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.

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.

## Quadrilaterals Class 9 Solutions - Important Formulae And Points

• 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.

## Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 8 NCERT solutions - Exercise: 8.1

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 : $3x=3\times 12=36\degree$

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

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

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

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

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

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.

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,

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

BO=OD (Given )

AO=AO (common)

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

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

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,

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

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)

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.

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,

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.

it bisects $\small \angle C$ also.

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.

ABCD is a rhombus.

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.

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.

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.

ABCD is a square

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.

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

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.

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)

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)

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)

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)

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.

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)

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)

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.

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.

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

Proof :

In ABED,

In BEFC,

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

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

From 2 and 4 , we get

From 1 and 3, we get

To prove : quadrilateral ACFD is a parallelogram

Proof :

In ABED,

In BEFC,

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

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

From 2 and 4 , we get

From 1 and 3, we get

From 5 and 6, we get

Thus, quadrilateral ACFD is a parallelogram

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)

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

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)

Hence, AECD is a parallelogram.

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, $\angle 3=\angle 4$ )

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

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 , $\small \angle 1=\angle 2$ )

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

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 $\small \Delta ABC\, \, and\, \, \, \Delta BAD$ ,

AB=AB (Common )

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

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

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

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

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

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

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

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

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, $\small PQ=SR$

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.

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.

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.

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.

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.

AB=CD (Given)

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

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

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.

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.

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.

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, $\small MD\perp AC$ .

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,

$\angle$ 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
• 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 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Quadrilaterals Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15

## 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:

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.

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

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