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A free printable book

Maths for classes 9 and 10

7 lessons, each with worked examples, a try-at-home activity and a practice sheet. The practice sheets are together after the lessons, and the answer keys are at the very back.

  • Maths
  • Classes 9-10
  • 7 lessons
  • 71 practice questions
  • Free to print

Contents

Free learning for kids, Gurukul Technicals Institute. Printed from engineergurukul.com/resources/kids/books/maths-9-10/

Contents

What is in this book

  1. 1Pythagoras theoremIn a right-angled triangle, the square on the longest side equals the sum of the squares on the other two sides.4 min
  2. 2Mean, median and modeSummarise a set of numbers with the mean, median and mode, and learn when each one gives the best picture of the data.4 min
  3. 3Rational and irrational numbersTell rational numbers from irrational ones, turn recurring decimals into fractions and find where √2 sits on the number line.5 min
  4. 4Polynomials and factorisationLearn degree and zeroes of a polynomial, use the identities for squares, and factorise quadratics by splitting the middle term.5 min
  5. 5Linear equations in two variablesSee why ax + by = c draws a straight line, and solve a pair of such equations by substitution, by elimination and by graphing.6 min
  6. 6Quadratic equationsSolve ax² + bx + c = 0 by factorising and by the quadratic formula, and use the discriminant to predict how many real roots there are.5 min
  7. 7Arithmetic progressionsRecognise a sequence with a constant difference, find any term with a + (n − 1)d and add up n terms with the sum formula.5 min

Every lesson here can also be read on its own page, with links to related lessons. All printable books.

Lesson 1 of 7

Pythagoras theorem

In a right-angled triangle, the square on the longest side equals the sum of the squares on the other two sides.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

Builders, ladders and map makers all use one famous rule about right-angled triangles. It lets you find a missing side without measuring it.

A right-angled triangle has one angle of 90°. The side opposite the right angle is the longest side and is called the hypotenuse. The other two sides are called the legs.

The Pythagoras theorem says: in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. If the legs are a and b and the hypotenuse is c, then c² = a² + b².

For the 3, 4, 5 triangle: 3² + 4² = 9 + 16 = 25 = 5². Picture a square drawn on each side. The two smaller squares together have the same area as the big square.

The rule also works backwards. If the sides of a triangle satisfy c² = a² + b², then the triangle has a right angle. Sets of whole numbers like 3, 4, 5 or 5, 12, 13 that fit the rule are called Pythagorean triples.

Squares on the sides of a 3-4-5 triangleA right-angled triangle with legs 3 and 4 and hypotenuse 5, with a square drawn on each side. The square on the leg of 3 has area 9, the square on the leg of 4 has area 16, and the square on the hypotenuse has area 25. 9 plus 16 equals 25. 3² = 9 4² = 16 5² = 25 3² + 4² = 5² 9 + 16 = 25 c² = a² + b² c is the longest side
The areas of the two small squares, 9 and 16, add up to the area of the big square, 25.

Worked examples

See how it works

  1. Find the hypotenuse

    The legs are 6 cm and 8 cm. c² = 6² + 8² = 36 + 64 = 100. So c = √100 = 10 cm.

  2. Find a leg

    The hypotenuse is 13 cm and one leg is 5 cm. The other leg b satisfies b² = 13² − 5² = 169 − 25 = 144. So b = √144 = 12 cm.

  3. A ladder against a wall

    A 10 m ladder has its foot 6 m from the wall. The height it reaches is h, where h² = 10² − 6² = 100 − 36 = 64. So h = 8 m.

  4. Is it a right triangle?

    The sides are 9, 12 and 15. The longest is 15. 9² + 12² = 81 + 144 = 225 and 15² = 225. They are equal, so yes, it is a right-angled triangle.

Try it at home

The knotted-string triangle

You will need

  • A piece of string about 60 cm long
  • A ruler
  • A pen
  • A book or notebook with a square corner

What to do

  1. Mark a point on the string every 5 cm so that there are 12 equal parts.
  2. Lay the string on a table in a triangle with sides of 3 parts, 4 parts and 5 parts. Hold the corners in place with your fingers.
  3. Place the corner of the book in the corner between the 3 side and the 4 side.
  4. Check: it fits exactly, so the angle is 90°. This is because 3² + 4² = 5².
  5. Try sides of 3, 3 and 6 parts or 2, 5 and 5 parts. Do they make a square corner?

