Grade 8 Mathematics Comprehensive Study Notes: Rational Numbers, Parallel Lines, Indices, and Triangle Geometry

Rational and Irrational Numbers

Review of Number Systems

• Natural Numbers: The set of counting numbers starting from $1$. Set representation: $\{1, 2, 3, 4, \dots\}$.
• Whole Numbers: The set of natural numbers including zero. Set representation: $\{0, 1, 2, 3, 4, \dots\}$.
• Integers: The set containing all whole numbers and negative natural numbers. Set representation: $\{\dots, -4, -3, -2, -1, 0, 1, 2, 3, 4, \dots\}$.
• Rational Numbers: Any number that can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are integers and $n \neq 0$. Examples of rational numbers include $\frac{-25}{3}, \frac{10}{-7}, -4, 0, 3, 8, \frac{32}{3}, \frac{67}{5}$.
• Infinite Density Property: Between any two rational numbers, there exist infinitely many rational numbers.

Showing Rational Numbers on a Number Line

• General Procedure: Draw a horizontal line with an origin point representing $0$. Positive rational numbers are placed to the right of $0$, and negative rational numbers are placed to the left of $0$.
• Subdividing Units: To plot a rational number with denominator $n$, each unit length on the number line must be divided into $n$ equal parts.
• Example Problem: Plot $\frac{7}{3}, 2, \frac{-2}{3}$ on a number line:
  • To represent $2$: Locate the mark $2$ units to the right of $0$
  • To represent $\frac{7}{3}$: Express $\frac{7}{3}$ as $2 + \frac{1}{3}$ or $7 \times \frac{1}{3}$. Divide each unit interval into $3$ equal parts. The point positioned at a distance of $1$ mark ($\frac{1}{3}\,\text{rd}$ unit) past $2$, or the $7\,\text{th}$ mark to the right of $0$, represents $\frac{7}{3}$.
  • To represent $\frac{-2}{3}$: Locate the $2\,\text{nd}$ mark to the left of $0$ (symmetrically opposite to $\frac{2}{3}$).

Practice Set 1.1 Solutions

• Question 1: Show the following numbers on a number line (separate line for each):
  • (1) $\frac{3}{2}, \frac{5}{2}, \frac{-3}{2}$ (Divide unit into $2$ equal parts)
  • (2) $\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}$ (Divide unit into $5$ equal parts)
  • (3) $\frac{-5}{8}, \frac{11}{8}$ (Divide unit into $8$ equal parts)
  • (4) $\frac{13}{10}, \frac{-17}{10}$ (Divide unit into $10$ equal parts)
• Question 2: Observe the given number line (points $A, B, C, D, O, E, F$) and answer:
  • (1) Which number is indicated by point $B$?
    Answer: Point $B$ is at the $10\,\text{th}$ division to the left of $0$ with units divided into $4$ parts, so it represents $\frac{-10}{4} = \frac{-5}{2}$.
  • (2) Which point indicates the number $1\frac{3}{4}$?
    Answer: $1\frac{3}{4} = \frac{7}{4}$. The $7\,\text{th}$ mark to the right of $0$ is point $C$
  • (3) State whether the statement, "the point $D$ denotes the number $\frac{5}{2}$" is true or false.
    Answer: True. Point $D$ is at the $10\,\text{th}$ division to the right, representing $\frac{10}{4} = \frac{5}{2}$.

Comparison of Rational Numbers

• Position Rule: For any pair of numbers on a number line, the number located further to the left is always smaller than the number to its right.
• Invariance Property: Multiplying both the numerator and denominator of a rational number by any non-zero integer $k$ does not alter its value:

$\frac{a}{b} = \frac{k \times a}{k \times b} \quad (k \neq 0)$

• Formal Comparison Rules for rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ where $b, d > 0$:

  • If $a \times d < b \times c$, then $\frac{a}{b} < \frac{c}{d}$.
  • If $a \times d = b \times c$, then $\frac{a}{b} = \frac{c}{d}$.
  • If $a \times d > b \times c$, then $\frac{a}{b} > \frac{c}{d}$.

