NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Edited By Ramraj Saini | Updated on Sep 27, 2023 10:40 PM IST

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Quadrilaterals class 9 questions and answers are discussed here. These NCERT solutions are developed by the expert at Careers360 taking care of latest CBSE syllabus 2023. the solutions are provided for all the NCERT problems in simple and easy to understand language covering step by step all the concepts which ultimately help during the exam.

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
Quadrilaterals Class 9 Questions And Answers PDF Free Download
Quadrilaterals Class 9 Solutions - Important Formulae And Points
Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)
Summary Of Class 9 Quadrilaterals NCERT Solutions
Important Concepts Learned in NCERT Maths Chapter 8 Class 9 – Quadrilaterals
Quadrilaterals Class 9 Solutions - Exercise Wise
NCERT Solutions For Class 9 Maths - Chapter Wise
NCERT Solutions For Class 9 - Subject Wise

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

In class 9 maths chapter 8 question answer, you will get to know about angles sum property, rhombus, parallelogram, square, rectangle, trapezium, kite, etc. The chapter is starting with recollecting the angle sum property of a quadrilateral. In total there are 2 exercises that consist of 37 questions. NCERT solutions for class 9 maths chapter 8 Quadrilaterals has the solutions to all the 37 questions. Here you will get NCERT solutions for class 9 Maths also.

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Quadrilaterals Class 9 Questions And Answers PDF Free Download

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Quadrilaterals Class 9 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, 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 mid-points 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.

Free download NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals for CBSE Exam.

Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 8 NCERT solutions - Exercise: 8.1

Q1 The angles of quadrilateral are in the ratio $\small 3:5:9:13$ . Find all the angles of the quadrilateral.

Answer:

Given : The angles of a quadrilateral are in the ratio $\small 3:5:9:13$ .
Let the angles of quadrilateral be $3x,5x,9x ,13x$ .

Sum of all angles is 360.

Thus, $3x+5x+9x +13x=360 \degree$

$\Rightarrow 30x=360 \degree$

$\Rightarrow x=12 \degree$

All four angles are : $3x=3\times 12=36\degree$

$5x=5\times 12=60 \degree$

$9x=9\times 12=108 \degree$

$13x=13\times 12=156 \degree$

Q2 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 $\triangle$ ABC and $\triangle$ BAD,

1640170405939

BC= AD (Opposite sides of parallelogram)

AC=BD (Given)

AB=AB (common)

$\triangle$ ABC $\cong$ $\triangle$ BAD (By SSS)

$\angle ABC=\angle BAD$ (CPCT)

and $\angle ABC+\angle BAD=180 \degree$ (co - interior angles)

$2\angle BAD=180 \degree$

$\angle BAD= 90 \degree$

Hence, it is a rectangle.

Q3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In $\triangle$ AOB and $\triangle$ AOD,

1640170440375

$\angle AOB=\angle AOD$ (Each $90 \degree$ )

BO=OD (Given )

AO=AO (common)

$\triangle$ AOB $\cong$ $\triangle$ AOD (By SAS)

AB=AD (CPCT)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Q4 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 $\angle COD=90 \degree$

Proof : In $\triangle$ BAD and $\triangle$ ABC,

1651648000412

$\angle BAD = \angle ABC$ (Each $90 \degree$ )

AD=BC (Given )

AB=AB (common)

$\triangle$ BAD $\cong$ $\triangle$ ABC (By SAS)

BD=AC (CPCT)

In $\triangle$ AOB and $\triangle$ COD,

$\angle$ OAB= $\angle$ OCD (Alternate angles)

AB=CD (Given )

$\angle$ OBA= $\angle$ ODC (Alternate angles)

$\triangle$ AOB $\cong$ $\triangle$ COD (By AAS)

AO=OC ,BO=OD (CPCT)

In $\triangle$ AOB and $\triangle$ AOD,

OB=OD (proved above)

AB=AD (Given )

OA=OA (COMMON)

$\triangle$ AOB $\cong$ $\triangle$ AOD (By SSS)

$\angle$ AOB= $\angle$ AOD (CPCT)