Ask a grown-up: A grown-up can hold the corners while you measure. No sharp tools are needed.

Fun fact

Ancient Indian texts called the Sulba Sutras, written more than 2,000 years ago, already describe this rule for building altars.

Lesson 2 of 7

Mean, median and mode

Summarise a set of numbers with the mean, median and mode, and learn when each one gives the best picture of the data.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

A cricketer scores different runs in different matches. How do we describe all of them with one number? Mean, median and mode are three ways to do this.

The mean is the average. Add up all the values and divide by how many there are. Mean = sum of values ÷ number of values.

The median is the middle value when the numbers are arranged in order. If there is an odd number of values, take the middle one. If there is an even number, take the mean of the two middle values.

The mode is the value that appears most often. A set can have more than one mode, or none if every value appears once.

The range is the biggest value minus the smallest value. It tells how spread out the data is.

The mean uses every number, so one very big or very small value can pull it. The median is not affected as much. The mode is useful for the most common size or choice, such as a shoe size.

Runs scored in 5 matchesA bar chart of the runs scored in 5 matches: 10, 40, 10, 30 and 35. The two bars of 10 are shaded gold to show the mode. A dashed line across the chart at 25 shows the mean.Runs scored in 5 matches 10 40 10 30 35 Match 1 Match 2 Match 3 Match 4 Match 5 mean = 25 Gold bars: the mode (10 runs twice)
Mean = 25 (dashed line), median = 30 (middle value when ordered), mode = 10 (the gold bars).

Worked examples

See how it works

  1. Mean

    Find the mean of 5, 8, 6, 9, 7. Sum = 5 + 8 + 6 + 9 + 7 = 35. There are 5 values. Mean = 35 ÷ 5 = 7.

  2. Median with an odd number of values

    Find the median of 12, 5, 9, 3, 7. In order: 3, 5, 7, 9, 12. The middle value is 7. So the median is 7.

  3. Median with an even number of values

    Find the median of 4, 8, 6, 10, 2, 12. In order: 2, 4, 6, 8, 10, 12. The middle two values are 6 and 8. Median = (6 + 8) ÷ 2 = 7.

  4. Mode, and all three together

    Shoe sizes of 7 children are 6, 7, 7, 8, 6, 7, 9. The size 7 appears 3 times, more than any other, so the mode is 7. For the runs 10, 40, 10, 30, 35: mean = 125 ÷ 5 = 25, in order 10, 10, 30, 35, 40 the median is 30, and the mode is 10.

Try it at home

Name-length survey

You will need

  • Paper and a pencil
  • Family members or friends who agree to take part

What to do

  1. Ask 7 people for their first name and count the letters in each. Write the numbers down.
  2. Find the mean: add the numbers and divide by 7.
  3. Arrange the numbers in order and find the median (the 4th number).
  4. Find the mode, the number that appears most often.
  5. Find the range. Draw a bar chart of your numbers.

Ask a grown-up: An adult can help choose people to ask. Only first names are needed. Do not ask for private information.

Lesson 3 of 7

Rational and irrational numbers

Tell rational numbers from irrational ones, turn recurring decimals into fractions and find where √2 sits on the number line.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

Every number you have met so far is a real number, and real numbers come in two families. Rational numbers can be written as fractions. Irrational numbers, like √2 and π, cannot, however hard you try.

A rational number is any number that can be written as p/q, where p and q are integers and q is not zero. So 3 (which is 3/1), −5/8, 0.75 (which is 3/4) and 0.333... (which is 1/3) are all rational.

When you write a rational number as a decimal, one of two things happens: the decimal stops (0.75, 2.5) or it repeats a block forever (0.333..., 0.272727...). A repeating decimal is written with a bar over the repeating block.

An irrational number is a real number that is not rational. Its decimal goes on forever without ever repeating a block. √2 = 1.41421356..., √3, √5 and π = 3.14159265... are irrational. Note that √4 = 2 and √9 = 3 are rational, because they are whole numbers.

Every real number has a place on the number line. √2 is the length of the diagonal of a square of side 1, so it can be drawn exactly with a compass, even though its decimal never ends.

Between any two rational numbers there are infinitely many more. One quick way to find a number between a and b is to take their average (a + b)/2.