• Negative Number Rules:

  • A negative rational number is always strictly less than any positive rational number.
  • If $a$ and $b$ are positive numbers such that $a < b$, then $-a > -b$.

• Worked Examples:

  • Compare $\frac{5}{4}$ and $\frac{2}{3}$:

$\frac{5}{4} = \frac{5 \times 3}{4 \times 3} = \frac{15}{12}$

$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

Since $\frac{15}{12} > \frac{8}{12}$, we have $\frac{5}{4} > \frac{2}{3}$.

*   **Compare $\frac{-7}{9}$ and $\frac{4}{5}$:**  

        A negative number is always less than a positive number, so $\frac{-7}{9} < \frac{4}{5}$. * Compare $\frac{-7}{3}$ and $\frac{-5}{2}$: * First compare positive values: $\frac{7}{3} = \frac{14}{6}$ and $\frac{5}{2} = \frac{15}{6}$. * Since $\frac{14}{6} < \frac{15}{6}$, we know $\frac{7}{3} < \frac{5}{2}$. * Applying negative signs reverses the inequality: $\frac{-7}{3} > \frac{-5}{2}$. * Compare $\frac{3}{5}$ and $\frac{6}{10}$:

$\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10} \implies \frac{3}{5} = \frac{6}{10}$

Practice Set 1.2 Solutions

• Compare the following numbers:
  • (1) $-7$ and $-2$
    Answer: $-7 < -2$
  • (2) $0$ and $\frac{-9}{5}$
    Answer: $0 > \frac{-9}{5}$
  • (3) $\frac{8}{7}$ and $0$
    Answer: $\frac{8}{7} > 0$
  • (4) $\frac{-5}{4}$ and $\frac{1}{4}$
    Answer: $\frac{-5}{4} < \frac{1}{4}$
  • (5) $\frac{40}{29}$ and $\frac{141}{29}$
    Answer: $\frac{40}{29} < \frac{141}{29}$
  • (6) $\frac{-17}{20}$ and $\frac{-13}{20}$
    Answer: $\frac{-17}{20} < \frac{-13}{20}$
  • (7) $\frac{15}{12}$ and $\frac{7}{16}$
    Answer: $\frac{15}{12} = \frac{60}{48}$ and $\frac{7}{16} = \frac{21}{48}$, so $\frac{15}{12} > \frac{7}{16}$.
  • (8) $\frac{-25}{8}$ and $\frac{-9}{4}$
    Answer: $\frac{-9}{4} = \frac{-18}{8}$. Since $-25 < -18$, $\frac{-25}{8} < \frac{-9}{4}$.
  • (9) $\frac{12}{15}$ and $\frac{3}{5}$
    Answer: $\frac{3}{5} = \frac{9}{15}$, so $\frac{12}{15} > \frac{3}{5}$.
  • (10) $\frac{-7}{11}$ and $\frac{-3}{4}$
    Answer: $\frac{-7}{11} = \frac{-28}{44}$ and $\frac{-3}{4} = \frac{-33}{44}$. Since $-28 > -33$, $\frac{-7}{11} > \frac{-3}{4}$.

Decimal Representation of Rational Numbers

• Terminating Decimal Form: Dividing the numerator of a rational number by its denominator terminates when the remainder becomes zero. Example: $\frac{7}{4} = 1.75$.
• Non-Terminating Recurring Decimal Form: When the division never yields a remainder of zero, a single digit or a block of digits repeats infinitely.
  • Single Repeating Digit: Marked with a dot above the repeating digit.
    • $\frac{7}{6} = 1.1666\dots = 1.1\dot{6}$
    • $\frac{5}{6} = 0.8333\dots = 0.8\dot{3}$
    • $\frac{-5}{3} = -1.666\dots = -1.\dot{6}$
  • Repeating Block of Digits: Marked with a horizontal bar over the block.
    • $\frac{22}{7} = 3.142857142857\dots = 3.\overline{142857}$
    • $\frac{23}{99} = 0.2323\dots = 0.\overline{23}$
  • Expressing Terminating Decimals as Recurring: Any terminating decimal can be written in recurring form by appending repeating zeros: $\frac{7}{4} = 1.75 = 1.75000\dots = 1.75\dot{0}$.

