Class 9 · 9th standard · 2026-27

Class 9 maths worksheets, free, with every step on the page

Three free Class 9 maths worksheets, with every question and every step printed on this page rather than behind a download. The middle one is the one that matters: a proof worksheet that teaches the format - figure, given, to prove, then steps each with a reason - rather than a list of theorems to memorise. A memorised proof scores on one question; the format scores on questions nobody has seen. Every exercise opens with one question worked through in full.

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3 worksheets · 85 questions · every answer on the page · no sign-up · aligned to NCERT's Class 9 maths book is Mathematics. Schools also use their own texts, so these sheets are written to the ideas rather than to one book's chapter order.

What these practise

Degree, zeroes and the value of a polynomial
The remainder theorem, substituting the value that makes the divisor zero
The factor theorem, and finding an unknown coefficient with it
Factorising cubics completely
The congruence criteria - SSS, SAS, ASA, AAS, RHS - and why AAA is not one
Writing a proof: figure, given, to prove, steps with reasons
Using CPCT correctly, and the step that often follows it
Rational and irrational numbers, judged after simplifying
Rationalising a denominator with the conjugate
Recurring decimals converted to fractions, and fractional exponents

Last updated 2026-07-29

Free, with answers

The worksheets themselves

Every question is on this page, and so is every answer. Nothing is behind a sign-up, and there is no file to download unless you want one - use the print button on any sheet to save it as a PDF.

practice · medium · 30 min

Polynomials

Class 9 Maths · Polynomials · 33 questions with answers

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Polynomials is the first chapter of Class 9 and it assumes Class 8 completely - expanding a bracket and factorising an expression have to be automatic before any of this works. Everything here rests on one small idea: to find the remainder when dividing by (x - a), substitute x = a. That single substitution replaces long division and unlocks the whole chapter.

States the degree and finds the value of a polynomial at a point
Uses the remainder theorem instead of long division
Uses the factor theorem to test a factor and to find an unknown coefficient
Factorises a cubic completely
Applies the expanded identities for three terms and for cubes

Exercise 1 of 3 · 11 questions

Answer each one.

Write the answer · Warm-up

Here’s one done for you

p(2) = 0 for p(x) = x² - 3x + 2, which means 2 is a zero of the polynomial. Notice what that also tells you: since p(2) = 0, the factor theorem says (x - 2) is a factor. Every 'find p(a)' question is quietly a factor question, and seeing that early makes the rest of the chapter much shorter.

1)What is the degree of p(x) = 5x³ - 2x + 7?
2)If p(x) = x² - 3x + 2, find p(2).
3)If p(x) = x³ + 2x² - x + 1, find p(-1).
4)Is x = 3 a zero of p(x) = x² - 5x + 6?
5)Find the zero of p(x) = 2x + 5.
6)What is the degree of p(x) = 7 - 2x + x⁴?
7)If p(x) = 2x² + x - 6, find p(-1).
8)If p(x) = x³ - 4x, find p(0) and p(2).
9)Is x = -2 a zero of p(x) = x² + 5x + 6?
10)Find the zero of p(x) = 3x - 7.
11)Write the coefficient of x² in p(x) = 5 - 3x² + x³.

Answers

1) 3 - the highest power of x present.
2) p(2) = 4 - 6 + 2 = 0. So 2 is a zero of this polynomial.
3) p(-1) = -1 + 2 + 1 + 1 = 3.
4) Yes. p(3) = 9 - 15 + 6 = 0, and a zero is a value that makes the polynomial zero.
5) x = -5/2, from 2x + 5 = 0.
6) 4. The degree is the highest power of x, whatever order the terms are written in.
7) p(-1) = 2(1) + (-1) - 6 = -5.
8) p(0) = 0 and p(2) = 8 - 8 = 0, so 0 and 2 are both zeroes.
9) Yes. p(-2) = 4 - 10 + 6 = 0, and a zero is a value that makes p(x) = 0.
10) 3x - 7 = 0, so x = 7/3.
11) -3. The coefficient carries its sign.