Placing numbers on the number lineA number line from minus 2 to 3. A square of side 1 stands on the segment from 0 to 1 and its diagonal is swung down with a dotted arc to mark the point root 2, about 1.414. The points minus 3 over 2 and one half are also marked.Where does √2 sit on the number line? -2 -1 0 1 2 3 side 1 1 √2 √2 ≈ 1.414 (irrational) 1/2 (rational) -3/2 = -1.5 (rational)
Rational numbers like −3/2 and 1/2 and the irrational number √2 all have exact places on the number line.

Worked examples

See how it works

  1. Is 0.75 rational?

    Yes. 0.75 = 75/100 = 3/4. Any decimal that stops can be written over a power of ten, so it is rational.

  2. Turn 0.333... into a fraction

    Let x = 0.333... Multiply by 10: 10x = 3.333... Subtract the first line from the second: 10x − x = 3.333... − 0.333..., so 9x = 3 and x = 3/9 = 1/3.

  3. Turn 0.272727... into a fraction

    The repeating block has two digits, so multiply by 100. Let x = 0.272727..., then 100x = 27.272727... Subtract: 99x = 27, so x = 27/99 = 3/11.

  4. A rational number between 1/3 and 1/2

    Average them: (1/3 + 1/2)/2 = (2/6 + 3/6)/2 = (5/6)/2 = 5/12. Check: 1/3 = 4/12 and 1/2 = 6/12, and 5/12 sits between them.

Try it at home

Draw √2 with a square

You will need

  • A sheet of paper
  • A pencil and ruler
  • A length of string or a strip of paper
  • A scale marked in centimetres

What to do

  1. Draw a number line along the bottom of the page. Mark 0 and then 1, 2 and 3, using 4 cm for each unit.
  2. On the segment from 0 to 1, draw a square of side 4 cm standing on the line.
  3. Draw the diagonal from the point 0 to the opposite top corner of the square. This diagonal has length √2 units.
  4. Hold one end of the string at 0, stretch it along the diagonal and pinch it at the corner. Swing the pinched end down to the number line and mark the spot.
  5. Measure the mark with the scale: it should be about 5.66 cm from 0, which is 4 × 1.414. That point is √2.

Lesson 4 of 7

Polynomials and factorisation

Learn degree and zeroes of a polynomial, use the identities for squares, and factorise quadratics by splitting the middle term.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

A polynomial is an expression made from a variable, numbers, plus signs and whole-number powers, like 3x² + 5x − 2. Factorising means writing it as a product of simpler pieces, which makes solving equations easy.

The degree of a polynomial is its highest power. 4x³ − 2x + 7 has degree 3. A degree-1 polynomial is linear, degree 2 is quadratic and degree 3 is cubic. The numbers in front of the powers are the coefficients.

A zero of a polynomial p(x) is a value of x that makes p(x) equal to 0. For p(x) = x² − 3x + 2, p(1) = 1 − 3 + 2 = 0, so 1 is a zero. The factor theorem says: if p(a) = 0 then (x − a) is a factor of p(x).

Three identities you should know by heart: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and (a + b)(a − b) = a² − b². They work both ways: to expand and to factorise.

To factorise x² + bx + c, find two numbers whose product is c and whose sum is b, then split the middle term and group. For ax² + bx + c the two numbers must multiply to a × c.

Area model of (a + b) squaredA large square with side a plus b is divided into four regions: a square of area a squared at the top left, two rectangles of area ab, and a small square of area b squared at the bottom right.(a + b)² = a² + 2ab + b² a² ab ab b² a b a b Whole square: side a + b, area (a + b)² Four pieces: a² + ab + ab + b² = a² + 2ab + b² The two shaded rectangles are the 2ab.
The area of the big square, (a + b)², equals the four pieces added together: a² + ab + ab + b².

Worked examples

See how it works

  1. Split the middle term: x² + 5x + 6

    We need two numbers with product 6 and sum 5: they are 2 and 3. Write x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3). Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6.

  2. Factorise 6x² + 11x + 3

    Product a × c = 18, sum 11: the numbers are 9 and 2. So 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (3x + 1)(2x + 3).

  3. Use the factor theorem on x³ − 6x² + 11x − 6

    Try x = 1: 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. Dividing gives x² − 5x + 6, which factorises as (x − 2)(x − 3). So the cubic is (x − 1)(x − 2)(x − 3).

  4. A quick square with an identity

    98² = (100 − 2)² = 100² − 2 × 100 × 2 + 2² = 10000 − 400 + 4 = 9604. No long multiplication needed.

