\documentclass[10pt]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage[version=4]{mhchem} \usepackage{stmaryrd} \usepackage{graphicx} \usepackage[export]{adjustbox} \graphicspath{ {./images/} } examples exercises homework \section*{Solution} The base is a square with side length \(4 \mathrm{~cm}\), so: \[ \begin{aligned} A C^{2} & =4^{2}+4^{2} \\ & =32 \\ A C & =\sqrt{32} \\ & =4 \sqrt{2} \end{aligned} \] Hence \(E C=2 \sqrt{2} \quad\) (Leave in exact form to maintain accuracy.) Triangle \(V E C\) is right-angled, so: \[ \begin{aligned} x^{2} & =V E^{2}+E C^{2} \\ & =52+\left(2 \sqrt{2)^{2}}\right. \\ & =25+8 \\ & =33 \\ x & =\sqrt{33} \\ & \approx 5.74 \quad \text { (correct to 2 decimal places) } \end{aligned} \] The lenth of \(V C\) is \(5.74 \mathrm{~cm}\) correct to 2 decimal places. \section*{Applications of Pythagoras' theorem in three dimensions} \begin{itemize} \item Pythagoras' theorem can be used to find lengths in three-dimensional problems. \item Always draw a careful diagram identifying the appropriate right-angled triangle(s). \item To maintain accuracy, use exact values and only approximate using a calculator at the end of the problem if required. \end{itemize} \section*{Exercise 2G} 2 Find the length of the space diagonal of the rectangular prism whose length, width and height are: a \(12 \mathrm{~cm}, 9 \mathrm{~cm}, 8 \mathrm{~cm}\) b \(12 \mathrm{~cm}, 5 \mathrm{~cm}, 8 \mathrm{~cm}\) c \(10 \mathrm{~cm}, 4 \mathrm{~cm}, 7 \mathrm{~cm}\) d \(8 \mathrm{~cm}, 6 \mathrm{~cm}, 4 \mathrm{~cm}\) e \(7 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}\) f \(a \mathrm{~cm}, b \mathrm{~cm}, c \mathrm{~cm}\) 5 A builder needs to carry lengths of timber along a corridor in order to get them to where he is working. There is a right-angled bend in the corridor along the way. The corridor is \(3 \mathrm{~m}\) wide and the ceiling is \(2.6 \mathrm{~m}\) above the floor. What is the longest length of timber that the builder can take around the corner in the corridor? (Hint: Draw a diagram.) 7 For the rectangular prism shown opposite, \(E H=4 \mathrm{~cm}\) and \(H G=2 \mathrm{~cm}\). a Find the exact length of \(E G\), giving your answer as a surd in simplest form. b If \(A E=\frac{1}{2} E G\), find the exact value of \(A E\). \includegraphics[max width=\textwidth, center]{2023_11_09_1c5231c22d14cc915b41g-09(2)} c Find the length of: i \(B E\) ii \(B H\) d What type of triangle is triangle \(B E H\) ? e Show that if \(E H=2 a \mathrm{~cm}, H G=a \mathrm{~cm}\) and \(A E=\frac{1}{2} E G\), then the sides of the triangle \(B E H\) are in the ratio \(3: 4: 5\). \section*{Binomial denominators} Consider the expression \(\frac{1}{\sqrt{7}-\sqrt{5}}\). How can we write this as a quotient with a rational denominator? In the section on special products, we saw that: \[ \begin{aligned} (\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) & =(\sqrt{7})^{2}-(\sqrt{5})^{2} \\ & =7-5 \\ & =2 \text {, which is rational, } \end{aligned} \] so \[ \begin{aligned} \frac{1}{\sqrt{7}-\sqrt{5}} & =\frac{1}{\sqrt{7}-\sqrt{5}} \times \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} \\ & =\frac{\sqrt{7}+\sqrt{5}}{7-5} \\ & =\frac{\sqrt{7}+\sqrt{5}}{2} \end{aligned} \] Similarly, \[ \begin{aligned} (5 \sqrt{2}-4)(5 \sqrt{2}+4) & =(5 \sqrt{2})^{2}-4^{2} \\ & =34, \text { which again is rational, } \end{aligned} \] so \[ \begin{aligned} \frac{3}{5 \sqrt{2}-4} & =\frac{3}{5 \sqrt{2}-4} \times \frac{5 \sqrt{2}+4}{5 \sqrt{2}+4} \\ & =\frac{15 \sqrt{2}+12}{34} \end{aligned} \] Using the difference of two squares identity in this way is an important technique. \section*{Example 28} Simplify the following: \section*{Solution} a \(\frac{2 \sqrt{5}}{2 \sqrt{5}-2}=\frac{2 \sqrt{5}}{2 \sqrt{5}-2} \times \frac{2 \sqrt{5}+2}{2 \sqrt{5}+2}\) b \(\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{3 \sqrt{2}+2 \sqrt{3}} \times \frac{3 \sqrt{2}-2 \sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}\) \(=\frac{20+4 \sqrt{5}}{20-4}\) \(=\frac{(\sqrt{3}+\sqrt{2})(3 \sqrt{2}-2 \sqrt{3})}{18-12}\) \(=\frac{20+4 \sqrt{5}}{16}\) \(=\frac{(3 \sqrt{6}-6+6-2 \sqrt{6})}{6}\) \(=\frac{4(5+\sqrt{5})}{16}\) \(=\frac{\sqrt{6}}{6}\) \(=\frac{5+\sqrt{5}}{4}\) \section*{Binomial denominators} To rationalise a denominator which has two terms, we use the difference of two squares identity: \begin{itemize} \item In an expression such as \(\frac{3}{5+\sqrt{3}}\), multiply top and bottom by \(5-\sqrt{3}\). \item In an expression such as \(\frac{\sqrt{2}}{7-3 \sqrt{2}}\), multiply top and bottom by \(7+3 \sqrt{2}\). \end{itemize} The surd \(7-\sqrt{3}\) is called the conjugate of \(7+\sqrt{3}\) and \(7+\sqrt{3}\) is the conjugate of \(7-\sqrt{3}\). \section*{Exercise \(2 \mathrm{H}\)} 1 Simplify: a \((\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\) b \((2 \sqrt{5}-\sqrt{3})(2 \sqrt{5}+\sqrt{3})\) c \((7 \sqrt{2}+4 \sqrt{3})(7 \sqrt{2}-4 \sqrt{3})\) d \((4-\sqrt{3})(4+\sqrt{3})\) 2 Rationalise the denominator in each expression. a \(\frac{1}{\sqrt{6}+1}\) d \(\frac{2}{3-\sqrt{5}}\) g \(\frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\) h \(\frac{\sqrt{2}}{\sqrt{2}-\sqrt{3}}\) i \(\frac{\sqrt{2}}{2 \sqrt{5}+\sqrt{2}}\) j \(\frac{\sqrt{5}}{2 \sqrt{5}-\sqrt{3}}\) k \(\frac{\sqrt{5}}{3 \sqrt{2}+4 \sqrt{3}}\) l \(\frac{\sqrt{3}}{2 \sqrt{3}-3 \sqrt{6}}\) m \(\frac{2 \sqrt{6}}{4 \sqrt{2}-3 \sqrt{7}}\) n \(\frac{3 \sqrt{2}}{6 \sqrt{3}+11 \sqrt{5}}\) o \(\frac{2 \sqrt{3}}{3 \sqrt{2}+10 \sqrt{3}}\) 3 Rationalise the denominators in these expressions and use the decimal approximations \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\) to evaluate them correct to 2 decimal places. 5 Simplify: 6 Simplify: a \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\) b \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\) \section*{Irrational numbers and surds} We first recall some facts about fractions and decimals. \section*{Fractions and decimals} We know that some fractions can be written as decimals that terminate. For example, \[ \frac{1}{4}=0.25, \frac{3}{16}=0.1875 \] Some fractions have decimal representations that do not terminate. For example, \(\frac{1}{3}=0.33333 \ldots\) which we write as \(0 . \dot{3}\). The dot above the 3 indicates that the digit 3 is repeated forever. Some fractions have decimal representations that eventually repeat. For example, \(\frac{1}{6}=0.1666 \ldots\) which is written as \(0.1 \dot{6}\). Other fractions have decimal representations with more than one repeating digit. For example, \(\frac{1}{11}=0.090909 \ldots\) which we write as \(0 . \dot{0} \dot{9}\) with a dot above