt{\frac{11 \times 4}{11 \times 9}}$

$\displaystyle = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Answer: $\displaystyle \frac{2}{3}$

$17 = \sqrt{17} \times \sqrt{17}$

$\displaystyle \frac{17}{\sqrt{17}} = \frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17}} = \sqrt{17}$

Answer: $\sqrt{17}$

LHS $= \sqrt{100} + \sqrt{49} = 10 + 7 = 17 = \sqrt{289}$

RHS $= \sqrt{100 + 49} = \sqrt{149}$

$\therefore$ LHS $\ne$ RHS

Answer: LHS $= 17 = \sqrt{289}$, RHS $= \sqrt{149}$; LHS $\ne$ RHS

(i) LHS $= \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$

RHS $= a + b = 12 + 16 = 28$

$20 \ne 28$, so $\sqrt{a^2 + b^2} \ne a + b$

(ii) LHS $= \sqrt{(12 + 16)^2} = \sqrt{28^2} = 28$

RHS $= 12 + 16 = 28$

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

Answer: (i) $20 \ne 28$ (ii) $28 = 28$

$\displaystyle \sqrt{0.0529} = \sqrt{\frac{529}{10000}}$

$\displaystyle = \frac{\sqrt{529}}{\sqrt{10000}}$

$\displaystyle = \frac{23}{100}$

$= 0.23$

Answer: $0.23$

$\displaystyle \sqrt{0.001296} = \sqrt{\frac{1296}{1000000}}$

$\displaystyle = \frac{\sqrt{1296}}{\sqrt{1000000}}$

$36^2 = 1296$, so $\sqrt{1296} = 36$

$\displaystyle = \frac{36}{1000}$

$= 0.036$

Answer: $0.036$

$\displaystyle \sqrt{11.56} = \sqrt{\frac{1156}{100}} = \frac{\sqrt{1156}}{\sqrt{100}}$

$\sqrt{1156}$ by the long division method:

Pair the digits from the right: $\overline{11}\;\overline{56}$

$3 \times 3 = 9 \le 11$: first digit $3$, and $11 - 9 = 2$

Bring down $56$: we get $256$. New divisor: $3 \times 2 = 6$; $64 \times 4 = 256$ ($65 \times 5 = 325$ is too big); $256 - 256 = 0$

$\sqrt{1156} = 34$

$\displaystyle \therefore \sqrt{11.56} = \frac{34}{10} = 3.4$

Answer: $3.4$