Practice Set 1.3 Solutions

• Write the following rational numbers in decimal form:
  • (1) $\frac{9}{37}$
    Answer: $0.\overline{243}$
  • (2) $\frac{18}{42}$
    Answer: $\frac{18}{42} = \frac{3}{7} = 0.\overline{428571}$
  • (3) $\frac{9}{14}$
    Answer: $0.6428571428571\dots = 0.6\overline{428571}$
  • (4) $\frac{-103}{5}$
    Answer: $-20.6$
  • (5) $\frac{-11}{13}$
    Answer: $-0.\overline{846153}$

Irrational Numbers

• Definition: Numbers that cannot be expressed in the rational form $\frac{m}{n}$ (where $m, n \in \mathbb{Z}, n \neq 0$) are called irrational numbers. The decimal form of an irrational number is non-terminating and non-recurring.
• Constructing $\sqrt{2}$ on a Number Line:
  1. Draw a number line. Let point $O$ represent $0$ and point $A$ represent $1$, so $OA = 1\,\text{unit}$.
  2. Construct line $l$ perpendicular to the number line at point $A$
  3. Mark point $P$ on line $l$ such that $AP = 1\,\text{unit}$.
  4. Join $OP$ to form right-angled $\triangle OAP$. By Pythagoras Theorem:

$[l(OP)]^2 = [l(OA)]^2 + [l(AP)]^2 = 1^2 + 1^2 = 1 + 1 = 2$

$l(OP) = \sqrt{2}\,\text{units}$

5. Draw an arc centered at $O$ with radius $OP$. The arc intersects the number line at point $Q$ on the right, representing $\sqrt{2}$.
6. Symmetrically mark point $R$ to the left of $0$ at distance $OQ$; point $R$ represents $-\sqrt{2}$.

• Constructing $\sqrt{3}$ on a Number Line:
  1. Point $Q$ represents $\sqrt{2}$, so $l(OQ) = \sqrt{2}\,\text{units}$.
  2. Construct a perpendicular line passing through point $Q$
  3. Mark point $R$ on this perpendicular line such that $l(QR) = 1\,\text{unit}$.
  4. Join $OR$ to form right-angled $\triangle ORQ$. By Pythagoras Theorem:

$[l(OR)]^2 = [l(OQ)]^2 + [l(QR)]^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$

$l(OR) = \sqrt{3}\,\text{units}$

5. Draw an arc centered at $O$ with radius $OR$. The arc intersects the number line at point $C$, representing $\sqrt{3}$.

• Real Numbers: The complete set of numbers corresponding to every point on the number line. It encompasses all rational numbers and all irrational numbers.
• Properties of $\pi$: The mathematical constant $\pi$ is an irrational number. Rational values such as $\frac{22}{7}$ or $3.14$ are close approximations used for calculation, but they are not equal to $\pi$
• Other Examples of Irrationals: Numbers such as $3\sqrt{2}$, $7 + \sqrt{2}$, $3 - \sqrt{2}$ are irrational. (Proof by contradiction: if $3\sqrt{2}$ were rational, then $\frac{3\sqrt{2}}{3} = \sqrt{2}$ would be rational, which is impossible).

Practice Set 1.4 Solutions

• Question 1: Complete the activity to show $\sqrt{3}$ on the number line:
  • Point $Q$ on the number line shows the number $\mathbf{\sqrt{2}}$.
  • Line perpendicular through $Q$ has point $R$ at unit distance ($l(QR) = 1$).
  • By Pythagoras theorem:

$[l(OR)]^2 = [l(OQ)]^2 + [l(QR)]^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$

$l(OR) = \mathbf{\sqrt{3}}$

*   Draw an arc with center $O$ and radius $OR$ to intersect at point $C$, showing $\sqrt{3}$.