both of the repeating digits. Another example is \(0.12 \dot{3} 45 \dot{6}=0.1234563456 \ldots\) \section*{Converting decimals to fractions} It is easy to write terminating decimals as fractions by using a denominator which is a power of 10 . For example, \[ 0.14=\frac{14}{100}=\frac{7}{50} \] Decimals that have a repeated sequence of digits can also be written as fractions. For example, \[ 0.12 \dot{3}=\frac{41}{333} \text { and } 0.67 \dot{1} \dot{2}=\frac{443}{660} \] A method for doing this is shown in the next two examples. \section*{Example 29} \section*{Example 30} \section*{Rational numbers} A rational number is a number that can be written as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). All integers are rational numbers. For example, \[ -3=\frac{-3}{1} \] \section*{Example 31} Explain why each of these numbers is rational. a \(5 \frac{3}{7}\) b 2.1237 c \(0 . \dot{2} \dot{4}\) \section*{Solution} We will express each number in the form \(\frac{p}{q}\). a \(5 \frac{3}{7}=\frac{38}{7}\) b \(2.1237=2 \frac{1237}{10000}\) \[ =\frac{21237}{10000} \] (continued over page) \[ \begin{aligned} S & =0 . \dot{2} \dot{4} \\ & =0.24242 \ldots \\ 100 S & =24.24242 \ldots \quad \text { (Multipy by 100.) } \\ 100 S & =24+S \\ 99 S & =24 \\ \text { Hence } S & =\frac{24}{99} \\ & =\frac{8}{33} \\ 0 . \dot{2} \dot{4} & =\frac{8}{33} \\ \text { Thus } \quad & \end{aligned} \] Each number has been expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers, so each number is rational. \section*{Irrational numbers} Mathematicians up to about \(600 \mathrm{BCE}\) thought that all numbers were rational. However, when we apply Pythagoras' theorem, we encounter numbers such as \(\sqrt{2}\) that are not rational. The number \(\sqrt{2}\) is an example of an irrational number. An irrational number is one that is not rational. Hence an irrational number cannot be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Nor can it be written as a terminating or repeating decimal. In \(300 \mathrm{BCE}\), Euclid proved that making the assumption that \(\sqrt{2}\) is rational leads to a contradiction. Hence \(\sqrt{2}\) was shown to be irrational. The proof is outlined here. Assume \(\sqrt{2}=\frac{p}{q}\) where \(p\) and \(q\) are integers with highest common factor 1 . Square both sides of this equation to obtain \(2=\frac{p^{2}}{q^{2}}\). We can write \(p^{2}=2 q^{2}\). Hence \(p^{2}\) is even and thus \(p\) is even. We can now write, \(2=\frac{4 k^{2}}{q^{2}}\) For some whole number \(k\). From this, show \(q\) is also even which is a contradiction of our assumption that the highest common factor is 1 . The decimal expansion of \(\sqrt{2}\) goes on forever but does not repeat. The value of \(\sqrt{2}\) can be approximated using a calculator. The same is true of other irrational numbers such as \(\sqrt{3}, \sqrt{14}\) and \(\sqrt{91}\). Try finding these numbers on your calculator and see what you get. Other examples of irrational numbers include: \[ \sqrt{3}, \sqrt[3]{2}, \pi, \pi^{3} \text { and } \sqrt{\pi} \] There are infinitely many rational numbers because every whole number is rational. We can also easily write down infinitely many other fractions. For example, \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\) There are infinitely many irrational