NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

In NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 introduces students to the fundamental properties of quadrilaterals, particularly focusing on the angle sum property. In this exercise, students learn that the sum of the interior angles of any quadrilateral is always 360 degrees. Through a series of problems, they apply this concept to calculate unknown angles, analyze different quadrilaterals, and understand the logical steps behind solving angle-related questions.

These solutions help simplify complex questions, ensuring that learners build a strong foundation for further geometry topics in higher classes. The NCERT solutions are designed to be clear and precise, guiding students step by step toward accurate answers and deeper understanding.

1.0Download NCERT Solutions of Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1 Free PDF

Get simple and accurate solutions for NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1. When you download the NCERT Solutions PDF, you will improve your understanding of quadrilaterals in order to efficiently approach problems before the exam.

Key Concepts of Exercise 8.1

• Definition and properties of a quadrilateral
• Interior angles of quadrilaterals
• Angle sum property: The sum of the interior angles of any quadrilateral is 360°
• Being able to solve for the unknown angles given the relevant information
• Understanding quadrilaterals based on angle properties and side properties

2.0NCERT Exercise Solutions Class 9 Chapter 8 Quadrilaterals : All Exercises 

Here is a quick overview and access to solutions for all exercises from Chapter 8, Quadrilaterals .

3.0NCERT Class 9 Maths Chapter 8 Exercise 8.1 : Detailed Solutions

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Sol. Given : ABCD is a parallelogram with diagonal AC = diagonal BD.

To prove : ABCD is a rectangle.

Proof : In triangle ABC and BAD:

AB=AB [Common side]

AC=BD [Given]

AD=BC [Opposite sides of a parallelogram are equal]

Therefore, ΔABC ≅ ΔBAD [By SSS congruency rule]

This implies ∠DAB = ∠CBA [By C.P.C.T. (Corresponding Parts of Congruent Triangles)] ... (i)

Since AD∥BC and transversal AB cuts them, the sum of the interior angles on the same side of the transversal is 180°.

∠DAB + ∠CBA = 180° ... (ii)

From equations (i) and (ii), since ∠DAB = ∠CBA, we can substitute:

∠DAB + ∠DAB = 180°

2∠DAB = 180°

∠DAB = 90°

So, ABCD is a parallelogram with one of its angles equal to 90°.

Hence, ABCD is a rectangle.

2. Show that the diagonals of a square are equal and bisect each other at right angles.

Sol. Given: ABCD is a square.

To Prove :

(i) AC = BD

(ii) AC and BD bisect each other at right angles.

Proof:

(i) To prove AC = BD:

In ΔABC and ΔBAD:

AB = BA [Common side]

BC = AD [Opposite sides of square ABCD are equal]

∠ABC = ∠BAD [Each = 90° (since ABCD is a square)]

Therefore, ΔABC ≅ ΔBAD [By SAS Congruence Rule]

Thus, AC = BD ... (i) [By C.P.C.T.]

(ii) To prove that diagonals bisect each other:

In ΔAOD and ΔCOB:

AD = CB [Opposite sides of square ABCD are equal]

∠OAD = ∠OCB [Alternate angles, as AD∥BC and transversal AC intersects them]

∠ODA = ∠OBC [Alternate angles, as AD∥BC and transversal BD intersects them]

Therefore, ΔAOD ≅ ΔCOB [By ASA Congruence Rule]

Thus, OA = OC and OB = OD ... (ii) [By C.P.C.T.]

So, O is the midpoint of AC and BD.

To prove that diagonals bisect each other at right angles:

In ΔAOB and ΔCOB:

AB = CB [Given, sides of a square are equal]

OA = OC [From (ii)]

OB = OB [Common side]

Therefore, ΔAOB ≅ ΔCOB [By SSS Congruence Rule]

Thus, ∠AOB = ∠BOC [By C.P.C.T.]

But ∠AOB + ∠BOC = 180° [Linear pair of angles]

Since ∠AOB = ∠BOC (proved earlier), substitute:

∠AOB + ∠AOB = 180°

2∠AOB = 180°

∠AOB = 180°/2 = 90°

Therefore, ∠AOB = ∠BOC = 90°.

Hence, AC and BD bisect each other at right angles.

3. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also

(ii) ABCD is a rhombus.

Sol. Given: Diagonal AC bisects ∠A of the parallelogram ABCD.

To prove:

(i) AC bisects ∠C also.

(ii) ABCD is a rhombus.

Proof:

(i) Since ABCD is a parallelogram, AB∥DC and AC intersects them.

Therefore, ∠BAC = ∠DCA [Alternate interior angles] ... (i)

Similarly, since AD∥BC and AC intersects them.

Therefore, ∠DAC = ∠BCA [Alternate interior angles] ... (ii)

Given that AC bisects ∠A, so ∠BAC = ∠DAC ... (iii)

From (i), (ii), and (iii), we have:

∠DCA = ∠BAC = ∠DAC = ∠BCA

This means ∠DCA = ∠BCA.