$\angle$ AOB+ $\angle$ AOD = $180 \degree$

2. $\angle$ AOB = $180 \degree$

$\angle$ AOB = $90 \degree$

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

Q5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer:

Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, $\angle$ COD = $90 \degree$

To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

Thus, AB=BC=CD=DA

In $\triangle$ BAD and $\triangle$ ABC,

AD=BC (proved above )

AB=AB (common)

BD=AC

$\triangle$ BAD $\cong$ $\triangle$ ABC (By SSS)

$\angle BAD = \angle ABC$ (CPCT)

$\angle$ BAD+ $\angle$ ABC = $180 \degree$ (Co-interior angles)

2. $\angle$ ABC = $180 \degree$

$\angle$ ABC = $90 \degree$

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

Q6 (i) Diagonal AC of a parallelogram ABCD bisects $\small \angle A$ (see Fig. $\small 8.19$ ). Show that

it bisects $\small \angle C$ also.

1640170496650

Answer:

Given: $\angle$ DAC= $\angle$ BAC ................1

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

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

From equation 1,2 and 3, we get

$\angle$ ACD= $\angle$ BCA...................4

Hence, diagonal AC bisect angle C also.

Q6 (ii) Diagonal AC of a parallelogram ABCD bisects $\small \angle A$ (see Fig. $\small 8.19$ ). Show that

ABCD is a rhombus.

1640170530307

Answer:

Given: $\angle$ DAC= $\angle$ BAC ................1

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

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

From equation 1,2 and 3, we get

$\angle$ ACD= $\angle$ BCA...................4

From 2 and 4, we get

$\angle$ ACD= $\angle$ DAC

In $\triangle$ ADC,

$\angle$ ACD= $\angle$ 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.

Q7 ABCD is a rhombus. Show that diagonal AC bisects $\small \angle A$ as well as $\small \angle C$ and diagonal BD bisects $\small \angle B$ as well as $\small \angle D$ .

Answer:

1640170558677

In $\triangle$ ADC,

AD = CD (ABCD is a rhombus)

$\angle$ 3= $\angle$ 1.................1(angles opposite to equal sides are equal )

$\angle$ 3= $\angle$ 2.................2 (alternate angles)

From 1 and 2, we have

$\angle$ 1= $\angle$ 2.................3

and $\angle$ 1= $\angle$ 4.................4 (alternate angles)

From 1 and 4, we get

$\angle$ 3= $\angle$ 4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

1640170587935

In $\triangle$ ADB,

AD = AB (ABCD is a rhombus)

$\angle$ 5= $\angle$ 7.................6(angles opposite to equal sides are equal )

$\angle$ 7= $\angle$ 6.................7 (alternate angles)

From 6 and 7, we have

$\angle$ 5= $\angle$ 6.................8

and $\angle$ 5= $\angle$ 8.................9(alternate angles)

From 6 and 9, we get

$\angle$ 7= $\angle$ 8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Q8 (i) ABCD is a rectangle in which diagonal AC bisects $\small \angle A$ as well as $\small \angle C$ . Show that:

ABCD is a square

Answer:

1640170617153

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

To prove: ABCD is a square.

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

$\angle$ 3= $\angle$ 4 ................2(given )

From 1 and 2, $\angle$ 1= $\angle$ 3.....................................3

In $\triangle$ ADC,

$\angle$ 1= $\angle$ 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.

Q8 (ii) ABCD is a rectangle in which diagonal AC bisects $\small \angle A$ as well as $\small \angle C$ . Show that:

diagonal BD bisects $\small \angle B$ as well as $\small \angle D$ .