Try it at home

Cut-out proof of (a + b)²

You will need

  • A sheet of paper
  • A pencil and ruler
  • Scissors

What to do

  1. Draw a square of side 10 cm. Mark 7 cm along the top edge and 7 cm down the left edge.
  2. Draw a vertical and a horizontal line through those marks. The big square is now cut into four pieces.
  3. Label the pieces: a 7 by 7 square (a²), two 7 by 3 rectangles (ab each) and a 3 by 3 square (b²).
  4. Cut them out and add up the areas: 49 + 21 + 21 + 9 = 100, which is the area of the whole 10 by 10 square.
  5. That is the identity (a + b)² = a² + 2ab + b² with a = 7 and b = 3.

Ask a grown-up: Please take care with the scissors.

Lesson 5 of 7

Linear equations in two variables

See why ax + by = c draws a straight line, and solve a pair of such equations by substitution, by elimination and by graphing.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

An equation like x + y = 4 has two unknowns, and on its own it has endless solutions. But two such equations together usually pin down exactly one pair of values. The picture behind this is two straight lines crossing.

A linear equation in two variables has the form ax + by + c = 0, with a and b not both zero. Each solution is a pair (x, y). For x + y = 4 the pairs (0, 4), (1, 3), (2, 2) and (4, 0) all work, and so do infinitely many others.

If you plot every solution on a graph you get a straight line. That is why the equation is called linear. Two points are enough to draw the line, but plotting a third is a good check.

A pair of linear equations can be solved by elimination: add or subtract the equations so that one variable cancels. Or by substitution: write one variable in terms of the other and substitute it into the second equation.

Graphically, the two lines either cross at one point (one solution, a consistent pair), are parallel (no solution, an inconsistent pair) or are the same line (infinitely many solutions). Compare the ratios a1/a2, b1/b2 and c1/c2 to decide which case you have without drawing.

Two lines crossing at one pointA graph with x and y axes. The line x plus y equals 4 slopes down to the right and the line x minus y equals 2 slopes up to the right. They cross at the point (3, 1), which is marked with a dot.Solving x + y = 4 and x − y = 2 by graph x y 1 2 3 4 5 6 1 2 3 4 5 x + y = 4 x − y = 2 (3, 1) Both equations are true only at (3, 1): 3 + 1 = 4 and 3 − 1 = 2.
Each equation is a straight line. The pair of equations is solved where the two lines meet: (3, 1).

Worked examples

See how it works

  1. Elimination: x + y = 4 and x − y = 2

    Add the two equations: 2x = 6, so x = 3. Put x = 3 into the first: 3 + y = 4, so y = 1. The solution is (3, 1). Check in the second: 3 − 1 = 2. Correct.

  2. Substitution: 2x + 3y = 12 and x = y + 1

    Replace x by y + 1 in the first equation: 2(y + 1) + 3y = 12, so 2y + 2 + 3y = 12, 5y = 10 and y = 2. Then x = 2 + 1 = 3. Check: 2 × 3 + 3 × 2 = 12.

  3. A word problem

    Five pencils and seven pens cost 50 rupees; seven pencils and five pens cost 46 rupees. Let a pencil cost x and a pen cost y: 5x + 7y = 50 and 7x + 5y = 46. Adding gives 12x + 12y = 96, so x + y = 8. Subtracting the first from the second gives 2x − 2y = −4, so x − y = −2. Adding these two: 2x = 6, x = 3 and y = 5. A pencil costs 3 rupees and a pen 5 rupees.

  4. How many solutions?

    x + 2y = 4 and 2x + 4y = 12: the ratios are a1/a2 = 1/2, b1/b2 = 1/2 but c1/c2 = 4/12 = 1/3. The first two match and the third does not, so the lines are parallel and there is no solution.

Try it at home

Plot two lines on a paper grid

You will need

  • A sheet of ruled or squared paper
  • A pencil and ruler
  • Two coloured pencils if you have them

What to do

  1. Draw an x-axis and a y-axis and number them from 0 to 6 using one square per unit.
  2. For x + y = 4, make a table: x = 0 gives y = 4, x = 2 gives y = 2, x = 4 gives y = 0. Plot the three points and join them with a ruler.
  3. For x − y = 2, make a table: x = 2 gives y = 0, x = 4 gives y = 2, x = 6 gives y = 4. Plot and join them in the other colour.
  4. Find the point where the two lines cross and read its coordinates. It should be (3, 1).
  5. Check: 3 + 1 = 4 and 3 − 1 = 2. Both equations are true at the crossing point.