Thus, AC bisects ∠C.

(ii) From the relations derived in (i), we have ∠DAC = ∠DCA.

In ΔADC, sides opposite to equal angles are equal.

Therefore, AD = CD.

Also, ABCD is a parallelogram. So, opposite sides are equal: AD=BC and AB=CD.

Combining these, we get: AB = CD = AD = BC.

Hence, ABCD is a rhombus.

4. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

(i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

Sol. Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

To prove :

(i) ABCD is a square.

(ii) diagonal BD bisects ∠B as well as ∠D.

Proof:

(i) Since ABCD is a rectangle, AB∥DC and transversal AC intersects them.

So, ∠BAC = ∠DCA [Alternate interior angles]

Given that AC bisects ∠A, so ∠BAC = ∠DAC.

From the above two statements, ∠DCA = ∠DAC.

In ΔADC, sides opposite to equal angles are equal.

So, DA = CD.

But AB=CD and DA=BC (Opposite sides of a rectangle).

Therefore, AB = BC = CD = DA. (All sides are equal)

Also, ∠A=∠B=∠C=∠D=90° (Since ABCD is a rectangle).

Hence, ABCD is a square.

(ii) In ΔBAD and ΔBCD:

BA = BC [Since ABCD is a square, all sides are equal]

AD = CD [Since ABCD is a square, all sides are equal]

BD = BD [Common side]

Therefore, ΔBAD ≅ ΔBCD [By SSS congruence rule]

Thus, ∠ABD = ∠CBD [By C.P.C.T.] (BD bisects ∠B)

And ∠ADB = ∠CDB [By C.P.C.T.] (BD bisects ∠D)

Hence, diagonal BD bisects ∠B as well as ∠D.

5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ. Show that:

(i) ΔAPD≅ΔCQB

(ii) AP=CQ

(iii) ΔAQB≅ΔCPD

(iv) AQ=CP

(v) APCQ is a parallelogram

Sol. Given: ABCD is a parallelogram. P and Q are points on diagonal BD such that DP=BQ.

(i) In ΔAPD and ΔCQB:

DP = BQ [Given]

AD = CB [Opposite sides of parallelogram ABCD are equal]

∠ADP = ∠CBQ [Alternate interior angles, as AD∥BC and transversal BD intersects them]

Therefore, ΔAPD ≅ ΔCQB [By SAS congruence criteria]

(ii) By CPCT from ΔAPD ≅ ΔCQB, we have AP = CQ.

(iii) In ΔAQB and ΔCPD:

BQ = DP [Given (or DP=BQ)]

AB = CD [Opposite sides of parallelogram ABCD are equal]

∠ABQ = ∠CDP [Alternate interior angles, as AB∥DC and transversal BD intersects them]

Therefore, ΔAQB ≅ ΔCPD [By SAS congruence criteria]

(iv) By CPCT from ΔAQB ≅ ΔCPD, we have AQ = CP.

(v) From (ii) AP=CQ and from (iv) AQ=CP.

Since both pairs of opposite sides are equal, APCQ is a parallelogram.

6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

(i) ΔAPB≅ΔCQD

(ii) AP = CQ

Sol. Given : ABCD is a parallelogram and AP⊥BD, CQ⊥BD (perpendiculars from vertices A and C on diagonal BD respectively).

To prove :

(i) ΔAPB≅ΔCQD

(ii) AP=CQ

Proof:

(i) In ΔAPB and ΔCQD:

AB = CD [Opposite sides of parallelogram ABCD are equal]

∠ABP = ∠CDQ [Alternate interior angles, as AB∥DC and transversal BD intersects them]

∠APB = ∠CQD [Each = 90°, as AP and CQ are perpendiculars]

Therefore, ΔAPB ≅ ΔCQD [By AAS (Angle-Angle-Side) Congruence Rule]

(ii) By C.P.C.T. from ΔAPB ≅ ΔCQD, we have AP = CQ.

7. ABCD is a trapezium in which AB∥CD and AD=BC. Show that (figure)

(i) ∠A=∠B

(ii) ∠C=∠D

(iii) ΔABC≅ΔBAD

(iv) diagonal AC= diagonal BD

Sol. Given : 

ABCD is a trapezium with AB∥CD and AD=BC.

To Prove:

(i) ∠A=∠B

(ii) ∠C=∠D

(iii) ΔABC≅ΔBAD

(iv) Diagonal AC = Diagonal BD

Construction : Draw CE parallel to AD, intersecting extended AB at E.

Proof:

(i) Since AECD is a parallelogram (by construction CE∥AD and CD∥AE).

Therefore, AD = EC [Opposite sides of parallelogram]

But AD = BC [Given]

So, BC = EC.