Exercise 2 of 3 · 11 questions

Use the remainder or factor theorem. Do not use long division.

Write the answer · Guided

Here’s one done for you

For the remainder on dividing by (2x - 1), set 2x - 1 = 0 rather than reaching for x = 1. That gives x = 1/2, and substituting gives 3/2. Children who have learned the theorem as 'put x = a for (x - a)' get this one wrong every time. The rule is really 'substitute the value that makes the divisor zero', which covers every case including this one.

1)Find the remainder when x³ - 3x² + 4x - 5 is divided by (x - 2).
2)Find the remainder when x³ + 3x² + 3x + 1 is divided by (x + 1).
3)Is (x - 1) a factor of x³ - 2x² + x?
4)Find k if (x - 2) is a factor of x³ + kx² + 2x - 4.
5)Find the remainder when 4x³ - 12x² + 14x - 3 is divided by (2x - 1).
6)Find the remainder when x³ + x² - 2x + 1 is divided by (x - 2).
7)Find the remainder when 2x³ - 3x² + 4x - 5 is divided by (x - 1).
8)Is (x - 1) a factor of x³ - 6x² + 11x - 6?
9)Find k if (x + 1) is a factor of x³ + 3x² + kx + 4.
10)Find the remainder when x⁴ - x² + 2 is divided by (x + 1).
11)Show that (x - 3) is a factor of x³ - 6x² + 11x - 6.

Answers

1) p(2) = 8 - 12 + 8 - 5 = -1. The remainder is -1.
2) p(-1) = -1 + 3 - 3 + 1 = 0. The remainder is 0, so (x + 1) is a factor.
3) Yes. p(1) = 1 - 2 + 1 = 0, so by the factor theorem (x - 1) is a factor.
4) p(2) = 8 + 4k + 4 - 4 = 8 + 4k. For (x - 2) to be a factor this must be zero, so k = -2.
5) Set 2x - 1 = 0, giving x = 1/2. Then p(1/2) = 1/2 - 3 + 7 - 3 = 3/2. The remainder is 3/2.
6) p(2) = 8 + 4 - 4 + 1 = 9. The remainder is 9.
7) p(1) = 2 - 3 + 4 - 5 = -2. The remainder is -2.
8) Yes. p(1) = 1 - 6 + 11 - 6 = 0, so by the factor theorem (x - 1) is a factor.
9) p(-1) = -1 + 3 - k + 4 = 0, so 6 - k = 0 and k = 6.
10) p(-1) = 1 - 1 + 2 = 2. The remainder is 2.
11) p(3) = 27 - 54 + 33 - 6 = 0, so by the factor theorem (x - 3) is a factor.

Exercise 3 of 3 · 11 questions

Factorise completely. Find one factor by the factor theorem first.

Write the answer · Practice

Here’s one done for you

For x³ - 3x² - 9x - 5, try small values first: p(1) = -16, p(-1) = -1 - 3 + 9 - 5 = 0. So (x + 1) is a factor. Dividing gives x² - 4x - 5, which factorises to (x + 1)(x - 5), so the full answer is (x + 1)²(x - 5). Two habits make cubics reliable: try ±1 and ±2 before anything else, since exam cubics almost always have a small integer zero, and check the answer by expanding it back.

1)x³ - 2x² - x + 2
2)2y³ + y² - 2y - 1
3)x³ - 3x² - 9x - 5
4)x³ - 23x² + 142x - 120
5)Expand (2x + 3y)³.
6)x³ - 6x² + 11x - 6
7)x³ + 6x² + 11x + 6
8)Expand (x + 2y)³.
9)Expand (3a - b)³.
10)Factorise 8x³ + 27y³.
11)Factorise x³ - 64.

