Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 8 Quadrilaterals Important Questions and Answers.

I. Multiple Choice Questions :

Choose the correct answer from the given options.

Question 1.

Which of the following is not true for a parallelogram?

(a) Opposite sides are equal.

(b) Opposite angles are equal.

(c) Opposite angles are bisected by the diagonals.

(d) Diagonals bisect each other.

Answer:

(c) Opposite angles are bisected by the diagonals.

Question 2.

Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is :

(a) 90°

(b) 95°

(c) 105°

(d) 120°

Answer:

(d) 120°

Question 3.

ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is :

(a) 40°

(b) 45°

(c) 50°

(d) 60°

Answer:

(c) 50°

Question 4.

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonal is :

(a) 55°

(b) 50°

(c) 40°

(d) 25°

Answer:

(b) 50°

Question 5.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is square only if:

(a) ABCD is a rhombus.

(b) diagonals of ABCD are equal.

(c) diagonals of ABCD are perpendicular to each other.

(d) diagonals of ABCD are equal and perpendicular to each other.

Answer:

(d) diagonals of ABCD are equal and perpendicular to each other.

Question 6.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if:

(a) PQRS is a parallelogram.

(b) PQRS is a rectangle.

(c) the diagonals of PQRS are perpendicular to each other.

(d) the diagonals of PQRS are equal.

Answer:

(c) the diagonals of PQRS are perpendicular to each other.

Question 7.

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD form:

(a) a square

(b) a rhombus

(c) a rectangle

(d) any other parallelogram

Answer:

(c) a rectangle

Question 8.

D and E are mid-points of the sides AB and AC of ΔABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then ΔEQP is:

(a) a square

(b) a rectangle

(c) a rhombus

(d) a parallelogram

Answer:

(d) a parallelogram

Question 9.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rhombus if:

(a) ABCD is a parallelogram.

(b) ABCD is a rhombus.

(c) the diagonals of ABCD are equal.

(d) the diagonals of ABCD are perpendicular to each other.

Answer:

(c) the diagonals of ABCD are equal.

Question 10.

In the given figure, ABCD is a parallelogram and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then

(a) AF = \(\frac{3}{2}\)AB

(b) AF = 2AB

(c) AF = 3AB

(d) AP^{2} = 2AB^{2}

Answer:

(b) AF = 2AB

Question 11.

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?

(a) 45°

(b) 55°

(c) 60°

(d) 75°

Answer:

(c) 60°

Question 12.

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB = ?

(a) 46°

(b) 60°

(c) 90°

(d) 30°

Answer:

(c) 90°

Question 13.

In the given figure, AD is a median of ΔABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?

(a) \(\frac{1}{2}\)AC

(b) \(\frac{1}{3}\)AC

(c) \(\frac{2}{3}\)AC

(d) \(\frac{3}{4}\)AC

Answer:

(b) \(\frac{1}{3}\)AC

Question 14.

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30° and ∠AOB = 70°. Then ∠DBC = ?

(a) 40°

(b) 35°

(c) 45°

(d) 50°

Answer:

(a) 40°

Question 15.

In the given figure, ABCD is a rhombus. Then:

(a) AC^{2} + BD^{2} = AB^{2}

(b) AC^{2} + BD^{2} = 2AB^{2}

(c) AC^{2} + BD^{2} = 4AB^{2}

(d) 2(AC^{2} + BD^{2}) = 3AB^{2}

Answer:

(c) AC^{2} + BD^{2} = 4AB^{2}

II. Fill in the Blanks :

Question 1.

If the angles of a quadrilateral are (4x)°, (7x)°, (15x)° and (10x)°. Then the smallest angle of this quadrilateral is _________.

Answer:

40

Question 2.

In figure, PQRS is a parallelogram in which a pair of opposite angles is given. The value of x is _________.

Answer:

15

Question 3.

ABCD is rhombus such that ∠ACB = 40°, then ∠ADB is _________.

Answer:

50

Question 4.

