NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Have you noticed that the things surrounding you, like windows, doors, your notebook or tablet, all have a similar four-sided shape? Well, these four-sided shapes are called quadrilaterals. Quadrilaterals can be of various types, such as a Parallelogram, a Rhombus, a Trapezium, etc. From NCERT Class 9 Maths, the chapter Quadrilaterals contains the concepts of Properties of a Parallelogram, theorems related to angles and sides, the Mid-point Theorem, etc. These concepts will help the students grasp more advanced geometry concepts easily and enhance their problem-solving ability in real-world applications.

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1. Quadrilaterals Class 9 Questions And Answers PDF Free Download
2. Quadrilaterals Class 9 NCERT Solutions - Important Formulae And Points
3. Quadrilaterals Class 9 NCERT Solutions
4. Quadrilaterals Class 9 Solutions - Exercise Wise
5. NCERT Solutions For Class 9 Maths - Chapter Wise
6. Importance of Solving NCERT Questions of Class 9 Maths Chapter 8
7. NCERT Solutions For Class 9 - Subject Wise
8. NCERT Class 9 Books and Syllabus

This article on NCERT solutions for class 9 Maths Chapter 8 Quadrilaterals offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 9 Maths. Students who are in need of Quadrilaterals class 9 solutions will find this article very useful. It covers all the important Class 9 Maths Chapter 8 question answers. These Quadrilaterals class 9 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 9 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.

Quadrilaterals Class 9 Questions And Answers PDF Free Download

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Quadrilaterals Class 9 NCERT Solutions - Important Formulae And Points

Quadrilateral:

• Sum of all angles = 360°

Parallelogram:

• A diagonal of a parallelogram divides it into two congruent triangles.

• In a parallelogram, the diagonals bisect each other.

• In a parallelogram, opposite angles are equal.

• In a parallelogram, opposite sides are equal.

Square:

• Diagonals of a square bisect each other at right angles and are equal, and vice versa.

Triangle:

• A line through the midpoint of a side of a triangle parallel to another side bisects the third side (Midpoint theorem).

• The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half the third side.

Parallelogram Angle Bisectors:

• In a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.

• If a diagonal of a parallelogram bisects one of the angles of a parallelogram, it also bisects the second angle.

Rectangle:

• The angle bisectors of a parallelogram form a rectangle.

• Each of the four angles of a rectangle is a right angle.

Rhombus:

• The diagonals of a rhombus are perpendicular to each other.

Quadrilaterals Class 9 NCERT Solutions

Class 9 maths chapter 8 NCERT solutions - Exercise: 8.1
Page number: 110-111, Total questions: 7

Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof : In $\mathrm{△}$ ABC and $\mathrm{△}$ BAD,

BC= AD (Opposite sides of parallelogram)

AC=BD (Given)

AB=AB (common)

$\mathrm{△}$ ABC $\cong$ $\mathrm{△}$ BAD (By SSS)

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (CPCT)

and $\mathrm{\angle }ABC+\mathrm{\angle }BAD={180}^{\circ }$ (co - interior angles)

$2\mathrm{\angle }BAD={180}^{\circ }$

$\mathrm{\angle }BAD={90}^{\circ }$

Hence, it is a rectangle.

Q2. Show that the diagonals of a square are equal and bisect each other at right angles.

Answer:

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and $\mathrm{\angle }COD={90}^{\circ }$

Proof : In $\mathrm{△}$ BAD and $\mathrm{△}$ ABC,

$\mathrm{\angle }BAD=\mathrm{\angle }ABC$ (Each ${90}^{\circ }$ )

AD=BC (Given )

AB=AB (common)

$\mathrm{△}$ BAD $\cong$ $\mathrm{△}$ ABC (By SAS)

BD=AC (CPCT)

In $\mathrm{△}$ AOB and $\mathrm{△}$ COD,

$\mathrm{\angle }$ OAB= $\mathrm{\angle }$ OCD (Alternate angles)

AB=CD (Given )

$\mathrm{\angle }$ OBA= $\mathrm{\angle }$ ODC (Alternate angles)

$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ COD (By AAS)

AO=OC ,BO=OD (CPCT)

In $\mathrm{△}$ AOB and $\mathrm{△}$ AOD,

OB=OD (proved above)

