NCERT Solutions for Class 9 Maths Chapter 8 - Quadrilaterals

NCERT Solutions for Maths Class 9 Chapter 8, "Quadrilaterals" comprises two exercises with a total of 19 questions. Here's a detailed explanation of the types of questions included in each exercise:

Exercise 8.1:

This exercise consists of 12 questions that cover a wide range of concepts related to quadrilaterals. The questions require students to identify and recognize the properties of different types of quadrilaterals. Here are the different types of questions you can expect to find in this exercise:

1. Identification of Quadrilaterals: In this type of question, students are given a diagram of a quadrilateral and are asked to identify the type of quadrilateral, such as a square, rectangle, parallelogram, or trapezium.

2. Checking Properties: In these questions, students need to confirm a given characteristic of a quadrilateral. For instance, they might be asked to confirm that the opposite sides of a parallelogram are equal.

3. Application of Properties: In this type of question, students need to apply the properties of quadrilaterals to solve problems. For instance, they may be asked to find the perimeter or area of a given quadrilateral.

Exercise 8.2:

This section has seven questions that emphasize applying the properties of quadrilaterals. The questions are more intricate, demanding a thorough grasp of the concepts from the chapter. Here are the types of questions you can anticipate in this exercise:

1. Proving Properties: In these questions, students are asked to prove a given property of a quadrilateral using the properties they have learned in the chapter.

2. Applying Properties: Similar to Exercise 8.1, these questions need students to use quadrilateral properties to solve problems. However, the questions in this exercise are more intricate, demanding a deeper understanding of the concepts.

3. Construction of Quadrilaterals: In some questions, students are asked to construct a quadrilateral based on certain given conditions, such as the length of the sides or the angles of the quadrilateral.

Access NCERT Answers for Class 9 Mathematics Chapter 8 – Quadrilaterals

1. The angles of the quadrilateral are in the ratio \[3:5:9:13\].Find all the angles of the quadrilateral.

Answer:

Given: The quadrilateral angles are in the ratio \[3:5:9:13\].

To find: Angles of a quadrilateral

Consider the angle's common ratio is \[x\].

Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.

The sum of a quadrilateral's inner angles is \[360^{\circ} \] ,

$\therefore 3 x+5 x+9 x+13 x=360^{\circ} $

$30 x=360^{\circ} $

$x=12^{\circ}$

As a result, the angles can be calculated as

$3 x=3 \times 12=36^{\circ} $

$5 x=5 \times 12=60^{\circ} $

$9 x=9 \times 12=108^{\circ} $

$13 x=13 \times 12=156^{\circ}$

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

Answer:

Given: Diagonals of the parallelogram are the same.

To prove: It is a rectangle.

Consider ABCD be the given parallelogram.

Now we need to show that ABCD is a rectangle, by proving that one of its interior angles is .

In \[\Delta ABC\]and \[\Delta DCB\],

AB = DC (side opposite to the parallelogram are equal)

BC = BC (in common)

AC = DB (Given)

\[\therefore \Delta ABC \cong \Delta DCB\] (By SSS Congruence rule)

\[ \Rightarrow \angle ABC{\text{ }} = \angle DCB\]

The sum of the measurements of angles on the same side of a transversal is known to be \[{180^o}\].

Hence, ABCD is a rectangle because it is a parallelogram with a \[{90^o}\] inner angle.

3. 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 at right angles.

To prove: It is a rhombus.

Consider ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle

i.e.\[OA{\text{ }} = {\text{ }}OC,{\text{ }}OB{\text{ }} = {\text{ }}OD\], and .

In order to prove ABCD a rhombus, we need to prove ABCD is the parallelogram and all sides of the ABCD are the same.

In \[\Delta AOD\] and \[\Delta COD\],

OA = OC (Diagonals of the parallelogram bisect each other)

\[\angle AOD{\text{ }} = \angle COD\] (Given)

OD = OD (Common)

\[\therefore \Delta AOD \cong \Delta COD\] (By SAS congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}CD\]… (1)

Similarly, it can be proved that

\[AD{\text{ }} = {\text{ }}AB\] and \[CD{\text{ }} = {\text{ }}BC\]… (2)

From Equations (1) and (2),

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\]

To conclude, ABCD is a parallelogram because the opposite sides of the quadrilateral ABCD are equal. ABCD is a rhombus because all of the sides of a parallelogram ABCD are equal.
 

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

Answer:

Given: A square is given.

To find: The diagonals of a square are the same and bisect each other at ${90^o}$

Consider ABCD be a square.