In the figure, ΔPQR is formed by joining the mid-points of sides BC, CA, and AB respectively. If AABC is an equilateral triangle with side 12 cm, then the length of PQ is _________.

Answer:

6 cm

Question 5.

In a triangle, the line segment joining the mid-points of any given two sides is to the third side and is _________.

Answer:

Parallel, half of it.

III. True/False:

State whether the following statements are True or False.

Question 1.

A quadrilateral is a parallelogram if its opposite sides are parallel.

Answer:

True

Question 2.

In a parallelogram opposite angles are equal.

Answer:

True

Question 3.

Every square is a rectangle.

Answer:

True

Question 4.

In a parallelogram, diagonals bisect each other.

Answer:

True

Question 5.

Every square is a rhombus.

Answer:

True

Question 6.

The sum of the angles of a quadrilateral is 180°.

Answer:

True

Question 7.

In a parallelogram, the diagonals interesect at right angles.

Answer:

False

IV. Match the Columns :

Question 1.

Match the column I with the column II.

Column I |
Column II |

(1) Angle bisectors of a parallelogram form a |
(i) parallelogram |

(2) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a |
(ii) rectangle |

(3) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a |
(iii) square |

(4) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a |
(iv) rhombus |

Answer:

Column I |
Column II |

(1) Angle bisectors of a parallelogram form a |
(ii) rectangle |

(2) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a |
(iii) square |

(3) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a |
(iv) rhombus |

(4) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is a |
(i) parallelogram |

V. Very Short Answer Type Questions :

Question 1.

Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35° then ∠B = 145°. Is it true? Also justify your answer.

Answer:

It is true that if ∠A = 35° then ∠B = 145°. It is given that ABCD is a quadrilateral and diagonals of quadrilateral ABCD bisect each other. Then, it should be a parallelogram.

Also, we know that ∠A and ∠B are adjacent angles of parallelogram ABCD ∠A + ∠B = 180°

⇒ 35° + 145° = 180°

Hence ∠B = 145°.

Question 2.

In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 60° then what is the value of ∠GFE?

Answer:

∠A = ∠C (Opposite angles of parallelogram ABCD are equal)

Also, AEFG is a parallelogram (Given)

⇒ ∠GFE = ∠A (Opposite angle of parallelogram are equal)

⇒ ∠GFE = 60°

VI. Short Answer Type Questions :

Question 1.

In quadrilateral ABCD, ∠A + ∠C = 140°, ∠A : ∠C = 1 : 3 and ∠B : ∠D = 5:6. Find the ∠A, ∠B, ∠C and ∠D.

Answer:

In quadrilateral ABCD,

∠A + ∠C = 140° .................... (1)

∠A : ∠C = 1 : 3 .................. (2)

and ∠B : ∠D = 5 : 6 ................ (3)

Let ∠A = x and ∠C = 3x.

We know that, sum of angles of a quadrilateral = 360°

∠A + ∠B + ∠C + ∠D = 360°

⇒ ∠B + ∠D = 360° - (∠A + ∠C)

⇒ ∠B + ∠D = 360° - (35° + 105°) = 220°

Since, ∠B : ∠D = 5 : 6

⇒ ∠B : ∠D = 5 : 6

⇒ 5x + 6x = 11x

⇒ 11x = 220°

x = 20°

∠B = 5x = 5 × 20 = 100°

∠D = 6x = 6 × 20 = 120°

Hence, ∠A = 35°, ∠B = 100°, ∠C = 105° and ∠D = 120°.

Question 2.

In the adjoining figure, PQRS is a rhombus, SQ and PR are the diagonals of the rhombus intersecting at point O. If ∠OPQ = 35°, then find the value of ∠ORS + ∠OQP.

Answer:

Given : PQRS is a rhombus

⇒ PQRS is a parallelogram

Since, PQ ∥ SR and PR is a transversal.

∴ ∠ORS = ∠OPQ = 35° (Alternate interior angles) ....................(1)

Also, diagonals of a rhombus bisect each other at right angle.