numbers too. For example, \(\sqrt{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{4}, \ldots\) Every number, whether rational or irrational, is represented by a point on the number line. Conversely, we can think of each point on the number line as a number. \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-16(1)} \end{center} \section*{Example 32} Arrange these irrational numbers in order of size on the number line. \(\sqrt{8}\) \(\sqrt{2}\) \(\sqrt[3]{60}\) \(\sqrt[4]{30}\) \section*{Solution} Find on approximation of each number to 2 decimal places using a calculator. \[ \begin{array}{ll} \sqrt{8} \approx 2.83 & \sqrt{2} \approx 1.14 \\ \sqrt[3]{60} \approx 3.91 & \sqrt[4]{30} \approx 2.34 \end{array} \] \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-16} \end{center} \section*{Surds} In Year 8, we looked at squares, square roots, cubes and cube roots. For example: \[ \begin{aligned} & 5^{2}=25 \quad \text { and } \quad \sqrt{25}=5 \\ & 5^{3}=125 \quad \text { and } \quad \sqrt[3]{125}=5 \end{aligned} \] The use of this notation can be extended. For example: \[ 5^{4}=625 \quad \text { and } \quad \sqrt[4]{625}=5 \] This is read as 'The fourth power of 5 is 625 and the fourth root of 625 is 5 '. We also have fifth roots, sixth roots and so on. In general, we can take the \(n\)th root of any positive number \(a\). The \(n\)th root of \(a\) is the positive number whose \(n\)th power is \(a\). The statement \(\sqrt[n]{a}=\mathrm{b}\) is equivalent to the statement \(b^{n}=a\). Your calculator will give you approximations to the \(n\)th root of \(a\). Note: For \(n\), a positive integer, \(0^{n}=0\) and \(\sqrt[n]{0}=0\). An irrational number which can be expressed as \(\sqrt[n]{a}\), where \(a\) is a positive whole number, is called a surd. For example, \(\sqrt{2}, \sqrt{7}\) and \(\sqrt[3]{5}\) are all examples of surds, while \(\sqrt{4}\) and \(\sqrt{9}\) are not surds since they are whole numbers. The number \(\pi\), although it is irrational, is not a surd. This is also difficult to prove. \section*{Example 33} Use Pythagoras' theorem to construct a line of length \(\sqrt{20}\). \section*{Solution} Find two perfect squares that sum to 20 . \(4^{2}+2^{2}=20\) Draw perpendicular line segments from a common point of lengths 2 units and 4 units. Connect their endpoints with a third line segment. \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-17} \end{center} \section*{Irrational numbers and surds} \begin{itemize} \item Every fraction can be written as a terminating, or eventually repeating, decimal. \item Every terminating, or eventually repeating, decimal can be written as a fraction. \item A rational number is a number which can be expressed as \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\). That is, a rational number is an integer or a fraction. \item There are numbers such as \(\pi\) and \(\sqrt{2}\) that are irrational (not rational). \item Each rational and each irrational number represents a point on the number line. \item Every point on the number line represents a rational or an irrational number. \item An irrational number which can be expressed as \(\sqrt[n]{a}\), where \(n\) and \(a\) are positive whole numbers, is called a surd. \(\pi\) is not a surd. \end{itemize} \section*{Exercise 2I} 1 Write each repeating decimal as a fraction. 