Consider the diagonals AC and BD intersect each other at a point O.

We must first show that the diagonals of a square are equal and bisect each other at right angles,

\[{\text{AC  =  BD, OA  =  OC, OB  =  OD}}\], and .

In \[\Delta ABC\]and \[\Delta DCB\],

\[AB{\text{ }} = {\text{ }}DC\] (Sides of the square are equal)

\[\angle ABC{\text{ }} = \angle DCB\] (All the interior angles are of the value \[{90^o}\])

\[BC{\text{ }} = {\text{ }}CB\](Common side)

\[\therefore \Delta ABC \cong \Delta DCB\](By SAS congruency)

\[\therefore AC{\text{ }} = {\text{ }}DB\](By CPCT)

Hence, the diagonals of a square are equal in length.

In \[\Delta AOB\]and\[\Delta COD\],

\[\angle AOB{\text{ }} = \angle COD\](Vertically opposite angles)

\[\angle ABO{\text{ }} = \angle CDO\](Alternate interior angles)

AB = CD (Sides of a square are always equal)

\[\therefore \Delta AOB \cong \Delta COD\](By AAS congruence rule)

\[\therefore AO{\text{ }} = {\text{ }}CO\] and \[OB{\text{ }} = {\text{ }}OD\](By CPCT)

As a result, the diagonals of a square are bisected.

In \[\Delta AOB\]and \[\Delta COB\],

Because we already established that diagonals intersect each other,

\[AO{\text{ }} = {\text{ }}CO\]

\[AB{\text{ }} = {\text{ }}CB\](Sides of a square are equal)

\[BO{\text{ }} = {\text{ }}BO\](Common)

\[\therefore \Delta AOB \cong \Delta COB\](By SSS congruency)

\[\therefore \angle AOB{\text{ }} = \angle COB\](By CPCT)

However,  (Linear pair)

As a result, the diagonals of a square are at right angles to each other.

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

Answer:

Given: The diagonals of a quadrilateral are equal and bisect at ${90^o}$

To find: It is a square

Consider the quadrilateral ABCD, in which the diagonals AC and BD cross at point O.

The diagonals of ABCD are equal and bisect each other at right angles, which is a given.

Therefore, \[AC{\text{ }} = {\text{ }}BD,{\text{ }}OA{\text{ }} = {\text{ }}OC,{\text{ }}OB{\text{ }} = {\text{ }}OD\], and .

We need to prove that ABCD is a parallelogram, \[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\], and one of its interior angles being .

In \[\Delta AOB\] and \[\Delta COD\],

\[AO{\text{ }} = {\text{ }}CO\] (Diagonals bisect each other)

\[OB{\text{ }} = {\text{ }}OD\] (Diagonals bisect each other)

\[\angle AOB{\text{ }} = \angle COD\] (Vertically opposite angles)

\[\therefore \Delta AOB \cong \Delta COD\] (SAS congruence rule)

$AB = CD$ (By CPCT) ... (1)

And,\[\angle OAB{\text{ }} = \angle OCD\]. (By CPCT)

However, for lines AB and CD, they are alternate interior angles, and alternate interior angles are equal only when the two lines are parallel.

\[AB{\text{ }}||{\text{ }}CD\]... (2)

From (1) and (2), we obtain

ABCD is a parallelogram.

In \[\Delta AOD{\text{ , }}\Delta COD\],

\[AO{\text{ }} = {\text{ }}CO\] (Diagonals bisect each other)

\[\angle AOD{\text{ }} = \angle COD\] (Given that each is )

\[OD{\text{ }} = {\text{ }}OD\] (Common)

\[\therefore \Delta AOD \cong \Delta COD\] (SAS congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}DC\]... (3)

However, \[AD{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\](Opposite sides of the parallelogram ABCD)

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}DA\]

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In \[\Delta ADC\]and \[\Delta BCD\],

\[AD{\text{ }} = {\text{ }}BC\] (Already proved)

\[AC{\text{ }} = {\text{ }}BD\] (Given)

\[DC{\text{ }} = {\text{ }}CD\] (Common)

\[\therefore \Delta ADC \cong \Delta BCD\] (SSS Congruence rule)

\[\therefore \angle ADC{\text{ }} = \angle BCD\] (By CPCT)

However, \[\angle ADC{\text{ }} + \angle BCD{\text{ }} = {\text{ }}180^\circ \](Co-interior angles)

\[\begin{array}{*{20}{l}} {\angle ADC{\text{ }} + \angle ADC{\text{ }} = {\text{ }}180^\circ } \\ {2\angle ADC{\text{ }} = {\text{ }}180^\circ } \\ {\therefore \angle ADC{\text{ }} = {\text{ }}90^\circ } \end{array}\]

A right angle is one of the quadrilateral ABCD's inner angles.

