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Express $\dfrac{3}{2-\sqrt3}$ in the form $a+b\sqrt3$.

Rationalising a binomial denominator

Surds

Simplify $\sqrt{75}-2\sqrt{12}+\sqrt{27}$.

Simplifying like surds

<p>The diagram shows a prism in which its cross-section is a regular hexagon of side $ \left(5 + \sqrt{3} \right)$ cm.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find an expression for the cross-sectional area of the prism in the form $q\sqrt{3} + p$ cm$^2$, where&nbsp;$p$ and $q$ are integers.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Given that the volume of the prism is $3 \left(32\sqrt{3} + 138 \right)$ cm$^3$, find the height of the prism in the form $ \left(a\sqrt{3} + b \right)$ cm, where&nbsp;$a$ and $b$ are integers.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a prism in which its cross-section is a regular hexagon of side $ \left(5 + \sqrt{3} \right)$ cm.

2024 GESS PRELIMS P1 Q11

<p>Without using a calculator, find the values of $a$ and $b$ for which the solution of the equation $x\sqrt{24} = x\sqrt{3} + \sqrt{6}$ is $\frac{a + \sqrt{b}}{7}$.<span class="inline-marks">[5]</span></p>

Without using a calculator, find the values of $a$ and $b$ for which the solution of the equation $x\sqrt{24} = x\sqrt{3} + \sqrt{6}$ is $\frac{a + \sqrt{b}}{7}$.

Equation Involving Surds

2024 FMSS PRELIMS P1 Q2

<p>Two cylinders are such that the ratio of their heights is $\sqrt{7} : 1$.<br>The height of the smaller cylinder is $\frac{2\sqrt{7} - 1}{(2 - \sqrt{7})^2}$ cm.<br>Without using a calculator, find the height of the larger cylinder, expressing your answer in the form $(a + b\sqrt{7})$ cm, where&nbsp;$a$ and $b$ are integers.<span class="inline-marks">[4]</span></p>

Two cylinders are such that the ratio of their heights is $\sqrt{7} : 1$.
The height of the smaller cylinder is $\frac{2\sqrt{7} - 1}{(2 - \sqrt{7})^2}$ cm.
Without using a calculator, find the height of the larger cylinder, expressing your answer in the form $(a + b\sqrt{7})$ cm, where $a$ and $b$ are integers.

Applications of Surds

2024 CGSS P1 Q1

<p>A triangle has a base of $\left(\frac{1}{\sqrt{2}} - \frac{\sqrt{32}}{2} + \frac{16}{\sqrt{48}}\right)$ m and a height of $h$ m. Given that the area of the triangle is $\left(2\sqrt{2} - \sqrt{3}\right)$ m$^2$, find, without using calculator, the value of $h$ in the form $\frac{a\sqrt{6} + b}{5}$ where&nbsp;$a$ and $b$ are integers.<span class="inline-marks">[5]</span></p>

A triangle has a base of $\left(\frac{1}{\sqrt{2}} - \frac{\sqrt{32}}{2} + \frac{16}{\sqrt{48}}\right)$ m and a height of $h$ m. Given that the area of the triangle is $\left(2\sqrt{2} - \sqrt{3}\right)$ m$^2$, find, without using calculator, the value of $h$ in the form $\frac{a\sqrt{6} + b}{5}$ where $a$ and $b$ are integers.

2024 CCHY P1 Q2

<p>It is given that the solution of the equation $x\sqrt{24} = x\sqrt{3} + \sqrt{6}$ is $\frac{a + \sqrt{b}}{7}$. <br>Find the values of the integers $a$ and $b$.</p>

It is given that the solution of the equation $x\sqrt{24} = x\sqrt{3} + \sqrt{6}$ is $\frac{a + \sqrt{b}}{7}$. 
Find the values of the integers $a$ and $b$.

<div class="graph-block" data-graph-config="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" data-width="300" data-height="250" style="width:300px;height:250px;max-width:100%;display:block;margin:0 auto;"></div>
<p>The diagram above shows a right-angled triangle $ABC$, where&nbsp;$AB=\sqrt{2}$ cm, and $BC=\sqrt{2}+\sqrt{3}$ cm.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-content"><span class="list-item-text">Find the exact value of $\tan \angle ACB$ in the form of $a+\sqrt{b}$, where&nbsp;$a$ and $b$ are integers.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-content"><span class="list-item-text">Hence, find the exact value of $\sec^2 \angle ACB$ in the form $m+n\sqrt{k}$, where&nbsp;$m$ and $n$ are integers.</span><span class="inline-marks">[2]</span></div></li></ol>

The diagram above shows a right-angled triangle $ABC$, where $AB=\sqrt{2}$ cm, and $BC=\sqrt{2}+\sqrt{3}$ cm.

