CBSE Class 9 Maths Important Questions 2026: Chapter-Wise with Solutions

Question: Rationalise the denominator of $\frac{1}{\sqrt{5} + \sqrt{2}}$.

Solution:
Multiply numerator and denominator by the conjugate $\sqrt{5} - \sqrt{2}$:

Question: Simplify $\frac{2^{5/2} \times 2^{1/2}}{2^{3/2}}$.

Solution:
Using $a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$:

Question: Check whether $(x + 1)$ is a factor of $x^3 + x^2 + x + 1$.

Solution:
By the Factor Theorem, $(x + 1)$ is a factor if $p(-1) = 0$.

Since $p(-1) = 0$, yes, $(x + 1)$ is a factor of $x^3 + x^2 + x + 1$.

Question: Factorise $27x^3 + 125y^3$.

Solution:
Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = 3x$ and $b = 5y$:

Question: In which quadrant or on which axis do the following points lie? (i) $(-3, 5)$ (ii) $(4, -2)$ (iii) $(0, -7)$ (iv) $(-1, -4)$

Solution:
(i) $(-3, 5)$: $x < 0, y > 0$ -- Quadrant II
(ii) $(4, -2)$: $x > 0, y < 0$ -- Quadrant IV
(iii) $(0, -7)$: $x = 0$ -- On the y-axis (negative y-axis)
(iv) $(-1, -4)$: $x < 0, y < 0$ -- Quadrant III

Question: Find four different solutions of the equation $2x + y = 7$.

Solution:
Rearrange: $y = 7 - 2x$.

Four solutions: $(0, 7)$, $(1, 5)$, $(2, 3)$, $(3, 1)$.

All these points lie on a straight line when plotted on a graph.

Question: If $A$, $B$, and $C$ are three points on a line, and $B$ lies between $A$ and $C$, prove that $AC - BC = AB$.

Solution:
Since $B$ lies between $A$ and $C$:

Subtracting $BC$ from both sides (Axiom: if equals are subtracted from equals, the remainders are equal):

Hence proved: $AC - BC = AB$.

Question: In the figure, if $AB \parallel CD$, $\angle ABE = 120°$, and $\angle DCE = 110°$, find $\angle BEC$.

Solution:
Draw a line $EF$ through $E$ parallel to $AB$ (and hence parallel to $CD$).

Since $EF \parallel AB$ and $BE$ is a transversal:

Since $EF \parallel CD$ and $CE$ is a transversal:

Question: In $\triangle ABC$, $AB = AC$. The bisectors of $\angle B$ and $\angle C$ meet at $O$. Prove that $OB = OC$.

Solution:
In $\triangle ABC$, $AB = AC$ (given).

Therefore $\angle ABC = \angle ACB$ (angles opposite equal sides).

Since $BO$ bisects $\angle B$: $\angle OBC = \frac{1}{2}\angle ABC$.
Since $CO$ bisects $\angle C$: $\angle OCB = \frac{1}{2}\angle ACB$.

Since $\angle ABC = \angle ACB$:

In $\triangle OBC$:
$\angle OBC = \angle OCB$ implies $OB = OC$ (sides opposite equal angles are equal).

Hence proved.

Question: $ABCD$ is a quadrilateral in which $P$, $Q$, $R$, $S$ are the mid-points of sides $AB$, $BC$, $CD$, $DA$ respectively. Show that $PQRS$ is a parallelogram.

Solution:
Join diagonal $AC$.

In $\triangle ABC$, $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$.
By the Mid-Point Theorem:

In $\triangle ACD$, $S$ is the midpoint of $AD$ and $R$ is the midpoint of $CD$.
By the Mid-Point Theorem:

From $(i)$ and $(ii)$:
$PQ \parallel SR$ and $PQ = SR$.

Since one pair of opposite sides is both equal and parallel, $PQRS$ is a parallelogram.

Question: In a triangle $ABC$, $E$ is the midpoint of median $AD$. Show that $\text{Area}(\triangle BED) = \frac{1}{4} \text{Area}(\triangle ABC)$.

Solution:
$AD$ is a median of $\triangle ABC$, so $D$ is the midpoint of $BC$.

A median divides a triangle into two triangles of equal area:

Now, in $\triangle ABD$, $E$ is the midpoint of $AD$. So $BE$ is a median of $\triangle ABD$.

Therefore:

From $(i)$ and $(ii)$:

Hence proved.

Question: Two chords $AB$ and $CD$ of a circle are each $8$ cm long. If the radius of the circle is $5$ cm, find the distance of each chord from the centre.

Solution:
Let $O$ be the centre and let $OM \perp AB$.

The perpendicular from the centre bisects the chord:

In right $\triangle OMA$:

Since $AB = CD = 8$ cm (equal chords), and equal chords are equidistant from the centre, the distance of $CD$ from the centre is also $3$ cm.

Question: Construct a triangle $ABC$ in which $BC = 7$ cm, $\angle B = 75°$, and $AB + AC = 13$ cm.

Solution steps:
1. Draw $BC = 7$ cm.
2. At $B$, construct $\angle CBX = 75°$.
3. Cut $BD = 13$ cm (equal to $AB + AC$) on ray $BX$.
4. Join $DC$.
5. Construct the perpendicular bisector of $DC$. Let it meet $BD$ at $A$.
6. Join $AC$.

$\triangle ABC$ is the required triangle.

Why this works: $A$ lies on the perpendicular bisector of $DC$, so $AD = AC$. Therefore $BD = BA + AD = BA + AC = 13$ cm, which is the given condition.

Question: A field is in the shape of a quadrilateral $ABCD$ where $AB = 9$ m, $BC = 40$ m, $CD = 28$ m, $DA = 15$ m, and $\angle B = 90°$. Find the area of the field.

Solution:
Since $\angle B = 90°$, use Pythagoras in $\triangle ABC$:

Area of $\triangle ABC$ (right triangle):

For $\triangle ACD$ with sides $AC = 41$, $CD = 28$, $DA = 15$:

Total area $= 180 + 126 = 306$ m$^2$.

Question: A metallic sphere of radius $4.2$ cm is melted and recast into the shape of a cylinder of radius $6$ cm. Find the height of the cylinder.

Solution:
Volume of sphere $=$ Volume of cylinder (metal is conserved).

The height of the cylinder is $2.744$ cm.

Question: A cone has radius $7$ cm and height $24$ cm. Find its total surface area.

Solution:
Slant height:

Question: The marks of $30$ students are given below. Find the mean.

Solution:

$\sum f_i x_i = 4(10) + 6(20) + 10(30) + 7(40) + 3(50)$
$= 40 + 120 + 300 + 280 + 150 = 890$

$\sum f_i = 30$

The mean marks are $29.67$ (approximately $29.\overline{6}$).

Question: A die is thrown $200$ times and the following outcomes are recorded.

Find the probability of getting (i) a prime number, (ii) a number greater than $4$.

Solution:
Total trials $= 200$.

(i) Prime numbers on a die: $2, 3, 5$.
Frequency $= 36 + 32 + 28 = 96$.

(ii) Numbers greater than $4$: $5, 6$.
Frequency $= 28 + 34 = 62$.