As a result, we can see that ABCD is a parallelogram, \[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}AD\]and one of its interior angles is .

Hence , ABCD is a square.
 

6. Diagonal AC of a parallelogram ABCD is bisecting \[\angle A\](see the given figure). Show that

(i) It is bisecting \[\angle C\]also,

(ii) ABCD is a rhombus

Answer: 

Given: Diagonal AC of a parallelogram ABCD is bisecting \[\angle A\]

To find: (i) It is bisecting \[\angle C\] also,

(ii) ABCD is a rhombus

 (i) ABCD is a parallelogram.

\[\angle DAC{\text{ }} = \angle BCA\] (Alternate interior angles) ... (1)

And, \[\angle BAC{\text{ }} = \angle DCA\](Alternate interior angles) ... (2)

However, it is given that AC is bisecting  \[\angle A\].

\[\angle DAC{\text{ }} = \angle BAC\]... (3)

From Equations (1), (2), and (3), we obtain

\[\angle DAC{\text{ }} = \angle BCA{\text{ }} = \angle BAC{\text{ }} = \angle DCA\]... (4)

\[\angle DCA{\text{ }} = \angle BCA\]

Hence, AC is bisecting \[\angle C\].

(ii) From Equation (4), we obtain

\[\angle DAC{\text{ }} = \angle DCA\]

\[DA{\text{ }} = {\text{ }}DC\] (Side opposite to equal angles are equal)

However, \[DA{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of a parallelogram)

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}DA\]

As a result, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC is bisecting \[\angle A\] as well as \[\angle C\] and diagonal BD is bisecting \[\angle B\] as well as \[\angle D\].

Given: ABCD is a rhombus.

To find: Diagonal AC is bisecting \[\angle A\] as well as \[\angle C\] and diagonal BD is bisecting \[\angle B\] as well as \[\angle D\].

Answer:

Let us now join AC.

In \[\Delta ABC\],

\[BC{\text{ }} = {\text{ }}AB\] (Sides of the rhombus are equal to each other)

\[\angle 1{\text{ }} = \angle 2\] (Angles opposite to equal sides of a triangle are equal)

However, \[\angle 1{\text{ }} = \angle 3\] (Parallel lines AB and CD have different interior angles.)

\[\angle 2{\text{ }} = \angle 3\]

Hence, AC is bisecting \[\angle C\].

Also, \[\angle 2{\text{ }} = \angle 4\] (Alternate interior angles for \[||\] lines BC and DA)

\[\angle 1{\text{ }} = \angle 4\]

Hence, AC bisects \[\angle A\].

Also, it can be proved that BD is bisecting \[\angle B\] and \[\angle D\] as well.

8. ABCD is a rectangle in which diagonal AC bisects \[\angle A\] as well as \[\angle C\]. Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects \[\angle B\] as well as \[\angle D\].

Answer:

Given: ABCD is a rectangle where the diagonal AC bisects \[\angle A\] as well as \[\angle C\].

To find: (i) ABCD is a square

              (ii) Diagonal BD bisects \[\angle B\] as well as \[\angle D\].

It is given that ABCD is a rectangle.\[\angle A{\text{ }} = \angle C\]\[\]

$ \Rightarrow \dfrac{1}{2}\angle A = \dfrac{1}{2}\angle C$ (AC bisects \[\angle A\] and \[\angle C\])

$ \Rightarrow \angle DAC = \dfrac{1}{2}\angle DCA$

CD = DA (Sides that are opposite to the equal angles are also equal)

Also, \[DA{\text{ }} = {\text{ }}BC\] and \[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of the rectangle are same)

\[AB{\text{ }} = {\text{ }}BC{\text{ }} = {\text{ }}CD{\text{ }} = {\text{ }}DA\]

ABCD is a rectangle with equal sides on all sides.

Hence, ABCD is a square.

(ii) Let us now join BD.

In \[\Delta BCD\],

\[BC{\text{ }} = {\text{ }}CD\] (Sides of a square are equal to each other)

\[\angle CDB{\text{ }} = \angle CBD\] (Angles opposite to equal sides are equal)

However, \[\angle CDB{\text{ }} = \angle ABD\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\])

\[\angle CBD{\text{ }} = \angle ABD\]

BD bisects \[\angle B.\]

Also, \[\angle CBD{\text{ }} = \angle ADB\](Alternate interior angles for \[BC{\text{ }}||{\text{ }}AD\])

\[\angle CDB{\text{ }} = \angle ABD\]

BD bisects \[\angle D\] and \[\angle B\].