• Question 2: Show $\sqrt{5}$ on the number line:
  • Procedure: Mark $OA = 2\,\text{units}$ on the number line. Draw perpendicular $AP = 1\,\text{unit}$ at $A$. Hypotenuose $OP = \sqrt{2^2 + 1^2} = \sqrt{5}$. Draw arc centered at $O$ with radius $OP$ to locate $\sqrt{5}$.
• Question 3: Show $\sqrt{7}$ on the number line:
  • Procedure: First locate $\sqrt{5}$, construct perpendicular of $1\,\text{unit}$ to obtain $\sqrt{6}$, then construct perpendicular of $1\,\text{unit}$ at $\sqrt{6}$ to obtain hypotenuse $\sqrt{(\sqrt{6})^2 + 1^2} = \sqrt{7}$. Arc intersects number line at $\sqrt{7}$.

Parallel Lines and Transversal

Fundamental Definitions

• Parallel Lines: Lines in the same plane that do not intersect each other, no matter how far they are extended. Symbolic notation: $\text{line } l \parallel \text{line } m$
• Transversal: A line that intersects two or more coplanar lines at distinct points. If line $l$ intersects line $m$ and line $n$ at distinct points $M$ and $N$ respectively, then line $l$ is a transversal of line $m$ and line $n$

Angles Made by a Transversal

• When a transversal intersects two lines, $8$ distinct angles are created ($4$ angles at each intersection point).

• Angle Classifications:
  1. Corresponding Angles: A pair of angles whose arms on the transversal point in the same direction, and whose other arms lie on the same side of the transversal. There are $4$ pairs:
     • $\angle AMP$ and $\angle MNR$
     • $\angle PMN$ and $\angle RNT$
     • $\angle AMQ$ and $\angle MNS$
     • $\angle QMN$ and $\angle SNT$
  2. Interior Angles: A pair of angles that lie on the same side of the transversal and in the interior region between the two intersected lines. There are $2$ pairs:
     • $\angle PMN$ and $\angle MNR$
     • $\angle QMN$ and $\angle MNS$
  3. Alternate Angles: Pairs of angles lying on opposite sides of the transversal whose arms on the transversal point in opposite directions.
     • Interior Alternate Angles (angles inside the lines): $2$ pairs
       • $\angle PMN$ and $\angle MNS$
       • $\angle QMN$ and $\angle MNR$
     • Exterior Alternate Angles (angles outside the lines): $2$ pairs
       • $\angle AMP$ and $\angle SNT$
       • $\angle AMQ$ and $\angle RNT$

Practice Set 2.1 Solutions

• Question 1: In the given figure, fill in the blanks:
  • Corresponding angles:
            (1) $\angle p$ and $\mathbf{\angle w}$
            (2) $\angle q$ and $\mathbf{\angle x}$
            (3) $\angle r$ and $\mathbf{\angle y}$
            (4) $\angle s$ and $\mathbf{\angle z}$
  • Interior alternate angles:
            (5) $\angle s$ and $\mathbf{\angle x}$
            (6) $\angle w$ and $\mathbf{\angle r}$
• Question 2: Observe the labeled angles ($a, b, c, d, e, f, g, h$) and list:
  • (1) Interior alternate angles: $\angle c$ and $\angle e$; $\angle b$ and $\angle h$
  • (2) Corresponding angles: $\angle a$ and $\angle e$; $\angle b$ and $\angle f$; $\angle c$ and $\angle g$; $\angle d$ and $\angle h$
  • (3) Interior angles: $\angle c$ and $\angle h$; $\angle b$ and $\angle e$

Properties of Angles Formed by Two Parallel Lines and a Transversal

• Property of Corresponding Angles: Each pair of corresponding angles formed by two parallel lines and their transversal consists of congruent angles.