Lesson 6 of 7

Quadratic equations

Solve ax² + bx + c = 0 by factorising and by the quadratic formula, and use the discriminant to predict how many real roots there are.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

A quadratic equation has an x² term, and its graph is a U-shaped curve called a parabola. Solving the equation means finding where that curve crosses the x-axis. There can be two crossings, one touch, or none at all.

A quadratic equation is ax² + bx + c = 0 with a not equal to 0. Its solutions are called roots. Because the highest power is 2, there are at most two roots.

Method 1, factorising: write the left side as a product of two brackets. If (x − 2)(x − 3) = 0 then one bracket must be zero, so x = 2 or x = 3.

Method 2, the quadratic formula, which always works: x = (−b ± √(b² − 4ac)) / 2a. The part under the root, D = b² − 4ac, is called the discriminant.

The discriminant tells you the nature of the roots before you solve. If D is positive there are two different real roots. If D = 0 the two roots are equal (the parabola just touches the axis). If D is negative there are no real roots (the parabola never reaches the axis).

A parabola and its rootsA graph of y equals x squared minus 5x plus 6. The U-shaped curve comes down from the top left, crosses the x-axis at x equals 2 and x equals 3, dips just below the axis between them, and rises again to the top right.y = x² − 5x + 6 x y 1 2 3 4 5 2 4 6 roots: x = 2 and x = 3 The curve crosses the x-axis where y = 0. Those x values are the roots: 2 and 3. Lowest point: (2.5, −0.25)
The roots of x² − 5x + 6 = 0 are the x values where the parabola crosses the x-axis: 2 and 3.

Worked examples

See how it works

  1. By factorising: x² − 5x + 6 = 0

    Two numbers with product 6 and sum −5 are −2 and −3. So (x − 2)(x − 3) = 0 and the roots are x = 2 and x = 3. Check: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0.

  2. By factorising with a leading coefficient: 2x² + x − 6 = 0

    Product a × c = −12, sum 1: the numbers are 4 and −3. So 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2) = 0. The roots are x = 3/2 and x = −2.

  3. By the formula: x² + 4x + 1 = 0

    Here a = 1, b = 4, c = 1, so D = 16 − 4 = 12. Then x = (−4 ± √12) / 2 = (−4 ± 2√3) / 2 = −2 ± √3. The roots are about −0.27 and −3.73.

  4. A word problem

    The product of two consecutive positive whole numbers is 132. Call them n and n + 1: n(n + 1) = 132, so n² + n − 132 = 0. Numbers with product −132 and sum 1 are 12 and −11, so (n + 12)(n − 11) = 0. Since n is positive, n = 11 and the numbers are 11 and 12.

Try it at home

Plot a parabola by hand

You will need

  • Squared or ruled paper
  • A pencil and ruler

What to do

  1. Make a table for y = x² − 5x + 6 with x = 0, 1, 2, 2.5, 3, 4 and 5. You should get y = 6, 2, 0, −0.25, 0, 2, 6.
  2. Draw axes with x from 0 to 5 and y from −1 to 6, and plot the seven points.
  3. Join them with a smooth U-shaped curve (do not use the ruler for this part).
  4. Read off where the curve crosses the x-axis. Those two x values are the roots of x² − 5x + 6 = 0.
  5. Now try y = x² − 4x + 4 with the same x values. Notice the curve only touches the axis once, at x = 2, because D = 16 − 16 = 0.

Lesson 7 of 7

Arithmetic progressions

Recognise a sequence with a constant difference, find any term with a + (n − 1)d and add up n terms with the sum formula.

Open this lesson on its own page · Practice sheet · Answer key · Contents

Let us begin

The big idea

Stairs rise by the same height every step, and a savings plan can grow by the same amount every month. Sequences like these, where each term is the previous one plus a fixed number, are arithmetic progressions.

An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding the same number d, called the common difference, to the term before. In 3, 7, 11, 15, ... the first term a is 3 and d is 4. The difference d can be negative, as in 10, 7, 4, 1, ... where d = −3.

The nth term is a + (n − 1)d. You add d once to get the second term, twice to get the third term, so to reach the nth term you add d exactly n − 1 times.

The sum of the first n terms is S = n/2 × [2a + (n − 1)d]. If you know the last term l, the same sum is S = n/2 × (a + l): the number of terms times the average of the first and last term.

To test whether a list is an AP, subtract neighbouring terms. If every difference is the same, it is an AP. The list 2, 4, 8, 16 is not an AP because the differences 2, 4, 8 keep changing.