Answers

1) (x - 1)(x - 2)(x + 1). Since p(1) = 0, divide by (x - 1) to get x² - x - 2, which factorises further.
2) (y - 1)(y + 1)(2y + 1). Since p(1) = 0, divide by (y - 1) to get 2y² + 3y + 1.
3) (x + 1)²(x - 5). Since p(-1) = 0, divide by (x + 1) to get x² - 4x - 5 = (x + 1)(x - 5).
4) (x - 1)(x - 10)(x - 12). Since p(1) = 0, divide by (x - 1) to get x² - 22x + 120.
5) 8x³ + 36x²y + 54xy² + 27y³, using (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2x and b = 3y.
6) (x - 1)(x - 2)(x - 3). p(1) = 0 gives the first factor, and dividing leaves x² - 5x + 6.
7) (x + 1)(x + 2)(x + 3). p(-1) = 0 gives the first factor.
8) x³ + 6x²y + 12xy² + 8y³.
9) 27a³ - 27a²b + 9ab² - b³.
10) (2x + 3y)(4x² - 6xy + 9y²), using a³ + b³ = (a + b)(a² - ab + b²).
11) (x - 4)(x² + 4x + 16), using a³ - b³ = (a - b)(a² + ab + b²).

While your child works

The rule is 'substitute the value that makes the divisor zero', not 'put x = a'. That covers divisors like (2x - 1), which catch out children who learned the shortcut.
For cubics, try ±1 and ±2 first. Exam questions are built to have a small integer zero.
Always expand the factorised answer back. Fifteen seconds, and it catches every error.

Open the Polynomials worksheet on its own page

practice · challenging · 35 min

Geometry proof and congruence

Class 9 Maths · Triangles · 18 questions with answers

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This is the sheet that matters most in Class 9, and it teaches a format rather than a list of theorems. A memorised proof scores on the question it was memorised for. A child who owns the format - figure, given, to prove, then steps each with a reason - can attempt a theorem they have never seen and earn most of the marks.

Chooses the correct congruence criterion and says why it applies
Knows why AAA is not a congruence criterion
Writes a reason beside every statement in a proof
Uses CPCT correctly and only after congruence has been established
Sets out a full proof with figure, given and to prove

Exercise 1 of 3 · 11 questions

Which congruence criterion applies? If none does, say so and explain.

Write the answer · Warm-up

Here’s one done for you

AAA is not a congruence criterion, and the reason is worth saying rather than remembering: two equilateral triangles, one with sides of 2 cm and one with sides of 200 cm, have all three angles equal and are obviously not congruent. Angles fix shape; they say nothing about size. This is also the commonest wrong answer in the whole chapter.

1)Three pairs of sides are equal.
2)Two pairs of sides and the pair of angles between them are equal.
3)Two pairs of angles and the pair of sides between them are equal.
4)Two pairs of angles and a pair of sides not between them are equal.
5)Both triangles are right-angled, with equal hypotenuses and one equal side.
6)Three pairs of angles are equal.
7)Two pairs of sides and a non-included pair of angles are equal.
8)One pair of sides and two pairs of angles, the side lying between the two angles.
9)In two right triangles, one pair of legs and the pair of hypotenuses are equal.
10)Two pairs of sides are equal and the included angles are both 90 degrees.
11)All three pairs of sides are equal but the triangles face opposite ways.

Answers

1) SSS.
2) SAS. The angle must be the included one - the angle between the two sides.
3) ASA. The side must be the one between the two angles.
4) AAS. This works because knowing two angles fixes the third, which reduces it to ASA.
5) RHS, which applies only to right-angled triangles.
6) None. Equal angles make the triangles similar, not congruent - they can be the same shape at completely different sizes.
7) No criterion. SSA does not guarantee congruence - two different triangles can fit the same data.
8) ASA. The side is included between the two angles, which is what ASA requires.
9) RHS. Right angle, hypotenuse and one side.
10) SAS. The angle is included between the two known sides, and 90 degrees is still an included angle.
11) SSS. Congruence does not depend on orientation - one triangle can be turned or flipped onto the other.

Exercise 2 of 3 · 5 questions

The statements are given. Write the reason for each one.