∴ ∠SOR = 90° .............(2)

Now, in ΔSOR, ∠RSO + ∠SOR + ∠ORS = 180° (Angle sum property of triangles)

⇒ ∠RSO = 180° - ∠SOR - ∠ORS

⇒ ∠RSO = 180° - 90° - 35°

⇒ ∠RSO = 55°

⇒ ∠OQP = 55°

(Alternate interior angles, since PQ ∥ SR and QS is a transversal)

From equations (1) and (3), we get

∠ORS + ∠OQP = 35° + 55° = 90°

Question 3.

ABCD is a kite with AB = AD and CD = CR. Prove that the figure formed by joining the mid-points of the consecutive sides is a rectangle.

Answer:

We draw the figure as shown below.

Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA respectively. Join AC and BD. Which intersects each other at O.

Now, in ΔABC, PQ ∥ AC and PQ = \(\frac{1}{2}\)AC ............ (1) (By mid-point theorem)

and in ΔACD, RS ∥ AC and RS = \(\frac{1}{2}\)AC .................(2) (By mid-point theorem)

From equations (1) and (2) we get,

PQ ∥ RS and PQ = RS (Since, a pair of opposite sides equal and parallel)

So, PQRS is a parallelogram

Also, AB = AD (Given) ...................(1)

So, A lies on the perpendicular bisector of BD further

CB = CD (Given) ......................(2)

So, C lies on the perpendicular bisector of BD. .................(3)

From equations (2) and (3), AC is the perpendicular bisector of BD.

i.e. AC ⊥ BD ⇒ ∠AOD = 90°

Now, clearly HS ∥ OE and SE ∥ OH

So, SEOH is a parallelogram.

Hence, ∠ESH = ∠EOH = 90°

So, parallelogram PQRS is a rectangle.

Question 4.

ABCD is a rhombus such that ∠ACB = 40°, then find ∠ADB.

Answer:

Given : ABCD is a rhombus.

AB = CD = BC = AD

In ΔABC, AB = BC

∴ ∠ACB = ∠CAB = 40°

(Angles opposite to equal sides are equal)

Now, AB ∥ CD and AC is transversal.

∴ ∠DCA = ∠CAB = 40° (Alternate angles)

Now, ∠C = ∠BCD - ∠BCA + ∠DCA

= 40° + 40° = 80°

We know that, ∠D + ∠C = 180° (Co-interior angles of a rhombus)

∠D =180° - 80° = 100°

∴ ∠ADB = \(\frac{1}{2}\)∠D = \(\frac{100^{\circ}}{2}\) = 50°

Question 5.

In the adjoining figure, P is the mid-point of side BC of a parallelogram ABCD, such that ∠BAP = ∠DAP. Prove that AD = 2 CD.

Answer:

Given : ABCD is a parallelogram. P is the mid-point of BC,

i.e. BC = 2BP ⇒ AD = 2BP

∠BAP = ∠DAP .

AD = 2 CD

∠BAP = ∠DAP

∠DAP = ∠APB

∠BAP = ∠APB ⇒ BP = AB

(Sides opposite to equal angles of a triangle are equal)

AB = CD (Opposite sides of parallelogram)

BP = CD (∵ BP = AB) ...................(1)

BC = 2BP (Given)

⇒ AD = 2BP .................(2)

(∵ BC = AD, opposite sides of parallelogram)

From equations (1) and (2), we have :

AD = 2CD

Hence proved.

Question 6.

In quadrilateral ABCD, AP and BP are bisectors of ∠A and ∠B respectively, then find the value of x.

Answer:

From the fig.

∠1 = ∠2, ∠3 = ∠4 ...(1)

(AP and BP are bisectors of ∠A and ∠B.)

ABCD is quadrilateral,

So, ∠A + ∠B + ∠C + ∠D = 360° (Sum of all angles of a quad.)