In ΔBCE, since BC=EC, angles opposite to equal sides are equal.

Therefore, ∠CEB = ∠CBE.

Also, ∠DAE + ∠CEB = 180° (Co-Interior angles since AD∥CE and AE is transversal).

And ∠CBE + ∠CEB = 180° (Linear pair on line AE).

Since ∠DAE + ∠CEB = 180° and ∠CBE + ∠CEB = 180°, and ∠CEB = ∠CBE,

this implies ∠DAE = ∠CBE.

Therefore, ∠A = ∠B.

(ii) We know that the sum of adjacent angles on the non-parallel sides of a trapezium is 180°.

∠A + ∠D = 180°

∠B + ∠C = 180°

Since ∠A = ∠B (Proved in (i)), we can say:

∠A + ∠D = ∠A + ∠C

Therefore, ∠D = ∠C.

(iii) In ΔABC and ΔBAD:

AB = BA [Common side]

∠ABC = ∠BAD [Proved in (i), ∠B = ∠A]

BC = AD [Given]

Therefore, ΔABC ≅ ΔBAD [By SAS congruency rule]

(iv) Since ΔABC ≅ ΔBAD (Proved in (iii)),

Thus, AC = BD [By C.P.C.T.]

Hence, diagonal AC = diagonal BD.

8. The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Sol. Let the four angles of the quadrilateral be 3x, 5x, 9x and 13x.

The sum of all angles in a quadrilateral is 360°.

3x + 5x + 9x + 13x = 360°

30x = 360°

x = 360°/30

x = 12°

Hence, the angles of the quadrilateral are:

3x = 3 × 12° = 36°

5x = 5 × 12° = 60°

9x = 9 × 12° = 108°

13x = 13 × 12° = 156°

9. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Sol. Given : ABCD is a quadrilateral where diagonals AC and BD meet at O, such that AO=OC, OB=OD and AC⊥BD.

To Prove: Quadrilateral ABCD is a rhombus, i.e., AB=BC=CD=DA.

Proof: In ΔAOB and ΔAOD:

OB = OD [Given]

AO = AO [Common side]

∠AOB = ∠AOD [Each = 90°, since AC⊥BD]

Therefore, ΔAOB ≅ ΔAOD [By SAS Congruence Rule]

Thus, AB = AD [By C.P.C.T.] ... (i)

Similarly, consider ΔAOD and ΔCOD:

AO = AO [Common]

OD = OD [Common]

∠AOD = ∠COD [Each = 90°]

So, ΔAOD ≅ ΔCOD [SAS Rule]

Thus, AD = CD [C.P.C.T.] ... (ii)

Consider ΔBOC and ΔCOD:

OB = OD [Given]

OC = OC [Common]

∠BOC = ∠COD [Each = 90°]

So, ΔBOC ≅ ΔCOD [SAS Rule]

Thus, BC = CD [C.P.C.T.] ... (iii)

From (i), (ii), and (iii), we obtain: AB = AD = CD = BC.

Therefore, Quadrilateral ABCD is a rhombus.

10. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Sol. Given : The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.

To prove : Quadrilateral ABCD is a square.

Proof:

(i) To prove it is a parallelogram:

Given diagonals bisect each other (AO=OC, OB=OD).

In ΔAOD and ΔCOB:

OA = OC [Given]

OD = OB [Given]

∠AOD = ∠COB [Vertically Opposite Angles]

Therefore, ΔAOD ≅ ΔCOB [By SAS Rule]

Thus, AD = BC [C.P.C.T.]

And ∠ODA = ∠OBC [C.P.C.T.]

Since ∠ODA and ∠OBC are alternate interior angles and they are equal, AD∥BC.

Now, since AD=BC and AD∥BC, Quadrilateral ABCD is a parallelogram.

(ii) To prove all sides are equal (rhombus):

In ΔAOB and ΔAOD:

AO = AO [Common side]

OB = OD [Given]

∠AOB = ∠AOD [Each = 90° (Given diagonals bisect at right angles)]

Therefore, ΔAOB ≅ ΔAOD [By SAS Rule]

Thus, AB = AD [C.P.C.T.]

Since ABCD is a parallelogram and AB=AD, all sides are equal (AB=BC=CD=DA). So, ABCD is a rhombus.

(iii) To prove all angles are 90° (rectangle):

In ΔABC and ΔBAD:

AC = BD [Given]

BC = AD [Since ABCD is a parallelogram]

AB = BA [Common side]

Therefore, ΔABC ≅ ΔBAD [By SSS rule]

Thus, ∠ABC = ∠BAD [By C.P.C.T.]

Since AD∥BC (Opposite sides of parallelogram ABCD) and transversal AB intersects them.