Find the exact value of $\tan \angle ACB$ in the form of $a+\sqrt{b}$, where $a$ and $b$ are integers.

Hence, find the exact value of $\sec^2 \angle ACB$ in the form $m+n\sqrt{k}$, where $m$ and $n$ are integers.

(A Math) 2025 CCHY PRELIMS P1 Q1

<p>A closed circular cylinder has a volume of $ \left(66\sqrt{2} + 2\sqrt{3} \right)\pi$ cm$^3$, a radius of $ \left(\sqrt{2} + 2\sqrt{3} \right)$ cm and a height $h$ cm. Express $h$ in the form of $p\sqrt{2} + q\sqrt{3}$ where&nbsp;$p$ and $q$ are integers.<span class="inline-marks">[4]</span></p>

A closed circular cylinder has a volume of $\left(66\sqrt{2}+2\sqrt{3}\right)\pi\text{ cm}^3$, a radius of $\left(\sqrt{2}+2\sqrt{3}\right)\text{ cm}$ and a height $h\text{ cm}$. Express $h$ in the form of $p\sqrt{2}+q\sqrt{3}$ where $p$ and $q$ are integers.

2025 AHS PRELIMS P1 Q3

<ol class="abc-list"><li data-list-item-id="item-1782632548380-4rau6h5sh" data-list-item-path="item-1782632548380-4rau6h5sh"><div class="list-item-paragraph">Express $\frac{6\sqrt{2}-9}{\sqrt{2}-1}$in the form of $a+b\sqrt{2}$, where&nbsp;$a$ and $b$ are integers.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1782632549572-8jjvl9pod" data-list-item-path="item-1782632549572-8jjvl9pod"><div class="list-item-paragraph">$\frac{6\sqrt{2}-9}{\sqrt{2}-1}$ is a root of the equation $x^2 + px + q = 0$ where&nbsp;$p$ and $q$ are integers. Using the result in part <strong>(a)</strong>, find the value of $p$ and of $q$.<span class="inline-marks">[4]</span></div></li></ol>

Express $\frac{6\sqrt{2}-9}{\sqrt{2}-1}$ in the form of $a+b\sqrt{2}$, where $a$ and $b$ are integers.

$\frac{6\sqrt{2}-9}{\sqrt{2}-1}$ is a root of the equation $x^2+px+q=0$ where $p$ and $q$ are integers. Using the result in part (a), find the value of $p$ and of $q$.

2025 CGSS PRELIMS P1 Q1

<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770801443737-mxzzan9k6s.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA4MDE0NDM3MzctbXh6emFuOWs2cy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjI0MjMyLCJleHAiOjE3ODk2MjQ1MzJ9.zfqcm82HlfPMBRUlD53-QAYnpleInuzpJoDzLqVSVZE" style="max-width: min(150px, 100%); height: auto; display: block;" alt="The diagram shows a rectangle $ABCD$ with side $AB = (2\sqrt{2} + \sqrt{3})$ cm." class="aligncenter custom-width" /></figure></div>
<p>The diagram shows a rectangle $ABCD$ in which $AB = (2\sqrt{2} + \sqrt{3})$ cm. Given that the area of the rectangle is $(9 + \sqrt{6})$ cm$^2$, find, without using a calculator,</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">the exact value of $BC$ in the form $(\sqrt{a} - \sqrt{b})$ cm, where&nbsp;$a$ and $b$ are integers.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">the exact value of $(AC)^2$ in the form $(c - \sqrt{d})$ cm$^2$, where&nbsp;$c$ and $d$ are integers.</span><span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a rectangle $ABCD$ in which $AB = (2\sqrt{2} + \sqrt{3})$ cm. Given that the area of the rectangle is $(9 + \sqrt{6})$ cm$^2$, find, without using a calculator,

the exact value of $BC$ in the form $(\sqrt{a} - \sqrt{b})$ cm, where $a$ and $b$ are integers.

the exact value of $(AC)^2$ in the form $(c - \sqrt{d})$ cm$^2$, where $c$ and $d$ are integers.