Write the answer · Guided

Here’s one done for you

This is the proof that angles opposite equal sides of a triangle are equal, and it is five lines long. Notice that every statement has exactly one reason - given, construction, common, a criterion, or CPCT. Those five reasons cover almost every Class 9 proof, and a child who knows they are the only options stops staring at a blank page.

1)In triangle ABC, AB = AC, and AD bisects angle A meeting BC at D. Statement 1: AB = AC.
2)Statement 2: angle BAD = angle CAD.
3)Statement 3: AD = AD.
4)Statement 4: triangle ABD is congruent to triangle ACD.
5)Statement 5: angle B = angle C.

Answers

1) Given.
2) By construction, since AD bisects angle A.
3) Common side to both triangles.
4) By SAS - two sides and the included angle.
5) By CPCT - corresponding parts of congruent triangles.

Exercise 3 of 3 · 2 questions

Write the full proof. Figure, given, to prove, then steps with reasons.

Write the answer · Practice

Here’s one done for you

The second proof is worth studying for its last two lines. Congruence gives you that the two angles are equal; it does not by itself give you a right angle. The linear pair does - equal angles adding to 180° must each be 90°. Class 9 proofs frequently end with one extra step like that after CPCT, and children who stop at CPCT lose exactly the mark the question was set for.

1)Prove that the diagonals of a parallelogram bisect each other.
2)In triangle ABC, AB = AC and AD is the median to BC. Prove that AD is perpendicular to BC.

Answers

1) Given: parallelogram ABCD with diagonals AC and BD meeting at O. To prove: AO = OC and BO = OD. Proof: AB = CD (opposite sides of a parallelogram are equal). Angle OAB = angle OCD (alternate angles, since AB is parallel to DC with transversal AC). Angle OBA = angle ODC (alternate angles, with transversal BD). Therefore triangle AOB is congruent to triangle COD (ASA). Therefore AO = OC and BO = OD (CPCT).
2) Given: triangle ABC with AB = AC, and D the mid-point of BC. To prove: AD is perpendicular to BC. Proof: AB = AC (given). BD = DC (D is the mid-point). AD = AD (common). Therefore triangle ABD is congruent to triangle ACD (SSS). Therefore angle ADB = angle ADC (CPCT). But these two angles form a linear pair, so they add to 180°. Each is therefore 90°, and AD is perpendicular to BC.

While your child works

Check the figure first. A proof attempted without a labelled diagram almost always fails, and the diagram itself carries marks.
Every statement needs one of five reasons: given, construction, common, a congruence criterion, or CPCT. Knowing the list is finite stops the blank page.
A partial proof with reasons earns partial marks. A conclusion with no reasons earns almost nothing. Say that once and mean it.

Open the Triangles worksheet on its own page

practice · medium · 25 min

Number systems

Class 9 Maths · Number systems · 34 questions with answers

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The number system chapter is short, entirely rule-based, and among the most reliably scored in the year - which makes it a good place to start. Rationalising in particular is mechanical once a child sees that it is just the difference-of-squares identity from Class 8, used to remove a root from the bottom of a fraction.

Decides whether a number is rational or irrational and gives the reason
Rationalises a denominator using the conjugate
Converts a recurring decimal into a fraction
Applies the laws of exponents to fractional powers

Exercise 1 of 3 · 12 questions

Rational or irrational? Give the reason.

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Here’s one done for you

√2 × √8 is rational, and that surprises most children. Combining the roots gives √16 = 4. The lesson is that a number is not irrational because it has a root sign in it - it is irrational because it cannot be written as a fraction. Simplify first, then judge.