$\angle AMP \cong \angle MNR, \quad \angle PMN \cong \angle RNB$

$\angle AMQ \cong \angle MNS, \quad \angle QMN \cong \angle SNB$

• Property of Alternate Angles: Each pair of alternate angles (interior or exterior) formed by two parallel lines and their transversal consists of congruent angles.
  • Interior Alternate: $\angle PMN \cong \angle MNS$, $\angle QMN \cong \angle MNR$
  • Exterior Alternate: $\angle AMP \cong \angle SNB$, $\angle AMQ \cong \angle RNB$
• Property of Interior Angles: Each pair of interior angles on the same side of the transversal formed by two parallel lines and their transversal consists of supplementary angles (their measures sum to $180^\circ$).

$m\angle PMN + m\angle MNR = 180^\circ$

$m\angle QMN + m\angle MNS = 180^\circ$

• Determining All Eight Angles: When two lines are parallel, if the measure of any single angle is known, all remaining $7$ angles can be uniquely calculated using properties of linear pairs, vertically opposite angles, corresponding angles, alternate angles, or interior angles.

Solved Examples

• Example 1: Given $\text{line } AB \parallel \text{line } PQ$, transversal $LM$, and $m\angle MNQ = 70^\circ$. Find $m\angle AON$.
  • Method 1: $m\angle ONP = m\angle MNQ = 70^\circ$ (Vertically opposite angles). Since $\angle AON$ and $\angle ONP$ are interior angles on the same side:

$m\angle AON + m\angle ONP = 180^\circ \implies m\angle AON = 180^\circ - 70^\circ = 110^\circ$

*   *Method 2:* $m\angle NOB = m\angle MNQ = 70^\circ$ (Corresponding angles). $\angle AON$ and $\angle NOB$ form a linear pair:

$m\angle AON + m\angle NOB = 180^\circ \implies m\angle AON + 70^\circ = 180^\circ \implies m\angle AON = 110^\circ$

• Example 2: Given $\text{line } m \parallel \text{line } n$, transversal $l$. If $m\angle b = (x + 15)^\circ$ and $m\angle e = (2x + 15)^\circ$, find $x$
  • Solution: $\angle f \cong \angle b$ (Corresponding angles), so $m\angle f = (x + 15)^\circ$. $\angle f$ and $\angle e$ form a linear pair:

$m\angle f + m\angle e = 180^\circ$

$(x + 15)^\circ + (2x + 15)^\circ = 180^\circ$

$3x + 30 = 180 \implies 3x = 150 \implies x = 50^\circ$

Practice Set 2.2 Solutions

• Question 1(1): In adjoining figure, $\text{line } m \parallel \text{line } n$ and transversal $p$. Interior angles are $3x$ and $x$. Find $x$
  • Solution: $3x + x = 180^\circ \implies 4x = 180^\circ \implies x = 45^\circ$. Correct option: (C) $45^\circ$.
• Question 1(2): In adjoining figure, $\text{line } a \parallel \text{line } b$ and transversal $l$. Angles $4x$ and $2x$ are given. Find $x$
  • Solution: $4x + 2x = 180^\circ \implies 6x = 180^\circ \implies x = 30^\circ$. Correct option: (D) $30^\circ$.
• Question 2: In figure, $\text{line } p \parallel \text{line } q$, transversals $t$ and $s$. Given angles $40^\circ$ and $70^\circ$. Find $x$ and $y$
  • Answers:
            For $x$: Angle linear pair to $x$ corresponds to $40^\circ$, so $x = 180^\circ - 40^\circ = 140^\circ$.
            For $y$: Angle vertically opposite to $70^\circ$ is interior to $y$, so $y = 180^\circ - 70^\circ = 110^\circ$.
• Question 3: In figure, $\text{line } p \parallel \text{line } q$ and $\text{line } l \parallel \text{line } m$. Given angle $80^\circ$. Find $a, b, c$
  • Answers: $a = 100^\circ$, $b = 80^\circ$, $c = 80^\circ$.
• Question 4: Line $a \parallel \text{line } b$, transversal $l$. Given angle $105^\circ$. Find $x, y, z$
  • Answers: $x = 105^\circ$ (Corresponding), $y = 105^\circ$ (Vertically opposite to $x$), $z = 75^\circ$ (Linear pair with $105^\circ$).
• Question 5: Line $p \parallel \text{line } l \parallel \text{line } q$. Angles $40^\circ$ and $30^\circ$ given. Find $x$
  • Answers: $x = 40^\circ + 30^\circ = 70^\circ$ (Sum of alternate interior angles).