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:

(i) \[\Delta APD \cong \Delta CQB\]

(ii) \[AP{\text{ }} = {\text{ }}CQ\]

(iii) \[\Delta AQB \cong \Delta CPD\]

(iv) \[AQ{\text{ }} = {\text{ }}CP\]

(v) APCQ is a parallelogram

Answer:

Given: A parallelogram is given.

To prove: (i) \[\Delta APD \cong \Delta CQB\]

                  (ii) \[AP{\text{ }} = {\text{ }}CQ\]

      (iii) \[\Delta AQB \cong \Delta CPD\]

      (iv) \[AQ{\text{ }} = {\text{ }}CP\]

      (v) APCQ is a parallelogram

(i) In \[\Delta APD\] and \[\Delta CQB\],

\[\angle ADP{\text{ }} = \angle CBQ\] (Alternate interior angles for \[BC{\text{ }}||{\text{ }}AD\])

\[AD{\text{ }} = {\text{ }}CB\] (Opposite sides of the parallelogram ABCD)

\[DP{\text{ }} = {\text{ }}BQ\] (Given)

\[\therefore \Delta APD \cong \Delta CQB\] (Using SAS congruence rule)

(ii) As we had observed that \[\Delta APD \cong \Delta CQB\],

\[\therefore AP{\text{ }} = {\text{ }}CQ\] (CPCT)

(iii) In \[\Delta AQB\] and \[\Delta CPD\],

\[\angle ABQ{\text{ }} = \angle CDP\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\])

\[AB{\text{ }} = {\text{ }}CD\] (Opposite sides of parallelogram ABCD)

\[BQ{\text{ }} = {\text{ }}DP\] (Given)

\[\therefore \Delta AQB \cong \Delta CPD\] (Using SAS congruence rule)

(iv) Since we had observed that \[\Delta AQB \cong \Delta CPD\],

\[\therefore AQ{\text{ }} = {\text{ }}CP\] (CPCT)

(v) From the result obtained in (ii) and (iv),

\[AQ{\text{ }} = {\text{ }}CP\] and

\[AP{\text{ }} = {\text{ }}CQ\]

APCQ is a parallelogram because the opposite sides of the quadrilateral are equal.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that

(i) \[\Delta APB \cong \Delta CQD\]

(ii) \[AP{\text{ }} = {\text{ }}CQ\]

Answer:

(i) In \[\Delta APB\]and \[\Delta CQD\],

\[\angle APB{\text{ }} = \angle CQD\] (Each 90°)

\[AB{\text{ }} = {\text{ }}CD\] (The opposite sides of a parallelogram ABCD)

\[\angle ABP{\text{ }} = \angle CDQ\] (Alternate interior angles for \[AB{\text{ }}||{\text{ }}CD\])

\[\therefore \Delta APB \cong \Delta CQD\] (By AAS congruency)

(ii) By using

\[\therefore \Delta APB \cong \Delta CQD\], we obtain

\[AP{\text{ }} = {\text{ }}CQ\] (By CPCT)

11. In \[\Delta ABC\] and\[\Delta DEF\], \[AB{\text{ }} = {\text{ }}DE,{\text{ }}AB{\text{ }}||{\text{ }}DE,{\text{ }}BC{\text{ }} = {\text{ }}EF\] and \[BC{\text{ }}||{\text{ }}EF\]. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that

(i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iii) \[AD{\text{ }}||{\text{ }}CF\] and AD = CF

(iv)Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) \[\Delta ABC \cong \Delta DEF\].

Answer:

Given: \[\Delta ABC\] and\[\Delta DEF\] are two triangles.

To prove: Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iii) \[AD{\text{ }}||{\text{ }}CF\] and AD = CF

(iv)Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) \[\Delta ABC \cong \Delta DEF\].

(i) It is given that \[AB{\text{ }} = {\text{ }}DE\] and \[AB{\text{ }}||{\text{ }}DE\].

A parallelogram is formed when two opposite sides of a quadrilateral are equal and parallel to one other.

As a result, ABED is a parallelogram in quadrilateral form.

(ii) Again, BC = EF and \[BC{\text{ }}||{\text{ }}EF\]

Therefore, quadrilateral  BCEF  is a parallelogram.