1)√25
2)√7
3)0.3333...
4)π
5)2 + √3
6)√2 × √8
7)√16
8)√11
9)0.101001000100001...
10)22/7
11)√5 × √5
12)3 - √2

Answers

1) Rational - it equals 5, which can be written as 5/1.
2) Irrational - 7 is not a perfect square, so its root cannot be written as a fraction.
3) Rational - a recurring decimal, equal to 1/3.
4) Irrational - its decimal expansion is non-terminating and non-recurring. Note that 22/7 is an approximation, not π itself.
5) Irrational - a rational number added to an irrational one is always irrational.
6) Rational - it equals √16, which is 4.
7) Rational. √16 = 4, which can be written as 4/1.
8) Irrational. 11 is not a perfect square, so its root is non-terminating and non-repeating.
9) Irrational. The decimal never terminates and never settles into a repeating block.
10) Rational. It is a ratio of two integers, and it is an approximation of π rather than π itself.
11) Rational. √5 × √5 = 5.
12) Irrational. A rational number minus an irrational number is always irrational.

Exercise 2 of 3 · 10 questions

Rationalise the denominator.

Write the answer · Guided

Here’s one done for you

For 1/(2 + √3), multiply top and bottom by (2 - √3). The denominator becomes (2 + √3)(2 - √3) = 4 - 3 = 1, so the answer is simply 2 - √3. This is the Class 8 identity (a + b)(a - b) = a² - b² doing all the work. Once a child recognises that, rationalising stops being a separate technique and becomes an application of something they already know.

1)1/√5
2)1/(√7 - √6)
3)5/(√3 - 1)
4)1/(2 + √3)
5)7/(3 + 2√2)
6)1/√3
7)1/(√5 + √2)
8)3/(2 - √3)
9)1/(√7 + 2)
10)4/(3 - √5)

Answers

1) √5/5. Multiply top and bottom by √5.
2) √7 + √6. Multiply by the conjugate (√7 + √6); the denominator becomes 7 - 6 = 1.
3) 5(√3 + 1)/2. Multiply by (√3 + 1); the denominator becomes 3 - 1 = 2.
4) 2 - √3. Multiply by (2 - √3); the denominator becomes 4 - 3 = 1.
5) 21 - 14√2. Multiply by (3 - 2√2); the denominator becomes 9 - 8 = 1.
6) √3/3. Multiply top and bottom by √3.
7) (√5 - √2)/3. Multiply by the conjugate: the denominator becomes 5 - 2 = 3.
8) 3(2 + √3) = 6 + 3√3. The denominator becomes 4 - 3 = 1.
9) (√7 - 2)/3. The denominator becomes 7 - 4 = 3.
10) (3 + √5). The denominator becomes 9 - 5 = 4, and 4(3 + √5)/4 = 3 + √5.

Exercise 3 of 3 · 12 questions

Simplify, or convert as asked.

Write the answer · Practice

Here’s one done for you

To convert 0.474747... multiply by 100 rather than by 10, because two digits repeat. Then 100x - x = 47, so x = 47/99. The rule follows from that rather than needing to be memorised: multiply by 10 for one repeating digit, 100 for two, 1000 for three - enough to line the repeating part up so it cancels.

1)2^(2/3) × 2^(1/3)
2)(3^(1/5))^5
3)64^(1/2) × 64^(1/3)
4)Express 0.666... as a fraction.
5)Express 0.474747... as a fraction.
6)Simplify 1/(3 - √2) + 1/(3 + √2).
7)5^(1/2) × 5^(1/2)
8)(2^3)^(1/3)
9)27^(2/3)
10)Express 0.777... as a fraction.
11)Express 0.363636... as a fraction.
12)Simplify 1/(2 - √3) + 1/(2 + √3).

Answers

1) 2. Adding the powers gives 2^1.
2) 3. Multiplying the powers gives 3^1.
3) 32. That is 8 × 4, or equivalently 64^(5/6).
4) 2/3. Let x = 0.666...; then 10x = 6.666...; subtracting gives 9x = 6, so x = 2/3.
5) 47/99. Let x = 0.474747...; then 100x = 47.4747...; subtracting gives 99x = 47.
6) 6/7. The two fractions combine to give ((3 + √2) + (3 - √2))/(9 - 2) = 6/7.
7) 5. The powers add: 1/2 + 1/2 = 1.
8) 2. The powers multiply: 3 × 1/3 = 1.
9) 9. 27^(1/3) = 3, and 3² = 9.
10) 7/9. Let x = 0.777..., then 10x = 7.777..., so 9x = 7.
11) 36/99 = 4/11. Let x = 0.3636..., then 100x = 36.3636..., so 99x = 36.
12) 4. The two denominators are conjugates, so the terms become (2 + √3) + (2 - √3).