∴ ∠1 + ∠2 + ∠3 + ∠4 + 130° + 60° = 360°

⇒ 2∠1 + 2∠3 + 190° = 360° [From (1)]

⇒ 2∠1 + 2∠3 = 360° - 190° = 170°

⇒ ∠1 + ∠3 = \(\frac{170^{\circ}}{2}\) = 85° ...(2)

In ΔAPB, ∠1 + x + ∠3 = 180° (Angle sum property of a A)

⇒ ∠1 + ∠3 + x = 180°

⇒ 85° + x = 180°

∴ x = 180° - 85° = 95°

Hence, x = 95°

Question 7.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Answer:

Given : Parallelogram ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Now, BEDF is a quadrilateral, in which

∠BED = ∠BFD = 90°

∴ ∠FBE = 360° - (∠FDE + ∠BED + ∠BFD)

(Angle sum property of a quadrilateral) = 360° - (60° + 90° + 90°) = 360° - 240° = 120°

Since, ABCD is a parallelogram.

∠ADC = 120° [∵ ∠D = ∠B]

Now, ∠A + ∠B = 180° (Co-interior angles of a parallelogram)

∴ ∠A = 180° - ∠B = 180° - 120° (∵ ∠FBE = ∠B)

⇒ ∠A = 60°

Also, ∠C = ∠A = 60°

(∵ Opposite angles of a parallelogram are equal.)

Hence, angles of the parallelogram are 60°, 120°, 60° and 120°, respectively.

Question 8.

PQ and RS are two equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other.

Answer:

It is given that PQ = RS and PQ ∥ RS.

Since, a pair of opposite side of a quadrilateral PQSR is equal and parallel.

Therefore, PQSR is a parallelogram.

So, PR = QS and PR ∥ QS

Now, PR ∥ QS

Therefore, ∠RPQ + ∠PQS = 180°

(Interior angles on the same side of the transversal)

i.e. ∠RPQ + ∠PQM + ∠MQS = 180°

Also, PN ∥ QM (By Construction)

Therefore, ∠NPQ + ∠PQM = 180°

i.e. ∠NPR + ∠RPQ + ∠PQM = 180° ...........(3)

So, ∠NPR = ∠MQS [From equation (2) and (3)] ...........(4)

Similarly ∠NRP = ∠MSQ, ..........(5)

In ΔPNR and ΔQMR PR = QS [Using equation (1)]

∠NPR = ∠MQS [Using equation (4)]

∠NRP = ∠MSQ [Using equation (5)]

∠PNR = ΔQMS

So, PN = QM and NR = MS

PN = QM and PN ∥ QM, we have PQMN is a parallelogram.

(Since, a pair of opposite side of a quadrilateral equal and parallel)

So, NM = PQ and NM ∥ PQ.

Hence proved.

Question 9.

In the given figure, ABCD is a square and EF ∥ BD. M is the mid-point of EF. Prove that AM bisects ∠BAD.

Answer:

Given : ABCD is a square and BD is a diagonal.

∠CBD = ∠CDB = x 90° = 45°

(∵ Diagonal of a square bisect each angle at the vertex)

Also, EF ∥ BD (Given)

So, ∠CEF = ∠CBD = 45° (Corresponding angles)

and ∠CFE = ∠CDB = 45° . (Corresponding angles)

⇒ CE = CF (∵ Sides opposite to equal angles are equal.)

⇒ BC - CE = CD - CF (∵ BC = CD)

⇒ BE = DF ................(1)

Now, in AABE and AADF,

AB = AD (Adjacent sides of a square)

∠ABE = ∠ADF (Each 90°)

BE = DF [From equation (1)]

So, ΔABE ≅ ΔADF (By SAS congruence rule)

Then, AE = AF (By CPCT) .........(2)

and ∠BAE = ∠DAF .............(3)

Now, in ΔAEM and ΔAFM,

AE = AF [From equation (2)]

ME = MF (M is mid-point of EF)

AM = AM (Common side)

ΔAEM ≅ ΔAFM (By SSS congruence rule)

So, ∠EAM = ∠FAM (By CPCT) ...........(4)

On adding equations (3) and (4), we get

∠BAE + ∠EAM = ∠DAF + ∠FAM

⇒ ∠BAM = ∠DAM

i.e. AM bisects ∠BAD.