An arithmetic progression as a staircaseFive bars standing side by side with heights 3, 7, 11, 15 and 19. Each bar is 4 units taller than the one before, and the gap of plus 4 is written between neighbouring bars.3, 7, 11, 15, 19, ... a = 3, d = 4 3 7 11 15 19 +4 +4 +4 +4 1st term 2nd term 3rd term 4th term 5th term nth term = a + (n − 1)d = 3 + (n − 1) × 4
The AP 3, 7, 11, 15, 19 climbs like stairs: every step adds the common difference d = 4.

Worked examples

See how it works

  1. The 20th term of 3, 7, 11, ...

    Here a = 3 and d = 4. The 20th term is 3 + (20 − 1) × 4 = 3 + 76 = 79.

  2. The sum of those 20 terms

    The first term is 3 and the 20th is 79, so S = 20/2 × (3 + 79) = 10 × 82 = 820.

  3. Which term of 5, 8, 11, ... is 50?

    We need 5 + (n − 1) × 3 = 50, so (n − 1) × 3 = 45, n − 1 = 15 and n = 16. So 50 is the 16th term.

  4. A savings plan

    Riya saves 100 rupees in January and 20 rupees more each month than the month before. In 12 months she saves S = 12/2 × [2 × 100 + 11 × 20] = 6 × (200 + 220) = 6 × 420 = 2520 rupees.

Try it at home

Build an AP with coins

You will need

  • About 30 coins or buttons
  • Paper and a pencil

What to do

  1. Make a row of 2 coins, then a row of 5, then 8, then 11. Each row has 3 more coins than the one before.
  2. Write the sequence: 2, 5, 8, 11. Name a and d. Predict the next two rows and then build them to check.
  3. Use the formula to find the 10th row without building it: 2 + 9 × 3 = 29.
  4. Count all the coins in the first four rows: 2 + 5 + 8 + 11 = 26. Now use the sum formula: 4/2 × (2 + 11) = 2 × 13 = 26. The formula matches the count.

Ask a grown-up: Keep small coins and buttons away from young children.

Part two

Practice sheets

One sheet for each of the 7 lessons. Write your name and the date at the top, answer in the spaces, then check with the answer keys at the back.

Practice sheet

Pythagoras theorem

Practice sheet: Pythagoras theorem (Classes 9-10, Maths)

  1. What is the longest side of a right-angled triangle called?
  2. The legs of a right triangle are 3 cm and 4 cm. Find the hypotenuse.
  3. The legs are 5 cm and 12 cm. Find the hypotenuse.
  4. The legs are 9 cm and 12 cm. Find the hypotenuse.
  5. The hypotenuse is 17 cm and one leg is 8 cm. Find the other leg.
  6. Is a triangle with sides 6 cm, 8 cm and 10 cm right-angled?
  7. Is a triangle with sides 4 cm, 5 cm and 6 cm right-angled?
  8. A rectangle is 24 cm long and 7 cm wide. Find the length of its diagonal.
  9. A 5 m ladder has its foot 3 m from a wall. How high up the wall does it reach?
  10. Ravi walks 12 m west and then 5 m north. How far is he from where he started, in a straight line?
  11. A square has side 10 cm. Find its diagonal, correct to one decimal place (take √2 = 1.414).

Practice sheet

Mean, median and mode

Practice sheet: Mean, median and mode (Classes 9-10, Maths)

  1. Find the mean of 2, 4, 6, 8.
  2. Find the median of 3, 9, 5.
  3. Find the mode of 2, 3, 3, 5, 3, 7, 5.
  4. Find the mean of 10, 15, 20, 25, 30.
  5. Find the median of 8, 2, 10, 4, 6, 12.
  6. Find the range of 14, 9, 21, 5.
  7. The mean of 5 numbers is 12. What is their sum?
  8. Marks of 5 students are 40, 50, 35, 50, 45. Find the mean, median and mode.
  9. The mean of 4 numbers is 15. Three of them are 12, 18 and 10. Find the fourth number.
  10. The mean of 6 numbers is 8. A seventh number, 22, is added. What is the new mean?