(iii) As we had observed that ABED and BEFC are parallelograms, therefore

AD = BE and \[AD{\text{ }}||{\text{ }}BE\]

(Opposite sides of a parallelogram are equal and parallel)

And, BE = CF and \[BE{\text{ }}||{\text{ }}CF\]

(Opposite sides of a parallelogram are equal and parallel)

\[\therefore AD{\text{ }} = {\text{ }}CF\] and \[AD{\text{ }}||{\text{ }}CF\]

(iv) We know it's a parallelogram since one set of opposing sides (AD and CF) of quadrilateral ACFD are equal and parallel to one other.

(v) Due to the fact that ACFD is a parallelogram, the pair of opposing sides will be equal and parallel to one another.

\[\therefore AC{\text{ }}||{\text{ }}DF\] and \[AC{\text{ }} = {\text{ }}DF\]

(vi) In\[\Delta ABC\] and \[\Delta DEF\],

\[AB{\text{ }} = {\text{ }}DE\] (Given)

\[BC{\text{ }} = {\text{ }}EF{\text{ }}\](Given)

\[AC{\text{ }} = {\text{ }}DF\] (ACFD is a parallelogram)

\[\therefore \Delta ABC \cong \Delta DEF\] (By SSS congruence rule)

12. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) \[\Delta ABC \cong \Delta BAD\]

(iv) 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.

To find: (i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) \[\Delta ABC \cong \Delta BAD\]

(iv) diagonal AC = diagonal BD

Let us extend AB by drawing a line through C, which is parallel to AD, intersecting AE at point

E. It is clear that AECD is a parallelogram.

(i) \[AD{\text{ }} = {\text{ }}CE\] (Opposite sides of parallelogram AECD)

However, \[AD{\text{ }} = {\text{ }}BC\] (Given)

Therefore, \[BC{\text{ }} = {\text{ }}CE\]

\[\angle CEB{\text{ }} = \angle CBE\] (Angle opposite to the equal sides are also equal)

Consideing parallel lines AD and CE. AE is the transversal line for them.

 (Angles on a same side of transversal)

 (Using the relation ∠CEB = ∠CBE) ... (1)

However,  (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain

\[\angle A{\text{ }} = \angle B\]

 (ii) \[AB{\text{ }}||{\text{ }}CD\]

 (Angles on a same side of the transversal)

Also, \[\angle C{\text{ }} + \angle B{\text{ }} = {\text{ }}180^\circ \](Angles on a same side of a transversal)

\[\therefore \angle A{\text{ }} + \angle D{\text{ }} = \angle C{\text{ }} + \angle B\]

However, \[\angle A{\text{ }} = \angle B\] (Using the result obtained in (i))

\[\therefore \angle C{\text{ }} = \angle D\]

 (iii) In \[\Delta ABC\] and \[\Delta BAD\],

\[AB{\text{ }} = {\text{ }}BA\] (Common side)

\[BC{\text{ }} = {\text{ }}AD\] (Given)

\[\angle B{\text{ }} = \angle A\] (Proved before)

\[\therefore \Delta ABC \cong \Delta BAD\] (SAS congruence rule)

(iv) We had seen that,

\[\Delta ABC \cong \Delta BAD\]

\[\therefore AC{\text{ }} = {\text{ }}BD\] (By CPCT)

Exercise (8.2)

2. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

(i) \[SR{\text{ }}||{\text{ }}AC\] and \[SR = \dfrac{1}{2}\;AC\]

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Answer

Given: ABCD is a quadrilateral

To prove: (i) \[SR{\text{ }}||{\text{ }}AC\] and \[SR = \dfrac{1}{2}\;AC\]

     (ii) PQ = SR

     (iii) PQRS is a parallelogram.

(i) In \[\Delta ADC\], S and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment connecting the midpoints of any two sides is parallel to and half of the third side.

\[\therefore SR{\text{ }}||{\text{ }}AC\] and \[SR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\]... (1)

(ii) In ∆ABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using midpoint theorem,

\[PQ{\text{ }}||{\text{ }}AC\]and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\]... (2)

Using Equations (1) and (2), we obtain

\[PQ{\text{ }}||{\text{ }}SR\] and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}SR\]... (3)

\[\therefore PQ{\text{ }} = {\text{ }}SR\]

 (iii) From Equation (3), we obtained

\[PQ{\text{ }}||{\text{ }}SR\] and  \[PQ{\text{ }} = {\text{ }}SR\]

Clearly, one pair of quadrilateral PQRS opposing sides is parallel and equal.

PQRS is thus a parallelogram.