AB=AD (Given )

OA=OA (COMMON)

$\mathrm{△}$ AOB $\cong$ $\mathrm{△}$ AOD (By SSS)

$\mathrm{\angle }$ AOB= $\mathrm{\angle }$ AOD (CPCT)

$\mathrm{\angle }$ AOB+ $\mathrm{\angle }$ AOD = ${180}^{\circ }$

2. $\mathrm{\angle }$ AOB = ${180}^{\circ }$

$\mathrm{\angle }$ AOB = ${90}^{\circ }$

Hence, the diagonals of a square are equal and bisect each other at right angles.

Q3. (i) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that it bisects $\mathrm{\angle }C$ also.

Answer:

Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC ................1

$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA.................2 (Alternate angles)

$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD .................3 (Alternate angles)

From equation 1,2 and 3, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA...................4

Hence, diagonal AC bisect angle C also.

Q3. (ii) Diagonal AC of a parallelogram ABCD bisects $\mathrm{\angle }A$ (see Fig. $8.19$ ). Show that ABCD is a rhombus.

Answer:

Given: $\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BAC ................1

$\mathrm{\angle }$ DAC= $\mathrm{\angle }$ BCA.................2 (Alternate angles)

$\mathrm{\angle }$ BAC= $\mathrm{\angle }$ ACD .................3 (Alternate angles)

From equations 1,2, and 3, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ BCA...................4

From 2 and 4, we get

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC

In $\mathrm{△}$ ADC,

$\mathrm{\angle }$ ACD= $\mathrm{\angle }$ DAC (proved above )

AD=DC (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

Q4. (i) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that: ABCD is a square.

Answer:

Given: ABCD is a rectangle with AB=CD and BC=AD $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 2 and $\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4.

To prove: ABCD is a square.

Proof : $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 4 .............1(alternate angles)

$\mathrm{\angle }$ 3= $\mathrm{\angle }$ 4 ................2(given )

From 1 and 2, $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3.....................................3

In $\mathrm{△}$ ADC,

$\mathrm{\angle }$ 1= $\mathrm{\angle }$ 3 (from 3 )

DC=AD (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

Q4. (ii) ABCD is a rectangle in which diagonal AC bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$ . Show that: Diagonal BD bisects $\mathrm{\angle }B$ as well as $\mathrm{\angle }D$.

Answer:

In $\mathrm{△}$ ADB,

AD = AB (ABCD is a square)

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 7.................1(angles opposite to equal sides are equal )

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 8.................2 (alternate angles)

From 1 and 2, we have

$\mathrm{\angle }$ 7= $\mathrm{\angle }$ 8.................3

and $\mathrm{\angle }$ 7 = $\mathrm{\angle }$ 6.................4(alternate angles)

From 1 and 4, we get

$\mathrm{\angle }$ 5= $\mathrm{\angle }$ 6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Q5. (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$

Answer:

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$.
To prove : $\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

Q5. (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $AP=CQ$

Answer:

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

To prove : $AP=CQ$

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

$AP=CQ$ (CPCT)

Q5. (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$

Answer:

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

To prove : $\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$

Proof :

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

Q5. (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: $AQ=CP$

Answer:

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

To prove : $AQ=CP$

Proof :

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

$AQ=CP$ (CPCT)

Q5. (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$ (see Fig. $8.20$ ). Show that: APCQ is a parallelogram

Answer:

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

To prove: APCQ is a parallelogram

Proof :

In $\mathrm{\Delta }APDand\mathrm{\Delta }CQB,$

DP=BQ (Given )

$\mathrm{\angle }$ ADP= $\mathrm{\angle }$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$ (By SAS)

$AP=CQ$ (CPCT)...............................................................1

Also,

In $\mathrm{\Delta }AQBand\mathrm{\Delta }CPD,$

DP=BQ (Given )

$\mathrm{\angle }$ ABQ= $\mathrm{\angle }$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$ (By SAS)

$AQ=CP$ (CPCT)........................................2

From equation 1 and 2, we get

$AP=CQ$

$AQ=CP$

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Q6. (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.21$ ). Show that $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$

Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$ ,

$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$ )

$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)

Q6. (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $8.21$ ). Show that $AP=CQ$

Answer:

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To prove : $AP=CQ$

Proof: In $\mathrm{\Delta }APBand\mathrm{\Delta }CQD$ ,

$\mathrm{\angle }$ APB= $\mathrm{\angle }$ CQD (Each ${90}^{\circ }$ )

$\mathrm{\angle }$ ABP= $\mathrm{\angle }$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$ (By SAS)

$AP=CQ$ (CPCT)

Q7. (i) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that $\mathrm{\angle }A=\mathrm{\angle }B$

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\angle }A=\mathrm{\angle }B$

Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.