Answer: 1651648175820

In $\triangle$ ADB,

AD = AB (ABCD is a square)

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

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

From 1 and 2, we have

$\angle$ 7= $\angle$ 8.................3

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

From 1 and 4, we get

$\angle$ 5= $\angle$ 6.................5

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

Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small \Delta APD\cong \Delta CQB$

1640170652426

Answer:

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

To prove : $\small \Delta APD\cong \Delta CQB$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small AP=CQ$

1640170672862

Answer:

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

To prove : $\small AP=CQ$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

$\small AP=CQ$ (CPCT)

Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small \Delta AQB\cong \Delta CPD$

1640170699729

Answer:

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

To prove : $\small \Delta AQB\cong \Delta CPD$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: $\small AQ=CP$

1640170719974

Answer:

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

To prove : $\small AQ=CP$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

$\small AQ=CP$ (CPCT)

Q9 (v) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$ (see Fig. $\small 8.20$ ). Show that: APCQ is a parallelogram

1640170738363

Answer:

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

To prove: APCQ is a parallelogram

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ (Given )

$\angle$ ADP= $\angle$ CBQ (alternate angles)

AD=BC (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$ (By SAS)

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

Also,

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ (Given )

$\angle$ ABQ= $\angle$ CDP (alternate angles)

AB=CD (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$ (By SAS)

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

From equation 1 and 2, we get

$\small AP=CQ$

$\small AQ=CP$

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

Q10 (i) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $\small 8.21$ ). Show that $\small \Delta APB\cong \Delta CQD$

1640170770148

Answer:

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

To prove : $\small \Delta APB\cong \Delta CQD$

Proof: In $\small \Delta APB\, \, and\, \, \Delta CQD$ ,

$\angle$ APB= $\angle$ CQD (Each $90 \degree$ )

$\angle$ ABP= $\angle$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$ (By SAS)

Q10 (ii) ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. $\small 8.21$ ). Show that $\small AP=CQ$

1640170798432

Answer:

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

To prove : $\small AP=CQ$

Proof: In $\small \Delta APB\, \, and\, \, \Delta CQD$ ,

$\angle$ APB= $\angle$ CQD (Each $90 \degree$ )

$\angle$ ABP= $\angle$ CDQ (Alternate angles)

AB=CD (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$ (By SAS)

$\small AP=CQ$ (CPCT)

Q11 (i) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that quadrilateral ABED is a parallelogram

1640170849147

Answer:

Given : In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ .

To prove : quadrilateral ABED is a parallelogram

Proof : In ABED,

AB=DE (Given)

AB||DE (Given )

Hence, quadrilateral ABED is a parallelogram.

Q11 (ii) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that quadrilateral BEFC is a parallelogram.

1640170880705

Answer:

Given: In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ .

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

BC=EF (Given)

BC||EF (Given )

Hence, quadrilateral BEFC is a parallelogram.

Q11 (iii) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that $\small AD\parallel CF$ and $\small AD=CF$

1640170909196

Answer:

To prove: $\small AD\parallel CF$ and $\small AD=CF$

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF

From 1 and 3, we get

AD=CF

Q11 (iv) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF.$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that quadrilateral ACFD is a parallelogram.

1640170928287

Answer:

To prove : quadrilateral ACFD is a parallelogram

Proof :

In ABED,

AD=BE.................1(ABED is a parallelogram)

AD||BE .................2(ABED is a parallelogram)

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

AD||CF...........................5

From 1 and 3, we get

AD=CF...........................6

From 5 and 6, we get

AD||CF and AD=CF

Thus, quadrilateral ACFD is a parallelogram

Q11 (v) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC = EF$ and $\small BC\parallel EF$ . Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that $\small AC=DF$

1640170956375

Answer:

In ACFD,

AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)

Q11 (vi) In $\small \Delta ABC$ and $\small \Delta DEF$ , $\small AB=DE,AB\parallel DE,BC=EF$ and $\small BC\parallel EF.$ Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. $\small 8.22$ ). Show that $\small \Delta ABC\cong \Delta DEF$

1640170982314

Answer:

In $\small \Delta ABC$ and $\small \Delta DEF$ ,

AB=DE (Given )

BC=EF (Given )

AC=DF ( proved in (v) part)

$\small \Delta ABC\cong \Delta DEF$ (By SSS rule)

Q12 (i) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that $\small \angle A=\angle B$

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

1640171017353

Answer:

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

To prove : $\small \angle A=\angle B$

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

In AECD,

AE||DC (Given)