While your child works

Simplify before judging whether a number is irrational. √2 × √8 looks irrational and is 4.
Rationalising is just (a + b)(a - b) = a² - b² from Class 8. Say that once and the technique stops feeling new.
This chapter is short and reliably scored. It is a good place to start the year, and a good place to regain confidence mid-year.

Open the Number systems worksheet on its own page

Why is the proof worksheet the important one?

Because proof is the actual change in Class 9, and almost no practice material treats it as a skill.

For eight years an answer was a value. From Class 9, an answer can be an argument - and an argument is marked step by step on its reasons rather than on its conclusion. A child who writes 'therefore the diagonals bisect each other' with nothing above it has written nothing an examiner can credit.

What makes this fixable is that the format is small. Figure, given, to prove, then steps - and every step's reason is one of five things: given, construction, common, a congruence criterion, or CPCT. A child who knows the list is finite stops staring at the page. The full argument is on our Class 9 maths tuition page.

How should we use these at home?

Twenty-five to thirty-five minutes each, once or twice a week. Every exercise opens with one question worked through in full.

In polynomials, check the divisor. The rule is 'substitute the value that makes the divisor zero', which is what makes (2x - 1) work.
In proof, check the figure first. A proof attempted without a labelled diagram nearly always fails.
In number systems, simplify before judging. √2 × √8 looks irrational and equals 4.
Let your child mark their own work. Every answer is on the page for exactly that reason.

Where do Class 9 children usually stall?

Three places, and only one of them is about difficulty.

Proof with no reasons. The conclusion is right and unsupported, and the marks are all in the support.
The factor theorem applied as 'put x = a'. It fails on divisors like (2x - 1), and the correct rule covers both cases.
Class 8 algebra assumed and missing. Polynomials is the first chapter of the year and it needs expanding and factorising to be automatic - see our Class 8 maths worksheets.

Questions parents ask

Class 9 Maths worksheets - your questions

Are these Class 9 maths worksheets free?

Yes. Every question and every answer is on this page, with no sign-up, no email wall and no account. The printable version puts the answers on a separate final sheet.

Do these include full model proofs?

Yes - two complete proofs written out with figure, given, to prove and a reason beside every step, plus an exercise that practises writing the reasons alone.

Why do maths marks fall in Class 9?

Usually because of proof rather than difficulty. An answer can now be an argument, marked step by step on its reasons, and a child who writes the right conclusion without justification scores very little.

What is the factor theorem?

If p(a) = 0 then (x - a) is a factor of p(x). It turns a division problem into one substitution, which is why it replaces long division almost everywhere in the polynomials chapter.

Why is AAA not a congruence criterion?

Because equal angles fix shape but not size. Two equilateral triangles with sides of 2 cm and 200 cm have identical angles and are obviously not congruent - they are similar.

Are these matched to NCERT?

They are written to the ideas Class 9 maths tests rather than to one book's chapter order, so they work whether your school uses NCERT or its own text.

How long should a Class 9 child spend on one sheet?

Twenty-five to thirty-five minutes. The proof sheet takes longest and is worth the time - it is the only one whose skill transfers to questions nobody has seen.

Can I print these worksheets?

Yes. They print cleanly on A4 with the answers on a separate final page, and nothing needs to be downloaded or signed up for.

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Class 9 maths worksheets, free, with every step on the page

The worksheets themselves

Polynomials

Geometry proof and congruence

Number systems

Class 9 Maths - the guide

Why is the proof worksheet the important one?

How should we use these at home?

Where do Class 9 children usually stall?

Class 9 Maths worksheets - your questions

Worksheets are free. So is the first class.

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