2 Write each repeating decimal as a fraction. e \(0 . \dot{6} 1 \dot{3}\) 3 Show that each number is rational by writing it as a fraction. a \(3 \frac{2}{3}\) b 5.15 c \(0 . \dot{4}\) d \(0 . \dot{1} \dot{5}\) e \(5 \frac{1}{7}\) f 1.3 5 By approximating correct to 2 decimal places, place these real numbers on the same number line. a \(\sqrt{7}\) b 2.7 c \(\sqrt[3]{18}\) d \(2 \pi\) e \(\sqrt{2}\) 6 Which of these numbers are surds? a 3 b \(\sqrt{5}\) c 4 d \(\sqrt{9}\) e \(\sqrt{7}\) f \(\sqrt{16}\) g \(\sqrt{10}\) h \(\sqrt{1}\) i \(\sqrt{3}\) j \(\sqrt{15}\) k 5 l \(\sqrt{25}\) 7 a Use the fact that \(1^{2}+2^{2}=5\) to construct a length \(\sqrt{5}\). b Use the fact that \(4^{2}+5^{2}=41\) to construct a length \(\sqrt{41}\). 8 How would you construct an interval of length: a \(\sqrt{73}\) ? b \(\sqrt{12}\) ? c \(\sqrt{21}\) ? \section*{Review exercise} 1 For each right-angled triangle, find the value of the pronumeral. a \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(3)} \end{center} b \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(1)} \end{center} 2 For each right-angled triangle, find the value of the pronumeral. a \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18} \end{center} b \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-18(2)} \end{center} 3 A support bracket is to be placed under a shelf, as shown in the diagram. If \(A B=A C=20 \mathrm{~cm}\), find, correct to the nearest millimetre, the length of \(B C\). \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-19(1)} \end{center} 4 A road runs in an east-west direction joining towns \(A\) and \(B\), which are \(40 \mathrm{~km}\) apart. \(A\) third town, \(C\), is situated \(20 \mathrm{~km}\) due north of \(B\). A straight road is built from \(C\), to the road between \(A\) and \(B\) and meets it at \(D\), which is equidistant from \(A\) and \(C\). Find the length of road \(C D\). 5 Circles of radius \(6 \mathrm{~cm}\) and \(3 \mathrm{~cm}\) are placed in a square, as shown in the diagram opposite. Find, correct to 2 decimal places: a \(E B\) c the length of the diagonal \(B D\) d the side length of the square b \(F D\) 6 Simplify: \includegraphics[max width=\textwidth, center]{2023_11_09_1c5231c22d14cc915b41g-19} a \(3 \sqrt{2}+2 \sqrt{2}\) b \(\sqrt{32}-\sqrt{18}\) 7 Simplify: a \(2 \sqrt{3} \times 5 \sqrt{6}\) b \(3 \sqrt{5} \times 2 \sqrt{10}\) 8 Simplify: a \(2 \sqrt{3}(3+\sqrt{3})\) b \(5 \sqrt{2}(3 \sqrt{2}-2)\) 9 Expand and simplify: a \((2 \sqrt{2}+1)(3 \sqrt{2}-2)\) b \((5 \sqrt{3}-2)(2 \sqrt{3}-1)\) 10 Simplify: a \(\sqrt{80}\) b \(\sqrt{108}\) c \(\sqrt{125}\) d \(\sqrt{72}\) e \(\sqrt{2048}\) f \(\sqrt{448}\) g \(\sqrt{800}\) h \(\sqrt{112}\) 11 Simplify: a \(\sqrt{45}+2 \sqrt{5}-\sqrt{80}\) b \(\sqrt{28}+2 \sqrt{63}-5 \sqrt{7}\) c \(\sqrt{44}+\sqrt{275}-4 \sqrt{11}\) d \(\sqrt{162}-\sqrt{200}+\sqrt{288}\) 12 Express with a rational denominator. a \(\frac{3}{\sqrt{11}}\) b \(\frac{1}{5 \sqrt{15}}\) c \(\frac{4}{7 \sqrt{7}}\) d \(\frac{3}{\sqrt{17}}\) e \(\frac{3}{\sqrt{3}}\) f \(\frac{3}{5 \sqrt{15}}\) g \(\frac{14}{\sqrt{7}}\) h \(\frac{11}{\sqrt{3}}\) 13 Express with a rational denominator. a \(\frac{3}{3-\sqrt{3}}\) b \(\frac{22}{2+3 \sqrt{5}}\) c \(\frac{24}{1-\sqrt{7}}\) d \(\frac{24}{1+\sqrt{17}}\) e \(\frac{3}{2-\sqrt{3}}\) f \(\frac{30}{1-\sqrt{11}}\) g \(\frac{15}{2-\sqrt{7}}\) h \(\frac{10}{2-\sqrt{3}}\) 14 a Expand and simplify \((2 \sqrt{5}-\sqrt{3})^{2}\). b Simplify \(\frac{2}{\sqrt{3}-2}+\frac{2}{\sqrt{3}}\), expressing your answer with a rational denominator. 