Practice sheet

Rational and irrational numbers

Practice sheet: Rational and irrational numbers (Classes 9-10, Maths)

  1. Is 7 a rational number? Why?
  2. Write 0.6 as a fraction in its lowest terms.
  3. Is √16 rational or irrational?
  4. Is √7 rational or irrational?
  5. Write 0.444... as a fraction.
  6. Write 0.181818... as a fraction.
  7. Find a rational number between 1/4 and 1/2.
  8. Which of these decimals is irrational: 0.125, 0.121212..., 0.1010010001...?
  9. Is the product √2 × √2 rational or irrational?
  10. Is π equal to 22/7?

Practice sheet

Polynomials and factorisation

Practice sheet: Polynomials and factorisation (Classes 9-10, Maths)

  1. What is the degree of 4x³ − 2x + 7?
  2. Is 1 a zero of p(x) = x² − 3x + 2? Is 2?
  3. Factorise x² + 7x + 12.
  4. Expand (x + 5)².
  5. Factorise 49 − y².
  6. Is (x − 1) a factor of x³ − 1?
  7. Find the zeroes of x² − 9.
  8. Expand (2x − 3)(x + 4).
  9. Use an identity to find 103².
  10. Factorise x² − 5x + 6.

Practice sheet

Linear equations in two variables

Practice sheet: Linear equations in two variables (Classes 9-10, Maths)

  1. Is (2, 1) a solution of 3x − y = 5?
  2. Write three solutions of x + y = 5.
  3. Solve x + y = 10 and x − y = 4.
  4. What shape is the graph of a linear equation in two variables?
  5. Solve by substitution: y = 2x and x + y = 9.
  6. Solve 2x + y = 7 and x − y = 2.
  7. How many solutions do x + 2y = 3 and 2x + 4y = 6 have?
  8. How many solutions do x + y = 3 and x + y = 7 have?
  9. The sum of two numbers is 20 and their difference is 4. Find them.
  10. Where does the line x − y = 2 cross the y-axis?

Practice sheet

Quadratic equations

Practice sheet: Quadratic equations (Classes 9-10, Maths)

  1. Is x² − 4 = 0 a quadratic equation? What are its roots?
  2. Solve x² − 7x + 12 = 0 by factorising.
  3. Solve x² + 2x − 8 = 0.
  4. Find the discriminant of 2x² − 3x + 1 = 0 and then its roots.
  5. What is the nature of the roots of x² + x + 1 = 0?
  6. Solve x² − 6x + 9 = 0. What do you notice about the roots?
  7. The square of a number minus three times the number is 10. Find the positive number.
  8. Solve x² = 49.
  9. Solve 3x² − 12 = 0.
  10. Write the quadratic formula.

Practice sheet

Arithmetic progressions

Practice sheet: Arithmetic progressions (Classes 9-10, Maths)

  1. What is the next term of 2, 5, 8, 11, ...?
  2. What is the common difference of 10, 7, 4, 1, ...?
  3. Find the 15th term of 3, 7, 11, ...
  4. Is 2, 4, 8, 16 an AP? Why or why not?
  5. Write the formula for the nth term of an AP.
  6. Find the sum of the first 10 terms of 1, 3, 5, 7, ...
  7. Which term of 7, 10, 13, ... is 61?
  8. Find the sum 1 + 2 + 3 + ... + 100.
  9. Three terms of an AP add up to 15. What is the middle term?
  10. An AP has first term 5 and common difference 2. Find the sum of its first 20 terms.

Part three

Answer keys

The answers to every practice sheet, in the same order as the lessons. Parents and teachers may like to keep these pages separately.

Answer key

Answers: Pythagoras theorem

  1. The hypotenuse.
  2. 5 cm (9 + 16 = 25).
  3. 13 cm (25 + 144 = 169).
  4. 15 cm (81 + 144 = 225).
  5. 15 cm (289 − 64 = 225, and √225 = 15).
  6. Yes. 36 + 64 = 100 = 10².
  7. No. 16 + 25 = 41, but 6² = 36.
  8. 25 cm (576 + 49 = 625).
  9. 4 m (25 − 9 = 16).
  10. 13 m (144 + 25 = 169).
  11. About 14.1 cm (diagonal = √(100 + 100) = √200 = 10√2 = 14.14).

Answer key

Answers: Mean, median and mode

  1. 5 (20 ÷ 4).
  2. 5 (in order 3, 5, 9).
  3. 3 (it appears three times).
  4. 20 (sum 100, divided by 5).
  5. 7 (in order 2, 4, 6, 8, 10, 12; the middle two are 6 and 8).
  6. 16 (21 − 5).
  7. 60 (12 × 5).
  8. Mean 44 (sum 220 ÷ 5), median 45 (in order 35, 40, 45, 50, 50), mode 50.
  9. 20 (the total must be 4 × 15 = 60, and 12 + 18 + 10 = 40, so the fourth is 20).
  10. 10 (old sum 6 × 8 = 48; new sum 48 + 22 = 70; 70 ÷ 7 = 10).