3. 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:

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

To find: Quadrilateral PQRS is a rectangle

In \[\Delta ABC\], P and Q are the mid-points of sides AB and BC respectively.

\[PQ{\text{ }}||{\text{ }}AC{\text{ , }}PQ{\text{ }} = {\text{ }}\dfrac{1}{2}AC\] (Using mid-point theorem) ... (1)

In \[\Delta ADC\],

R and S are the mid-points of CD and AD respectively.

\[RS{\text{ }}||{\text{ }}AC{\text{ , }}RS{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] (Using mid-point theorem) ... (2)

From Equations (1) and (2), we obtain

\[PQ{\text{ }}||{\text{ }}RS\] and \[PQ{\text{ }} = {\text{ }}RS\]

It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is equal and parallel to each other. At position O, the diagonals of rhombus ABCD should cross.

In quadrilateral OMQN,

\[MQ{\text{ }}\left| {\left| {{\text{ }}ON{\text{ }}({\text{ }}PQ{\text{ }}} \right|} \right|{\text{ }}AC)\]

\[QN{\text{ }}\left| {\left| {{\text{ }}OM{\text{ }}({\text{ }}QR{\text{ }}} \right|} \right|{\text{ }}BD)\]

Hence , OMQN is a parallelogram.

\[\begin{array}{*{20}{l}} {\therefore \angle MQN{\text{ }} = \angle NOM} \\ {\therefore \angle PQR{\text{ }} = \angle NOM} \end{array}\]

Since,  \[\angle NOM{\text{ }} = {\text{ }}90^\circ \] (Diagonals of the rhombus are perpendicular to each other)

\[\therefore \angle PQR{\text{ }} = {\text{ }}90^\circ \]

Clearly, PQRS is a parallelogram having one of its interior angles as .

So , PQRS is a rectangle.

3. 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:

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.

Let us join AC and BD.

In \[\Delta ABC\],

P and Q are the mid-points of AB and BC respectively.

\[\therefore PQ{\text{ }}||{\text{ }}AC\] and \[PQ{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\](Mid-point theorem) ... (1)

Similarly in \[\Delta ADC\],

\[SR{\text{ }}||{\text{ }}AC{\text{ , }}SR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AC\] (Mid-point theorem) ... (2)

Clearly, \[PQ{\text{ }}||{\text{ }}SR\] and \[PQ{\text{ }} = {\text{ }}SR\]

It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is equal and parallel to each other.

\[\therefore PS{\text{ }}||{\text{ }}QR{\text{ }},{\text{ }}PS{\text{ }} = {\text{ }}QR\] (Opposite sides of parallelogram) ... (3)

In \[\Delta BCD\], Q and R are the mid-points of side BC and CD respectively.

\[\therefore QR{\text{ }}||{\text{ }}BD{\text{ , }}QR{\text{ }} = {\text{ }}\dfrac{1}{2}BD\] (Mid-point theorem) ... (4)

Also, the diagonals of a rectangle are equal.

\[\therefore AC{\text{ }} = {\text{ }}BD\]…(5)

By using Equations (1), (2), (3), (4), and (5), we obtain

\[PQ{\text{ }} = {\text{ }}QR{\text{ }} = {\text{ }}SR{\text{ }} = {\text{ }}PS\]

So , PQRS is a rhombus

4. ABCD is a trapezium in which \[AB{\text{ }}||{\text{ }}DC\], 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 the given figure). Show that F is the mid-point of BC.

Answer:

Given: ABCD is a trapezium in which \[AB{\text{ }}||{\text{ }}DC\], 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.

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

Let EF intersect DB at G.

We know that a line traced through the mid-point of any side of a triangle and parallel to another side bisects the third side by the reverse of the mid-point theorem.

In \[\Delta ABD\],

\[EF{\text{ }}||{\text{ }}AB\] and E is the mid-point of AD.

Hence , G will be the mid-point of DB.

As \[EF{\text{ }}\left| {\left| {{\text{ }}AB{\text{ , }}AB{\text{ }}} \right|} \right|{\text{ }}CD\],

\[\therefore EF{\text{ }}||{\text{ }}CD\] (Two lines parallel to the same line are parallel)

In \[\Delta BCD\], \[GF{\text{ }}||{\text{ }}CD\] and G is the mid-point of line BD. So , by using converse of mid-point

theorem, F is the mid-point of BC.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.

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.