AD||CE (By cnstruction)

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 $\triangle$ BCE,

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

$\angle 2+\angle 3=180 \degree$ ...................4(linear pairs)

$\angle 1+\angle 4=180 \degree$ .....................5(Co-interior angles)

From 4 and 5, we get

$\angle 2+\angle 3=\angle 1+\angle 4$

$\therefore \angle 2=\angle 1 \rightarrow \angle B=\angle A$ (Since, $\angle 3=\angle 4$ )

Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that $\small \angle C=\angle D$

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

1640171044691

Answer:

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

To prove : $\small \angle C=\angle D$

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

$\angle 1+\angle D= 180 \degree$ (Co-interior angles)

$\angle 2+\angle C= 180 \degree$ (Co-interior angles)

$\therefore \angle 1+\angle D=\angle 2+\angle C$

Thus, $\small \angle C=\angle D$ (Since , $\small \angle 1=\angle 2$ )

Q12 (iii) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that $\small \Delta ABC\cong \Delta BAD$

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

1640171072699

Answer:

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

To prove : $\small \Delta ABC\cong \Delta BAD$

Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\angle ABC=\angle BAD$ (proved in (i) )

Thus, $\small \Delta ABC\cong \Delta BAD$ (By SAS rule)

Q12 (iv) ABCD is a trapezium in which $\small AB\parallel CD$ and $\small AD=BC$ (see Fig. $\small 8.23$ ). Show that diagonal AC $\small =$ diagonal BD

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

1640171092180

Answer:

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

To prove: diagonal AC $\small =$ diagonal BD

Proof: In $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,

BC=AD (Given )

AB=AB (Common )

$\angle ABC=\angle BAD$ (proved in (i) )

Thus, $\small \Delta ABC\cong \Delta BAD$ (By SAS rule)

diagonal AC $\small =$ diagonal BD (CPCT)

Class 9 maths chapter 8 question answer - Exercise : 8.2

Q1 (i) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal. Show that : $\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$

1640171132022

Answer:

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove :

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$

Proof: In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$

Q1 (ii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal. Show that : $\small PQ=SR$

1640171177941

Answer:

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove : $\small PQ=SR$

Proof : In $\triangle$ ACD,

S is mid point of DA. (Given)

R is mid point of DC. (Given)

By mid point theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is mid point of AB. (Given)

Q is mid point of BC. (Given)

By mid point theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2,we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$

Q1 (iii) ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal. Show that : PQRS is a parallelogram.

1640171205589

Answer:

Given : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$ ). AC is a diagonal.

To prove : PQRS is a parallelogram.

Proof : In PQRS,

Since,

$\small PQ\parallel SR$ and $\small PQ=SR$ .

So,PQRS is a parallelogram.

Q2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer:

1640171236054

Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In $\triangle$ ACD,

S is midpoint of DA. (Given)

R is midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is midpoint of AB. (Given)

Q is mid point of BC. (Given)

By mid point theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2,we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So,the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$ BCD,

Q is mid point of BC. (Given)

R is mid point of DC. (Given)

By mid point theorem,

$\small QR\parallel BD$

So, QN || LM ...........5

LQ || MN ..........6 (Since, PQ || AC)

From 5 and 6, we get

LMPQ is a parallelogram.

Hence, $\small \angle$ LMN= $\small \angle$ LQN (opposite angles of the parallelogram)

But, $\small \angle$ LMN= 90 (Diagonals of a rhombus are perpendicular)

so, $\small \angle$ LQN=90

Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.

Q3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Answer:

1640172120681

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof :

In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By midpoint theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2, we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$ BCD,

Q is the midpoint of BC. (Given)

R is the midpoint of DC. (Given)

By midpoint theorem,

$\small QR\parallel BD$ and $\small QR=\frac{1}{2}BD$ ...................5

AC = BD.......................6(diagonals )

From 2, 5 and 6, we get

PQ=QR

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

Q4 ABCD is a trapezium in which $\small AB\parallel DC$ , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. $\small 8.30$ ). Show that F is the mid-point of BC.