Tests for Parallel Lines

• Corresponding Angles Test: If a transversal intersects two coplanar lines such that a pair of corresponding angles is congruent, then the lines are parallel.
• Alternate Angles Test: If a transversal intersects two coplanar lines such that a pair of alternate angles is congruent, then the lines are parallel.
• Interior Angles Test: If a transversal intersects two coplanar lines such that a pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

Geometric Constructions of Parallel Lines

• Construction I: To draw a line parallel to a given line through an external point using set-squares:
  • Method 1: Draw line $l$ and mark point $P$ outside line $l$. Place set-square $A$ along line $l$. Place set-square $B$ touching set-square $A$. Slide set-square $B$ until its edge passes through point $P$. Draw line $m$ along the edge. Line $m \parallel \text{line } l$

*   *Method 2:* Draw line $l$ and external point $P$. Draw perpendicular segment $PM \perp \text{line } l$. Choose another point $N$ on line $l$ and construct perpendicular segment $NQ \perp \text{line } l$ such that $l(NQ) = l(MP)$. Draw line $m$ passing through points $P$ and $Q$. Line $m \parallel \text{line } l$

• Construction II: To draw a line parallel to a given line at a specified distance (e.g., $2.5\,\text{cm}$):
  • Steps: Draw line $l$. Select two points $A$ and $B$ on line $l$. Construct perpendicular lines at points $A$ and $B$. Mark points $P$ and $Q$ on these perpendiculars at a distance of $2.5\,\text{cm}$ from $A$ and $B$ respectively. Draw line $PQ$. Line $PQ$ is parallel to line $l$ at a distance of $2.5\,\text{cm}$.

Practice Set 2.3 Problems

• 1. Draw a line $l$. Take a point $A$ outside the line. Through point $A$, draw a line parallel to line $l$
• 2. Draw a line $l$. Take a point $T$ outside the line. Through point $T$, draw a line parallel to line $l$
• 3. Draw a line $m$. Draw a line $n$ which is parallel to line $m$ at a distance of $4\,\text{cm}$ from it.

Indices and Cube Root

Laws of Indices

• Index Form Concept: In the product $2 \times 2 \times 2 \times 2 \times 2 = 2^5$, $2$ is the base, $5$ is the index (exponent), and $2^5$ is the index form of the number.

• Summary of Integral Exponent Laws (for non-zero bases $a, b$ and integers $m, n$):

  1. $a^m \times a^n = a^{m+n}$
  2. $a^m \div a^n = a^{m-n}$
  3. $(a \times b)^m = a^m \times b^m$
  4. $a^0 = 1$
  5. $a^{-m} = \frac{1}{a^m}$
  6. $(a^m)^n = a^{mn}$
  7. $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$
  8. $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^m$

• Fill-in-the-Blank Index Exercises:

  • (i) $3^5 \times 3^2 = 3^{5+2} = 3^7$
  • (ii) $3^7 \div 3^9 = 3^{7-9} = 3^{-2}$
  • (iii) $(3^4)^5 = 3^{4 \times 5} = 3^{20}$
  • (iv) $5^{-3} = \frac{1}{5^3}$
  • (v) $5^0 = 1$
  • (vi) $5^1 = 5$
  • (vii) $(5 \times 7)^2 = 5^2 \times 7^2$
  • (viii) $\left(\frac{5}{7}\right)^3 = \frac{5^3}{7^3}$
  • (ix) $\left(\frac{5}{7}\right)^{-3} = \left(\frac{7}{5}\right)^3$

Meaning of Numbers with Rational Indices

Index of the Form $\frac{1}{n}$ ($n\,\text{th}$ Root)

• Square Root: To represent the square of $a$, the index is written as $2$ ($a^2$). To represent the square root of $a$, the index is written as $\frac{1}{2}$ ($a^{\frac{1}{2}}$) or using the radical sign $\sqrt{a}$.
  