ABCD is a parallelogram.

\[AB{\text{ }}||{\text{ }}CD\]

And hence, \[AE{\text{ }}||{\text{ }}FC\]

Again, AB = CD (Opposite sides of parallelogram ABCD)

\[\dfrac{1}{2}AB{\text{ }} = {\text{ }}\dfrac{1}{2}CD\]

\[AE{\text{ }} = {\text{ }}FC\] (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of the opposite sides (AE and CF) is parallel and same to each other. So , AECF is a parallelogram.

\[\therefore AF{\text{ }}||{\text{ }}EC\] (Opposite sides of a parallelogram)

In \[\Delta DQC\], F is the mid-point of side DC and \[FP{\text{ }}||{\text{ }}CQ\] (as \[AF{\text{ }}||{\text{ }}EC\]). So , by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.

\[\therefore DP{\text{ }} = {\text{ }}PQ\]... (1)

Similarly, in \[\Delta APB\], E is the mid-point of side AB and \[EQ{\text{ }}||{\text{ }}AP\] (as \[AF{\text{ }}||{\text{ }}EC\]).

As a result, the reverse of the mid-point theorem may be used to say that Q is the mid-point of PB.

\[\therefore PQ{\text{ }} = {\text{ }}QB\]... (2)

From Equations (1) and (2),

\[DP{\text{ }} = {\text{ }}PQ{\text{ }} = {\text{ }}BQ\]

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

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

Answer:

Given: ABCD is a quadrilateral

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

Let ABCD be a quadrilateral in which the midpoints of sides AB, BC, CD, and DA are P, Q, R, and S, respectively. PQ, QR, RS, SP, and BD are all joined

S and P are the mid-points of AD and AB, respectively, in \[\Delta ABD\]. As a result, using the mid-point theorem, it can be shown that

\[SP{\text{ }}||{\text{ }}BD{\text{ , }}SP{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}BD\]... (1)

Similarly in \[\Delta BCD\],

\[QR{\text{ }}||{\text{ }}BD{\text{ , }}QR{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}BD\]... (2)

From the Equations (1) and (2), we obtain

\[SP{\text{ }}||{\text{ }}QR{\text{ , }}SP{\text{ }} = {\text{ }}QR\]

One set of opposing sides of quadrilateral SPQR is equal and parallel to each other. SPQR is a parallelogram as a result.

The diagonals of a parallelogram are known to bisect each other.

As a result, PR and QS cut each other in half.

7. 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

(i) D is the mid-point of AC

(ii) MD $ \bot $ AC

(iii) \[CM{\text{ }} = {\text{ }}MA{\text{ }} = \dfrac{1}{2}AB\]

Answer:

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: (i) D is the mid-point of AC

(ii) MD $ \bot $ AC

(iii) \[CM{\text{ }} = {\text{ }}MA{\text{ }} = \dfrac{1}{2}AB\]

(i) In \[\Delta ABC\],

It is given that M is the mid-point of AB and \[MD{\text{ }}||{\text{ }}BC\].

Therefore, D is the mid-point of AC. (Converse of the mid-point theorem)

(ii) As \[DM{\text{ }}||{\text{ }}CB\] and AC is a transversal line for them, therefore,

 (Co-interior angles)

(iii) Join MC.

In \[\Delta AMD\] and \[\Delta CMD\],

\[AD{\text{ }} = {\text{ }}CD\] (D is the mid-point of side AC)

\[\angle ADM{\text{ }} = \angle CDM\] (Each )

DM = DM (Common)

\[\therefore \Delta AMD \cong \Delta CMD\] (By SAS congruence rule)

Therefore, \[AM{\text{ }} = {\text{ }}CM\](By CPCT)

However, \[{\text{ }}AM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\] (M is mid-point of AB)

Therefore, it is said that

\[CM{\text{ }} = {\text{ }}AM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

NCERT Solutions for Maths Class 9 Chapter 8 Quadrilateral - PDF Download

NCERT Solutions Class 9 Chapter 8 provides students with tricks and techniques to solve the problem in abstract and easier methods, helping the students master the skill of critical thinking and problem-solving at their early stages. An established support system like Vedantu's Free Download NCERT Solutions Class 9 Maths at their early stage of education strengthens their reasoning and logical abilities paving the way to great heights in the field of Mathematics. 

For upcoming exams, you can choose Chapter 8 - Quadrilaterals NCERT Solutions for Class 9 Maths in PDF. Additionally, you can access solutions for all the math chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability
  
  

NCERT Class 9 Maths Chapter 8 Quadrilaterals

In NCERT Class 9 Maths Chapter 8, you'll find a concise explanation of special quadrilaterals like Parallelogram, Rhombus, Rectangle, Square, and Trapezium. Vedantu's solutions delve into crucial basics, covering angles and diagonal relations for each quadrilateral. Clear understanding from NCERT Solutions Class 9 Quadrilaterals aids students in mastering these topics. 