1640172147378

Answer:

Given: ABCD is a trapezium in which $\small AB\parallel DC$ , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. $\small 8.30$ ).

To prove: F is the mid-point of BC.

In $\triangle$ ABD,

E is the midpoint of AD. (Given)

EG || AB (Given)

By converse of midpoint theorem,

G is the midpoint of BD.

In $\triangle$ BCD,

G is mid point of BD. (Proved above)

FG || DC (Given)

By converse of midpoint theorem,

F is the midpoint of BC.

Q5 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. $\small 8.31$ ). Show that the line segments AF and EC trisect the diagonal BD.

1640172188917

Answer:

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

Proof : In quadrilatweral ABCD,

AB=CD (Given)

$\frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AE=CF$ (E and F are midpoints of AB and CD)

In quadrilateral AECF,

AE=CF (Given)

AE || CF (Opposite sides of a parallelogram)

Hence, AECF is a parallelogram.

In $\triangle$ DCQ,

F is the midpoint of DC. (given )

FP || CQ (AECF is a parallelogram)

By converse of midpoint theorem,

P is the mid point of DQ.

DP= PQ....................1

Similarly,

In $\triangle$ ABP,

E is the midpoint of AB. (given )

EQ || AP (AECF is a parallelogram)

By converse of midpoint theorem,

Q is the midpoint of PB.

OQ= QB....................2

From 1 and 2, we have

DP = PQ = QB.

Hence, the line segments AF and EC trisect the diagonal BD.

Q6 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Answer:

1640172226300

Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.

To prove: the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Proof: In $\triangle$ ACD,

S is the midpoint of DA. (Given)

R is midpoint of DC. (Given)

By midpoint theorem,

$\small SR\parallel AC$ and $\small SR=\frac{1}{2}AC$ ...................................1

In $\triangle$ ABC,

P is the midpoint of AB. (Given)

Q is the midpoint of BC. (Given)

By midpoint theorem,

$\small PQ\parallel AC$ and $\small PQ=\frac{1}{2}AC$ .................................2

From 1 and 2, we get

$\small PQ\parallel SR$ and $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$ and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.

Thus, SQ and PR bisect each other.

Q7 (i) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that D is the mid-point of AC.

Answer:

1640172246282

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove :D is mid point of AC.

Proof: In $\triangle$ ABC,

M is mid point of AB. (Given)

DM || BC (Given)

By converse of mid point theorem,

D is the mid point of AC.

Q7 (ii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that $\small MD\perp AC$

Answer:

1651648435406

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.

To prove : $\small MD\perp AC$

Proof : $\angle$ ADM = $\angle$ ACB (Corresponding angles)

$\angle$ ADM= $90 \degree$ . ( $\angle$ ACB = $90 \degree$ )

Hence, $\small MD\perp AC$ .

Q7 (iii) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that $\small CM=MA=\frac{1}{2}AB$

Answer:

1651648482141

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove : $\small CM=MA=\frac{1}{2}AB$

Proof : In $\triangle$ ABC,

M is the midpoint of AB. (Given)

DM || BC (Given)

By converse of midpoint theorem,

D is the midpoint of AC i.e. AD = DC.

In $\triangle$ AMD and $\triangle$ CMD,

AD = DC (proved above)

$\angle$ ADM = $\angle$ CDM (Each right angle)

DM = DM (Common)

$\triangle$ AMD $\cong$ $\triangle$ CMD (By SAS)

AM = CM (CPCT)

But , $\small AM=\frac{1}{2}AB$

Hence, $\small CM=MA=\frac{1}{2}AB$ .