IP3 Class Test (May 2025) Q3

<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768306403007-0qxq3k1ulha.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgzMDY0MDMwMDctMHF4cTNrMXVsaGEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTYyNDIzMiwiZXhwIjoxNzg5NjI0NTMyfQ.eAS4aIIaf21a_f1ra_3dUjpzMnqnYnhXI7oAOjPtF-w" style="max-width: min(300px, 100%); height: auto; display: block;" alt="A triangular prism with base length (2√5 + 3) cm, vertical height (3√5 - 2) cm, and length H cm." class="aligncenter custom-width" /></figure></div>
<p>The volume of the triangular prism shown below is $\left( 159 + \frac{179}{2}\sqrt{5} \right)$ cm${}^3$.</p>
<p>Find, <strong>without using a calculator</strong>, in the form of $a + b\sqrt{5}$, </p>
<ol class="abc-list"><li><div class="list-item-paragraph">the cross-sectional area of the triangular prism,<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">the height, $H$ of the triangular prism.<span class="inline-marks">[3]</span></div></li></ol>

The volume of the triangular prism shown below is $\left( 159 + \frac{179}{2}\sqrt{5} \right)$ cm${}^3$.

Find, **without using a calculator**, in the form of $a + b\sqrt{5}$, 

2023 CSS AM S3 EOY Q4

<p>Without using a calculator, find the value of $a$ and of $b$ such that $\sqrt{a - b\sqrt{6}} = \frac{2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$.<span class="inline-marks">[4]</span></p>

Without using a calculator, find the value of $a$ and of $b$ such that $\sqrt{a - b\sqrt{6}} = \frac{2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$.

2023 Ahmad Ibrahim Sec AM S3 EOY Q2

<p>A cone has a radius of $(1+\sqrt{2})$ cm and a volume of $(7+4\sqrt{2})\pi$ cm${}^3$. <strong>Without using a calculator</strong>, find the height of the cone, in cm, in the form $(a+b\sqrt{2})$, where $a$ and $b$ are integers.</p>
<p>[The volume of a cone of radius $r$ and height $h$ is $\frac{1}{3}\pi r^2 h$.]<span class="inline-marks">[4]</span></p>

A cone has a radius of $(1+\sqrt{2})$ cm and a volume of $(7+4\sqrt{2})\pi$ cm${}^3$. **Without using a calculator**, find the height of the cone, in cm, in the form $(a+b\sqrt{2})$, where $a$ and $b$ are integers.

[The volume of a cone of radius $r$ and height $h$ is $\frac{1}{3}\pi r^2 h$.]

2023 ACSBR AM S3 EOY Q2 

<p>A square of area $ \left(11 + \sqrt{120} \right)\ \text{cm}^2$ has a length of $ \left(\sqrt{a} + \sqrt{b} \right)\ \text{cm}$, where&nbsp;$a &lt; b$.</p>
<p>Without using a calculator, find the values of $a$ and $b$.<span class="inline-marks">[5]</span></p>

A square of area $(11 + \sqrt{120})\text{ cm}^2$ has a length of $(\sqrt{a} + \sqrt{b})\text{ cm}$, where $a < b$.

Without using a calculator, find the values of $a$ and $b$.

Simplifying Surds

<p>Giving your answer in the form $\frac{c+d\sqrt{2}}{3}$, solve, without using a calculator,</p>
<p>\[x\sqrt{18}=3x+\sqrt{32}\]<span class="inline-marks">[5]</span></p>

Giving your answer in the form $\frac{c+d\sqrt{2}}{3}$, solve, without using a calculator, \[x\sqrt{18}=3x+\sqrt{32}.\]

<p>A surd in the form of $\sqrt{a+\sqrt{a+\sqrt{a+...}}}$ is called an <strong>infinite surd</strong>, where&nbsp;$a&gt;0$. </p>
<p>Let&nbsp;$\operatorname{F}\left( a \right)=\sqrt{a+\sqrt{a+\sqrt{a+...}}}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Evaluate $\operatorname{F}\left( 2 \right)$ and $\operatorname{F}\left( 6 \right)$.</div></li><li><div class="list-item-paragraph">Simplify $\operatorname{F}\left( a \right)$, leaving your answer in terms of $a$.</div></li></ol>

A surd in the form of $\sqrt{a+\sqrt{a+\sqrt{a+...}}}$ is called an **infinite surd**, where $a>0$. 

Let $\operatorname{F}\left( a \right)=\sqrt{a+\sqrt{a+\sqrt{a+...}}}$.

Higher Order Thinking Questions — Surds

<p>Simplify $\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}$.</p>

Simplify $\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}$.

<p>Simplify ${{\left( \sqrt{6}+\sqrt{7} \right)}^{94}}\times {{\left( 2\sqrt{42}-13 \right)}^{47}}$.</p>

Simplify ${{\left( \sqrt{6}+\sqrt{7} \right)}^{94}}\times {{\left( 2\sqrt{42}-13 \right)}^{47}}$.

<p>Simplify $\sqrt[4]{7+4\sqrt{3}}+\sqrt[4]{7-4\sqrt{3}}$.</p>

Simplify $\sqrt[4]{7+4\sqrt{3}}+\sqrt[4]{7-4\sqrt{3}}$.

Math Question Bank