The Class 9 Maths Chapter 8 concentrates on sections of the Chapter Quadrilateral. Section 1 is an introduction to the quadrilateral, and some of its general properties are introduced. Section 2 is the introduction to types of quadrilateral and their properties. Section 3 covers the properties of a parallelogram whose significance is discussed elaborately in Maths NCERT Solutions Class 9 Chapter 8. Section 4 includes mainly the Mid-point theorem and its wide range of applications.

Exercise 8.1

• Introduction to Quadrilaterals

A figure formed by joining four non-collinear points is called a quadrilateral. In other words, it is a four-sided polygon. It has four sides and four vertices. By joining the opposite vertices of the quadrilateral, we obtain a diagonal.

• Angle Sum Property of Quadrilateral

The sum of the angles of a quadrilateral is 360 degrees. Chapter 8 Maths Class 9 has a crystal clear explanation of the angle sum property of a quadrilateral through a well-expounded problem. Chapter 8 Class 9 Maths NCERT Solutions registers these concepts strong in young minds.

• Types of Quadrilaterals.

The parallelism of sides plays a vital role in the determination of the properties of the Quadrilaterals. As the knowledge application of concepts is the pillar of good scores in Maths, Maths NCERT Solutions Class 9 Chapter 8 is the best stop. If one pair of opposite sides are parallel, it is called a Trapezium. If both the pairs of opposite sides are parallel, then it is called a parallelogram. A parallelogram in which all the sides are equal is called Rhombus. A parallelogram in which all angles are equal and measure 90 degrees is called a rectangle. A Rectangle whose all sides are equal is called a square. A kite is a Quadrilateral whose pairs of adjacent sides are equal.

• Properties of Parallelogram

The properties of the parallelogram have great application when it comes to solving Polygon problems. NCERT Solutions Class 9 Chapter 8 explains the properties of parallelogram elaborately through the illustrated issues. 

Some properties of parallelograms are listed and are proficiently used in Chapter 8 Class 9 Maths NCERT Solutions. The opposite sides of the parallelogram are equal and parallel. The opposite angles are equal. The properties concerning the diagonals of the parallelogram have high significance that NCERT Class 9 Maths Chapter 8 applies them in most sums. The diagonals form congruent triangles with the sides of the parallelogram. The diagonals bisect each other. x The diagonals are angle bisectors of the vertices.

The Experts of Vedantu emphasize the importance of the properties of a parallelogram, as it being the fundamental shape of other quadrilaterals like Rectangle, Rhombus, and Square. Also, these shapes have higher usage in daily life than any other Quadrilateral. Maths NCERT Class 9 Chapter 8 also uses all the properties of a parallelogram in certain proofs and solutions of the problems in the most efficient manner.

Exercise 8.2

• The Mid-point Theorem

The line segment joining the mid-points of two sides of a triangle is parallel to the third side. Conversely, The line drawn through the midpoint of one side of a triangle, parallel to another side bisects the third side. NCERT Solutions for Class 9 Maths Chapter 8 provides the derailed solutions for all the problems based on mid-point theorems. The type of problems in which the line segments joining the midpoints of sides of quadrilaterals in various combinations is dealt with in detail. It also includes midpoints intersecting the diagonals of trapeziums and some of their extended versions. Added to it are right-angled triangles and the application of the mid-point theorem. NCERT Solutions for Class 9 Chapter 8 breaks down all these involuted problems simple to the ease of students.

Why are NCERT Solutions for Class 9 Math Chapter 8 - Quadrilaterals

• NCERT Solutions for Math Class 9 Chapter 8 “Quadrilaterals” is the finest material to understand the topics in the best way.

• The material carries all the information in detail and pointwise.

• All the cases of the material are solved in the simplest way which explains the term clearly.

• NCERT Solutions Class 9 Math Chapter 8  “Quadrilaterals” provides all the main topics underlined so that the student can focus on it.

Conclusion

Class 9 Maths Chapter 8 is an important chapter that lays the foundation for future mathematics. Vedantu's Class 9 Maths Chapter 8 Solutions is a comprehensive and informative resource that will help students to understand the concepts, solve problems, and improve their analytical skills.

Success in exams requires regular practice. Vedantu's Class 9 Maths Chapter 8 Solutions offers an extensive set of practice questions along with solutions, aiding students in thorough exam preparation.

Students can also download a free PDF of Vedantu's Class 9 Maths Chapter 8 Solutions for easy access and offline use.