  • Example: Square root of $25$ is \sqrt{25} = 25^{\frac{1}{2}} = 5$.\n* **Cube Root:** To represent the cube of a$, the index is written as$3$($a^3$). To represent the cube root of$a$, the index is written as$\frac{1}{3}$($a^{\frac{1}{3}}$) or$\sqrt[3]{a}. \n * Example: 4^3 = 64 \implies \sqrt[3]{64} = 64^{\frac{1}{3}} = 4$.
• General $n\,\text{th}$ Root: The $n\,\text{th}$ root of a number $a$ is expressed as $a^{\frac{1}{n}}$ or $\sqrt[n]{a}$.
  • $3^5 = 243 \implies 243^{\frac{1}{5}} = 3$ ($5\,\text{th}$ root of $243$ is $3$).
  • $128^{\frac{1}{7}} = 7\,\text{th}$ root of $128$.
  • $900^{\frac{1}{12}} = 12\,\text{th}$ root of $900$.
  • If $10^{\frac{1}{5}} = x$, then $x^5 = 10$

Practice Set 3.1 Solutions

• Question 1: Express the following numbers in index form:
  • (1) Fifth root of $13$
    Answer: $13^{\frac{1}{5}}$
  • (2) Sixth root of $9$
    Answer: $9^{\frac{1}{6}}$
  • (3) Square root of $256$
    Answer: $256^{\frac{1}{2}}$
  • (4) Cube root of $17$
    Answer: $17^{\frac{1}{3}}$
  • (5) Eighth root of $100$
    Answer: $100^{\frac{1}{8}}$
  • (6) Seventh root of $30$
    Answer: $30^{\frac{1}{7}}$
• Question 2: Write in the form "$n\,\text{th}$ root of $a$" for each:
  • (1) $81^{\frac{1}{4}}$
    Answer: Fourth root of $81$
  • (2) $49^{\frac{1}{2}}$
    Answer: Square root of $49$
  • (3) $15^{\frac{1}{5}}$
    Answer: Fifth root of $15$
  • (4) $512^{\frac{1}{9}}$
    Answer: Ninth root of $512$
  • (5) $100^{\frac{1}{19}}$
    Answer: Nineteenth root of $100$
  • (6) $6^{\frac{1}{7}}$
    Answer: Seventh root of $6$

Index of the Form $\frac{m}{n}$

• Using the power law $(a^m)^n = a^{mn}$, a rational exponent $\frac{m}{n}$ has two equivalent interpretations:
  1. $a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}} \implies n\,\text{th}$ root of $m\,\text{th}$ power of $a$
  2. $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m \implies m\,\text{th}$ power of $n\,\text{th}$ root of $a$
• Concrete Demonstrations:
  • For $8^{\frac{2}{3}}$:
    • $(8^2)^{\frac{1}{3}} = (64)^{\frac{1}{3}} = 4 \implies$ cube root of square of $8$
    • $(8^{\frac{1}{3}})^2 = (2)^2 = 4 \implies$ square of cube root of $8$
  • For $27^{\frac{4}{5}}$:
    • $(27^4)^{\frac{1}{5}} \implies$ fifth root of fourth power of $27$
    • $(27^{\frac{1}{5}})^4 \implies$ fourth power of fifth root of $27$

Practice Set 3.2 Solutions

• Question 1: Complete the following table:

• Question 2: Write the following numbers in the form of rational indices:
  • (1) Square root of 5th power of $121$
    Answer: $121^{\frac{5}{2}}$
  • (2) Cube of 4th root of $324$
    Answer: $324^{\frac{3}{4}}$
  • (3) 5th root of square of $264$
    Answer: $264^{\frac{2}{5}}$
  • (4) Cube of cube root of $3$
    Answer: $3^{\frac{3}{3}}$