Summary Of Class 9 Quadrilaterals NCERT Solutions

Following topics are covered in the chapter:

Angle Sum Property
Types of Quadrilaterals
Properties of a Parallelogram
Condition for a Quadrilateral to be a Parallelogram
The Mid-point Theorem

Important Concepts Learned in NCERT Maths Chapter 8 Class 9 – Quadrilaterals

A quadrilateral is a polygon with four sides and four angles.
The sum of the angles in a quadrilateral is 360 degrees.
The opposite sides of a parallelogram are parallel and congruent.
The opposite angles of a parallelogram are congruent.
The diagonals of a parallelogram bisect each other.
A rectangle is a parallelogram with four right angles.
The diagonals of a rectangle are congruent and bisect each other.
A square is a rectangle with four congruent sides.
The diagonals of a square are congruent and bisect each other at right angles.
A rhombus is a parallelogram with four congruent sides.
The diagonals of a rhombus bisect each other at right angles.
The area of a quadrilateral can be calculated by dividing it into triangles and using the formula for the area of a triangle.
The perimeter of a quadrilateral is the sum of the lengths of its sides.
The properties of quadrilaterals can be used to solve problems related to real-life situations.

Quadrilaterals Class 9 Solutions - Exercise Wise

Students can practice class 9 maths ch 8 question answer using the exercise line given below.

NCERT Solutions For Class 9 Maths - Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

NCERT Solutions For Class 9 - Subject Wise

How To Use NCERT Solutions For Class 9 Maths Chapter 8 Quadrilaterals

Read and memorize the properties related to all kinds of quadrilaterals.
Have a glance through some solved to understand the pattern of the solution to that particular question.
Now check your learning on the practice exercises problems.
If you stuck in any question then you can assist yourself using NCERT solutions for class 9 maths chapter 8 Quadrilaterals.
Following the above-written steps, you can get 100% out of the chapter.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. What are the important topics in Quadrilaterals ch 8 maths class 9?

Angle sum property of a quadrilateral, properties of the parallelogram, another condition for a quadrilateral to be a parallelogram, and mid-point theorem are the important topics covered in class 9th quadrilateral solution. students can practice these NCERT solutions for class 9 to command the concepts.

2. How does the NCERT solutions are helpful ?

NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also it will provide them new ways to solve the problems. These solutions are provided in a very detailed manner which will give them conceptual clarity.

3. In NCERT Solutions for Class 9 Maths Chapter 8, how does the concept of "quadrilaterals" get defined?

Quadrilateral class 9 solutions define a quadrilateral as a two-dimensional shape that has four sides or edges and four corners or vertices. Quadrilaterals are commonly recognized by their standard shapes, such as rectangle, square, trapezoid, and kite, but they can also have irregular and undefined shapes.

4. Where can I find the complete solutions of class 9 chapter 8 maths ?

Here you will get the detailed NCERT solutions for class 9 by clicking on the link. for ease students can study quadrilateral class 9 pdf both online and offline mode.

Also Read

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Books for Class 9 English 2024 - Download All Chapters Pdf

NCERT Books for Class 9 Social Science 2024 - Download PDF

NCERT Books for Class 9 2024 - Free PDFs All Subjects Download

NCERT Books for Class 9 Hindi 2024 - Download PDF

NCERT Books for Class 9 Science 2023-24 - Download PDF (All Chapters)

NCERT Books for Class 9 Maths 2023-24 - Download PDF

NCERT Syllabus for Class 9 Social Science 2024 - Download PDF

NCERT Syllabus for Class 9 2024 - Download All subjects Pdf

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NCERT Syllabus for Class 9 Science 2024 - Download Chapter Wise Pdf

NCERT Syllabus for class 9 English 2024 - Download Syllabus PDF

NCERT Syllabus for Class 9 Maths 2023 - Download Syllabus Pdf

NCERT Solutions for Class 9 - All Subjects PDF Download

NCERT Solutions For Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1 Number Systems

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

Quadrilaterals Class 9 Questions And Answers PDF Free Download

Quadrilaterals Class 9 Solutions - Important Formulae And Points

Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)

Summary Of Class 9 Quadrilaterals NCERT Solutions

Important Concepts Learned in NCERT Maths Chapter 8 Class 9 – Quadrilaterals

Quadrilaterals Class 9 Solutions - Exercise Wise

NCERT Solutions For Class 9 Maths - Chapter Wise

NCERT Solutions For Class 9 - Subject Wise

Frequently Asked Question (FAQs)

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