15 The diagram opposite shows part of a skate board ramp. (It is a prism whose cross-section is a rightangled triangle.) Use the information in the diagram to find: a \(B C\) b \(A C\) \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20} \end{center} 16 Find the length of the space diagonal of a cube with side length \(5 \mathrm{~cm}\). 17 A motorist departs from town \(B\), which is \(8 \mathrm{~km}\) due south from another town, \(A\), and drives due east towards town \(C\), which is \(20 \mathrm{~km}\) from \(B\). After driving a distance of \(x \mathrm{~km}\), he notices that he is the same distance away from both towns \(A\) and \(C\). \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20(2)} \end{center} a Express the motorist's distance from \(A\) in terms of \(x\) (that is, \(A D\) ). b Express the motorist's distance from \(C\) in terms of \(x\). c Find the distance he has driven from \(B\). 18 The diagram opposite shows the logo of a particular company. The large circle has centre \(O\) and radius \(12 \mathrm{~cm}\) and \(A B\) is a diameter. \(D\) is the centre of the middle-sized circle with diameter \(O B\). Finally, \(C\) is the centre of the smallest circle. a What is the radius of the circle with centre \(D\) ? b If the radius of the circle with centre \(C\) is \(r \mathrm{~cm}\), express these in terms of \(r\). i \(O C\) ii \(D C\) \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-20(1)} \end{center} c Find the value of \(r\). 19 In the diagram opposite, \(V A B C D\) is a square-based pyramid with \(A B=B C=C D=D A=10\) and \(V A=V B=V C=V D=10\). a What type of triangle is triangle \(V B C\) ? b If \(M\) is the midpoint of \(C B\), find the exact values of: i \(V M\) ii \(V E\), the height of the pyramid \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-21(1)} \end{center} 20 Write each repeating decimal as a fraction. a \(0 . \dot{8}\) b \(0.1 \dot{5}\) c \(0.7 \dot{2}\) d \(0.0 \dot{3}\) e \(0 . \dot{8} \dot{1}\) f \(0 . \dot{9} \dot{6}\) g \(0.10 \mathrm{i}\) h 0.0015 21 Write \(\frac{67}{29}\) as a continued fraction. (See Exercise 2I, Question 9.) 22 Write \(0 . \dot{4} 79281\) as a fraction. 23 How would you construct intervals of the following lengths? Discuss with your teacher. a \(\sqrt{3}\) b \(\frac{1}{\sqrt{2}}\) c \(\sqrt{8}\) d \(\frac{1}{\sqrt{8}}\) 24 Find the length of the long diagonal of the rectangular prism whose length, width and height are: a \(3+\sqrt{2}, 3-\sqrt{2}, 3 \sqrt{3}\) b \(5+\sqrt{3}, 5-\sqrt{3}, 2 \sqrt{2}\) 25 For \(x=\sqrt{2}=1\) and \(y=\sqrt{5}-2\), find in simplest form. a \(\frac{1}{x}\) b \(\frac{1}{x}+\frac{1}{y}\) c \(\frac{1}{x^{2}}+\frac{1}{y^{2}}\) d \(\frac{1}{x}-\frac{1}{y}\) \section*{Challenge exercise} 1 Prove that \(\sqrt[3]{2}\) is irrational. 2 In the triangle below, \(4^{2}+7^{2}=65>8^{2}\). Is the angle \(\theta\) greater or less than \(90^{\circ}\) ? Explain your answer. \begin{center} \includegraphics[max width=\textwidth]{2023_11_09_1c5231c22d14cc915b41g-21} \end{center} \end{document}