Therefore, ∠ABC + ∠BAD = 180° [Sum of consecutive interior angles on the same side of the transversal is 180°]

Since ∠ABC = ∠BAD, substitute:

∠ABC + ∠ABC = 180°

2∠ABC = 180°

∠ABC = 90°

Similarly, ∠BCD = ∠ADC = 90°.

Since ABCD is a rhombus with one angle 90°, it is a square.

Therefore, ABCD is a square.

11. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Sol Given : ABCD is a rhombus and AC and BD are its diagonals.

To prove :

(i) Diagonal AC bisects ∠A as well as ∠C.

(ii) Diagonal BD bisects ∠B as well as ∠D.

Proof:

(i) Consider ΔABC.

AB = BC (Sides of Rhombus are equal)

In ΔABC, angles opposite to equal sides are equal. So, ∠BAC = ∠BCA. (Let this be ∠1 = ∠4)

Since AB∥CD (opposite sides of rhombus) and AC is a transversal, ∠BAC = ∠DCA (Alternate interior angles). (So ∠1 = ∠3)

Since AD∥BC (opposite sides of rhombus) and AC is a transversal, ∠DAC = ∠BCA (Alternate interior angles). (So ∠2 = ∠4)

From these relations: ∠1 = ∠3 (from alt. int. angles with AB∥CD) and ∠2 = ∠4 (from alt. int. angles with AD∥BC).

Also, in ΔABC, since AB=BC, ∠1 = ∠4.

Combining these, we get ∠1 = ∠2 = ∠3 = ∠4.

Since ∠1 = ∠2, AC bisects ∠A.

Since ∠3 = ∠4, AC bisects ∠C.

(ii) Consider ΔABD.

AB = AD (Sides of Rhombus are equal)

In ΔABD, angles opposite to equal sides are equal. So, ∠ABD = ∠ADB. (Let this be ∠5 = ∠7)

Since AB∥CD and BD is a transversal, ∠ABD = ∠CDB (Alternate interior angles). (So ∠5 = ∠8)

Since AD∥BC and BD is a transversal, ∠ADB = ∠DBC (Alternate interior angles). (So ∠7 = ∠6)

From these relations: ∠5 = ∠8 (from alt. int. angles with AB∥CD) and ∠7 = ∠6 (from alt. int. angles with AD∥BC).

Also, in ΔABD, since AB=AD, ∠5 = ∠7.

Combining these, we get ∠5 = ∠6 = ∠7 = ∠8.

Since ∠5 = ∠6, BD bisects ∠B.

Since ∠7 = ∠8, BD bisects ∠D.

12. In ΔABC and ΔDEF, AB=DE, AB∥DE, BC=EF and BC∥EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that :

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD∥CF and AD=CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC=DF

(vi) ΔABC≅ΔDEF

Sol. Given : AB=DE, AB∥DE, BC=EF & BC∥EF.

To Prove:

(i) ABED is a parallelogram.

(ii) BEFC is a parallelogram.

(iii) AD∥CF and AD=CF.

(iv) ACFD is a parallelogram.

(v) AC=DF.

(vi) ΔABC≅ΔDEF.

Proof:

(i) In quadrilateral ABED:

We are given AB = DE and AB∥DE.

A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.

Therefore, ABED is a parallelogram.

(ii) In quadrilateral BEFC:

We are given BC = EF and BC∥EF.

A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.

Therefore, BEFC is a parallelogram.

(iii) From (i), since ABED is a parallelogram, its opposite sides are parallel and equal.

So, AD∥BE and AD = BE ... (a)

From (ii), since BEFC is a parallelogram, its opposite sides are parallel and equal.

So, CF∥BE and CF = BE ... (b)

From (a) and (b), since both AD and CF are parallel to BE, they are parallel to each other (lines parallel to the same line are parallel to each other). So, AD∥CF.

Also, since both AD and CF are equal to BE, they are equal to each other. So, AD=CF.

(iv) In quadrilateral ACFD:

From (iii), we have proved AD∥CF and AD=CF.

A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.

Therefore, ACFD is a parallelogram.

(v) From (iv), since ACFD is a parallelogram, its opposite sides are equal.

Therefore, AC=DF.

(vi) In ΔABC and ΔDEF:

AB = DE [Given]

BC = EF [Given]

AC = DF [Proved in (v)]

Therefore, ΔABC ≅ ΔDEF [By SSS congruency rule]

4.0Key Features and Benefits: Class 9 Maths Chapter 8 Exercise 8.1

• Step by step support: All solutions are explained to be fully understood 
• Understanding Concepts: It concentrates on get an understanding of the angle sum property of quadrilaterals 
• Prepared for exams: It will be very easy to use the solutions as they are modelled with CBSE's marking scheme and syllabus 
• Enhances logical reasoning: The solutions will encourage the students to build deductions and use their angle rules.
• Useful for Upper classes: It will provide a great foundation for Geometry in Class 10 and above.