Cube and Cube Root

• Definition of Cube: If a number is multiplied by itself three times, the resulting product is the cube of that number ($a \times a \times a = a^3$).
• Square Root Properties (Recall):
  • Every positive number has two square roots: one positive ($\sqrt{a}$) and one negative ($-\sqrt{a}$). Example: $\sqrt{16} = 4$ and $-\sqrt{16} = -4$.
  • The square root of zero is zero ($\sqrt{0} = 0$).
• Calculations of Cubes:
  • $17^3 = 17 \times 17 \times 17 = 4913$
  • $(-6)^3 = (-6) \times (-6) \times (-6) = -216$
  • $\left(\frac{-2}{5}\right)^3 = \left(\frac{-2}{5}\right) \times \left(\frac{-2}{5}\right) \times \left(\frac{-2}{5}\right) = \frac{-8}{125}$
  • $(1.2)^3 = 1.2 \times 1.2 \times 1.2 = 1.728$
  • $(0.02)^3 = 0.02 \times 0.02 \times 0.02 = 0.000008$
• Rules Regarding Sign and Decimals in Cubes:
  • Sign Relation: The cube of a positive number is always positive. The cube of a negative number is always negative.
  • Decimal Relation: The number of decimal places in the cube of a number is exactly three times the number of decimal places in the original number.
• Finding Cube Root by Factorisation Method:
  • Example 1: Find $\sqrt[3]{216}$.
    • Prime factorise $216$: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 3)^3 = 6^3$
    • Therefore, $\sqrt[3]{216} = 6$ or $216^{\frac{1}{3}} = 6$.
  • Example 2: Find $\sqrt[3]{-1331}$.
    • Factorise $1331$: $1331 = 11 \times 11 \times 11 = 11^3$
    • $-1331 = (-11) \times (-11) \times (-11) = (-11)^3$
    • Therefore, $\sqrt[3]{-1331} = -11$
  • Example 3: Find $\sqrt[3]{1728}$.
    • Factorise: $1728 = 8 \times 216 = (2^3) \times (6^3) = (2 \times 6)^3 = 12^3$
    • Therefore, $\sqrt[3]{1728} = 12$ (Note: $\sqrt[3]{-1728} = -12$).
  • Example 4: Find $\sqrt[3]{0.125}$.
    • Solution: $\sqrt[3]{0.125} = \sqrt[3]{\frac{125}{1000}} = \frac{\sqrt[3]{125}}{\sqrt[3]{1000}} = \frac{5}{10} = 0.5$

Practice Set 3.3 Solutions

• Question 1: Find the cube roots of the following numbers:
  • (1) $8000$
    Answer: $\sqrt[3]{8000} = 20$
  • (2) $729$
    Answer: $\sqrt[3]{729} = 9$
  • (3) $343$
    Answer: $\sqrt[3]{343} = 7$
  • (4) $-512$
    Answer: $\sqrt[3]{-512} = -8$
  • (5) $-2744$
    Answer: $\sqrt[3]{-2744} = -14$
  • (6) $32768$
    Answer: $32768 = 32^3 \implies \sqrt[3]{32768} = 32$
• Question 2: Simplify:
  • (1) $\sqrt[3]{\frac{27}{125}}$
    Answer: $\frac{\sqrt[3]{27}}{\sqrt[3]{125}} = \frac{3}{5}$
  • (2) $\sqrt[3]{\frac{16}{54}}$
    Answer: Simplify fraction first: $\frac{16}{54} = \frac{8}{27}$. Then $\sqrt[3]{\frac{8}{27}} = \frac{2}{3}$.
• Question 3: If $\sqrt[3]{729} = 9$, then $\sqrt[3]{0.000729} = ?$
  • Answer: $\sqrt[3]{0.000729} = \sqrt[3]{\frac{729}{1000000}} = \frac{9}{100} = 0.09$

Altitudes and Medians of a Triangle

Altitude of a Triangle

• Definition: The perpendicular segment drawn from a vertex of a triangle on the side opposite to it is called an altitude of the triangle.
• Notation: In $\triangle ABC$, segment $AP \perp \text{side } BC$, so $\text{seg } AP$ is an altitude on base $BC$