Answer key

Answers: Rational and irrational numbers

  1. Yes, because 7 = 7/1, a fraction of integers.
  2. 0.6 = 6/10 = 3/5.
  3. Rational, because √16 = 4.
  4. Irrational, because 7 is not a perfect square and the decimal of √7 never repeats.
  5. Let x = 0.444..., then 10x = 4.444..., so 9x = 4 and x = 4/9.
  6. Let x = 0.181818..., then 100x = 18.1818..., so 99x = 18 and x = 18/99 = 2/11.
  7. 3/8, the average: (1/4 + 1/2)/2 = (3/4)/2 = 3/8. Any number between them is fine.
  8. 0.1010010001..., because its pattern never repeats a fixed block.
  9. Rational: √2 × √2 = 2.
  10. No. 22/7 is only a rational approximation of π. π itself is irrational.

Answer key

Answers: Polynomials and factorisation

  1. 3, because the highest power is x³.
  2. Yes to both: p(1) = 1 − 3 + 2 = 0 and p(2) = 4 − 6 + 2 = 0.
  3. (x + 3)(x + 4), since 3 × 4 = 12 and 3 + 4 = 7.
  4. x² + 10x + 25.
  5. (7 − y)(7 + y), using a² − b² = (a − b)(a + b).
  6. Yes, because putting x = 1 gives 1 − 1 = 0.
  7. 3 and −3, because x² − 9 = (x − 3)(x + 3).
  8. 2x² + 8x − 3x − 12 = 2x² + 5x − 12.
  9. (100 + 3)² = 10000 + 600 + 9 = 10609.
  10. (x − 2)(x − 3), since (−2) × (−3) = 6 and (−2) + (−3) = −5.

Answer key

Answers: Linear equations in two variables

  1. Yes: 3 × 2 − 1 = 5.
  2. For example (0, 5), (1, 4) and (5, 0). Any pair that adds to 5 is correct.
  3. Adding gives 2x = 14, so x = 7 and y = 3.
  4. A straight line.
  5. x + 2x = 9, so x = 3 and y = 6.
  6. Adding gives 3x = 9, so x = 3; then y = 1.
  7. Infinitely many: the second equation is twice the first, so both describe the same line.
  8. None: the lines are parallel, because x + y cannot be both 3 and 7.
  9. 12 and 8: x + y = 20 and x − y = 4 give 2x = 24.
  10. On the y-axis x = 0, so −y = 2 and y = −2. The point is (0, −2).

Answer key

Answers: Quadratic equations

  1. Yes, degree 2. Roots: x = 2 and x = −2.
  2. (x − 3)(x − 4) = 0, so x = 3 or x = 4.
  3. (x + 4)(x − 2) = 0, so x = −4 or x = 2.
  4. D = 9 − 8 = 1. Roots: x = (3 ± 1)/4, so x = 1 or x = 1/2.
  5. D = 1 − 4 = −3, which is negative, so there are no real roots.
  6. (x − 3)² = 0, so x = 3. The two roots are equal because D = 36 − 36 = 0.
  7. x² − 3x − 10 = 0, so (x − 5)(x + 2) = 0. The positive number is 5.
  8. x = 7 or x = −7.
  9. x² = 4, so x = 2 or x = −2.
  10. x = (−b ± √(b² − 4ac)) / 2a.

Answer key

Answers: Arithmetic progressions

  1. 14, because d = 3.
  2. −3, since 7 − 10 = −3.
  3. 3 + 14 × 4 = 59.
  4. No. The differences are 2, 4 and 8, not the same each time.
  5. nth term = a + (n − 1)d.
  6. S = 10/2 × [2 × 1 + 9 × 2] = 5 × 20 = 100.
  7. 7 + (n − 1) × 3 = 61 gives n − 1 = 18, so n = 19: the 19th term.
  8. 100/2 × (1 + 100) = 50 × 101 = 5050.
  9. 5, because three terms in an AP are (m − d), m and (m + d), which add to 3m.
  10. S = 20/2 × [2 × 5 + 19 × 2] = 10 × 48 = 480.

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