Solutions — Surds 4D Worksheet
Year 10 Mathematics Core · Cambridge Ch 4, §4D · Mr Wong
ANSWER KEY
Warm-up
Q1. Simplify each fraction:
a $1$
b $\tfrac{1}{2}$
c $-2$
d $6$
Q2. Missing numbers:
a $\sqrt{3}$
b $10$
c $\sqrt{5}$ ($2\sqrt{5}\cdot\sqrt{5}=10$)
d $\sqrt{7}$ ($\sqrt{7}\cdot 3\sqrt{7}=21$)
Part A — Rationalise (single-term numerator)
Q3. Rationalise and simplify:
a $\dfrac{\sqrt{5}}{5}$
b $2\sqrt{2}$ ($\tfrac{4\sqrt{2}}{2}$)
c $\dfrac{\sqrt{6}}{2}$
d $\dfrac{\sqrt{14}}{5}$ ($\tfrac{2\sqrt{14}}{10}$)
Q4. Rationalise and simplify:
a $\dfrac{5\sqrt{3}}{6}$ ($\times\tfrac{\sqrt{3}}{\sqrt{3}}\;\to\;\tfrac{5\sqrt{3}}{2\cdot 3}$)
b $\dfrac{3\sqrt{10}}{8}$ ($\tfrac{3\sqrt{10}}{4\cdot 2}$)
c $\sqrt{3}$ ($\tfrac{6}{\sqrt{12}}=\tfrac{6}{2\sqrt{3}}=\tfrac{3}{\sqrt{3}}=\sqrt{3}$)
d $\dfrac{4\sqrt{21}}{7}$
Part B — Binomial numerator
Q5. Rationalise (distribute over both terms on top):
a $\dfrac{\sqrt{3}-3}{3}$
b $\dfrac{2\sqrt{5}+5}{5}$
c $\dfrac{3\sqrt{2}+2}{2}$
d $\dfrac{2\sqrt{7}-7}{7}$
Challenge
Q6. $\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{8}}$ — simplify $\sqrt{8}=2\sqrt{2}$ first:
·
$=\dfrac{1}{\sqrt{2}}+\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{4}=\dfrac{2\sqrt{2}+\sqrt{2}}{4}=$ $\dfrac{3\sqrt{2}}{4}$
Q7. Right triangle, legs $1$ cm and $\sqrt{3}$ cm:
a
Hypotenuse$^{2}=1^{2}+(\sqrt{3})^{2}=1+3=4$, so hyp $=2$ cm
b
$\dfrac{1}{\text{hyp}}=\dfrac{1}{2}$ — already rational, so no rationalisation needed. $=\tfrac{1}{2}$ (a 30°–60°–90° triangle: this is the well-known result $\sin 30°=\tfrac{1}{2}$)