Hence proved.

Question 10.

ABCD is a parallelogram in which P and Q are mid-points of opposite sides AB and DC (given in figure). If AQ intersects DP at S and BQ intersects CP at R, show that:

(i) APCQ is a parallelogram.

(ii) DPBQ is a parallelogram.

(iii) PSQR is a parallelogram.

Answer:

Given : ABCD is a parallelogram.

∴ AB ∥ DC, AB = DC

and AD ∥ BC, AD = BC

P and Q are mid-points of sides AB and CD.

(i) In quadrilateral APCQ,

AP ∥ QC [∵ AB ∥ DC] ...................(1)

Now, AP = \(\frac{1}{2}\)AB, QC = \(\frac{1}{2}\)DC .................. (2)

(∵ P and Q are mid-points of AB and CD, respectively.)

Since, AB = DC (Given)

⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)CD

⇒ AP = QC [Using equation (2)]

Thus, we have AP ∥ QC and AP = QC

Hence, APCQ is a parallelogram.

(ii) Similarly, in quadrilateral DPBQ,

BP ∥ QD and BP = QD

DPBQ is a parallelogram.

(iii) Since, DPBQ is a parallelogram,

∴ DP ∥ QB

⇒ SP ∥ QR .................(3)

and APCQ is also a parallelogram.

∴ AQ ∥ PC

SQ ∥ PR ..............(4)

From equations (3) and (4), we have :

SP ∥ QR and SQ ∥ PR

Hence, PSQR is also a parallelogram.

Hence proved.

Question 11.

In the following figure, ABCD is a rhombus. If ∠ABC = 68°, then determine ∠ACD.

Answer:

Given : ABCD is a rhombus.

⇒ ABCD is a parallelogram

⇒ ∠ADC = ∠ABC = 68° (v Opposite angles of a parallelogram are equal.)

∠ODC = \(\frac{1}{2}\)∠ADC ( Diagonal DB of rhombus ABCD bisects its ∠ABC and ∠ADC)

∠ODO = \(\frac{1}{2}\) × 68°= 34°

Now, in ∠OCD

∠OCD + ∠ODC + ∠COD = 180° (Angle sum property of a triangle)

∠OCD + 340 + 90° = 180° (∵ ∠COD = 90°, diagonals of a rhombus bisect each other at right angles)

⇒ ∠OCD + 124° = 180°

⇒ ∠OCD = 180° - 124°

⇒ ∠OCD = 56° or ∠ACD = 56°

Question 12.

In the given figure, ABCD is a square, if ∠PQR = 90° and PB - QC = DR, then prove that QB = RC, PQ = QR and ∠QPR = 45°.

Answer:

Given : ABCD is a square, ∠PQR = 90° and PB = QC = DR.

To prove : QB = RC, PQ = QR and ∠QPR = 450

Proof: Since, ABCD is a square.

BC = CD

But QC = DR (Given)

⇒ BC - QC = CD - DR

⇒ QB = RC ................. (1)

Now, in ΔPBQ and ΔQCR, PB = QC (Given)

BQ = CR [From equation (1)]

∠PBQ = ∠QCR = 90°

∴ ΔPBQ ≅ ΔQCR (By SAS congruence rule)

So, PQ = QR (By CPCT)

Also, ∠BPQ = ∠CQR and ∠BQP = ∠CRQ

Now, since APQR is an isosceles right angled triangle.

So, ∠PRQ + ∠QPR + ∠PQR = 180° (Angle sum property of triangle)

⇒ ∠QPR + ∠QPR + 90° = 180° (Since, ∠PQR is isosceles)

⇒ ∠QPR = 90°

∴ ∠QPR = 45°

Hence proved.