In AECD,

AE||DC (Given)

AD||CE (By construction)

Hence, AECD is a parallelogram.

AD=CE...............1(opposite sides of a parallelogram)

AD=BC.................2(Given)

From 1 and 2, we get

CE=BC

In $\mathrm{△}$ BCE,

$\mathrm{\angle }3=\mathrm{\angle }4$ .................3 (opposite angles of equal sides)

$\mathrm{\angle }2+\mathrm{\angle }3={180}^{\circ }$ ...................4(linear pairs)

$\mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }$ .....................5(Co-interior angles)

From 4 and 5, we get

$\mathrm{\angle }2+\mathrm{\angle }3=\mathrm{\angle }1+\mathrm{\angle }4$

$\therefore \mathrm{\angle }2=\mathrm{\angle }1\to \mathrm{\angle }B=\mathrm{\angle }A$ (Since, $\mathrm{\angle }3=\mathrm{\angle }4$ )

Q7. (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\mathrm{\angle }C=\mathrm{\angle }D$

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\angle }C=\mathrm{\angle }D$

Proof: Let $\mathrm{\angle }$ A be $\mathrm{\angle }$ 1, $\mathrm{\angle }$ ABC be $\mathrm{\angle }$ 2, $\mathrm{\angle }$ EBC be $\mathrm{\angle }$ 3, $\mathrm{\angle }$ BEC be $\mathrm{\angle }$ 4.

$\mathrm{\angle }1+\mathrm{\angle }D={180}^{\circ }$ (Co-interior angles)

$\mathrm{\angle }2+\mathrm{\angle }C={180}^{\circ }$ (Co-interior angles)

$\therefore \mathrm{\angle }1+\mathrm{\angle }D=\mathrm{\angle }2+\mathrm{\angle }C$

Thus, $\mathrm{\angle }C=\mathrm{\angle }D$ (Since , $\mathrm{\angle }1=\mathrm{\angle }2$ )

Q7. (iii) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove : $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$

Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (proved in (i) )

Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)

Q7. (iv) ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$ (see Fig. $8.23$ ). Show that diagonal AC $=$ diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Given: ABCD is a trapezium in which $AB\parallel CD$ and $AD=BC$

To prove: diagonal AC $=$ diagonal BD

Proof: In $\mathrm{\Delta }ABCand\mathrm{\Delta }BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\mathrm{\angle }ABC=\mathrm{\angle }BAD$ (proved in (i) )

Thus, $\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ (By SAS rule)

diagonal AC $=$ diagonal BD (CPCT)

Quadrilaterals Class 9 Solutions - Exercise Wise

Students can practice Class 9 Maths Chapter 8 question answers using the exercise link given below.

• Quadrilaterals Class 9 NCERT Exercise 8.1
• Quadrilaterals Class 9 NCERT Exercise 8.2

NCERT Solutions For Class 9 Maths - Chapter Wise

Importance of Solving NCERT Questions of Class 9 Maths Chapter 8

• Solving these NCERT questions will help students understand the basic concepts of Quadrilaterals easily.
• Students can practice various types of questions, which will improve their problem-solving skills.
• These NCERT exercises cover all the important topics and concepts so that students can be well-prepared for various exams.
• By solving these NCERT problems, students will get to know about all the real-life applications of Quadrilaterals.

NCERT Solutions For Class 9 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 9:

• NCERT Solutions for Class 9 Maths
• NCERT Solutions for Class 9 Science

NCERT Class 9 Books and Syllabus

Given below are some useful links for NCERT books and the NCERT syllabus for class 9:

• NCERT Books Class 9 Maths
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9