VII. Long Answer Type Questions:

Question 1.

Prove that the opposite angles of an isosceles trapezium are supplementary.

Answer:

Let ABCD be an isosceles trapezium AB ∥ DC and AD = BC.

Through C, draw CE ∥ DA, AB ∥ DC i.e., AE ∥ DC and CE ∥ DA (By construction)

AECD is a parallelogram AD = EC (Opp. sides of a ∥gm)

AD = BC (Given)

BC = EC

In ΔCEB, BC = EC

∠3 = ∠2 (Angles opp. equal sides are equal.) ...(1)

As DA ∥ CE and AE is a transversal.

∠1 = ∠2 (Corresponding ∠s) ...(2)

Also AE ∥ DC and CE is a transversal

∠5 = ∠2 (Alternate ∠s)

Now ∠A + ∠C = ∠1 + (∠5 + ∠4)

⇒ ∠A + ∠C = ∠2 + ∠2 + ∠4 [Using (2) and (3)]

∠A + ∠C = ∠2 + ∠3 + ∠4 [Using (1)]

∠A + ∠C = 180° (Sum of ∠s of a Δ = 180°)

Now ∠A + ∠B + ∠C + ∠D = 360°

∠B + ∠D + 180 = 360° (Sum of ∠s of quadrilateral)

⇒ ∠B + ∠D = 180° (∵ ∠A + ∠C= 180°)

Hence, the opposite angles of an isosceles trapezium are supplementary.

Hence proved.

Question 2.

P, Q,R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD, such that AC ⊥ BD. Prove that PQRS is a rectangle.

Answer:

In the figure:

AC ⊥ BD

∠COD = ∠AOD = ∠AOB = ∠COB = 90°

In ΔADC, by mid-point theorem

SR ∥ AC and SR = \(\frac{1}{2}\)AC

Similarly in ΔABC, PQ ∥ AC and PQ = \(\frac{1}{2}\)AC

From equations (1) and (2) PQ ∥ SR and PQ = SR = \(\frac{1}{2}\)AC

Similarly, SP ∥ RQ and SP = RQ = \(\frac{1}{2}\)BD

In quadrilateral EDFR, DE ∥ FR, OF ∥ ER

∠EOF = ∠ERF = 90°

Hence, PQRS is a rectangle.

Question 3.

E is mid-point of the median AD of AABC and BE is produced to meet AC at F. Show that AF = \(\frac{1}{2}\)AC.

Answer:

Through D, draw a line parallel to BF to meet AC at G.

In ΔADG, E is mid-point of side AD and DG ∥ BF i.e.

DG ∥ EF, therefore by converse of mid-point theorem,

F is mid-point of AG i. e. AF = FG. ..........(i)

In ABCF, D is mid-point of side BC and DG ∥ BF, therefore, by converse of mid-point theorem, G is mid-point of CF,

i.e. FG = GC .......(ii)

From (i) and (ii), we get:

AF = FG = GC ........(iii)

Now AC = AF + FG + GC = AF + AF + AF [Using (iii)]

⇒ AC = 3AF ⇒ AF = \(\frac{1}{3}\)AC

Hence Proved.

- RBSE Class 9 Maths Important Questions in Hindi & English Medium
- RBSE Class 9 Maths Notes in Hindi & English Medium Pdf Download
- RBSE Solutions for Class 9 Maths in Hindi Medium & English Medium
- RBSE Class 9 Maths Notes Chapter 1 संख्या पद्धति
- RBSE Class 10 Maths Important Questions Chapter 11 रचनाएँ
- RBSE Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5
- RBSE Class 10 Maths Important Questions Chapter 14 सांख्यिकी
- RBSE Class 9 Maths Important Questions Chapter 11 रचनाएँ
- RBSE Class 9 Maths Important Questions Chapter 3 निर्देशांक ज्यामिति
- RBSE Class 9 Maths Important Questions Chapter 11 Constructions
- RBSE Class 9 Maths Important Questions Chapter 14 सांख्यिकी