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 Solutions for Class 9 Maths Chapter 8– Quadrilaterals

Exercise 8.1

1. 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 $$\mathrm{\Delta }ABC$$and $$\mathrm{\Delta }DCB$$,

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

BC = BC (in common)

AC = DB (Given)

$$\therefore \mathrm{\Delta }ABC\cong \mathrm{\Delta }DCB$$ (By SSS Congruence rule)

$$\Rightarrow \mathrm{\angle }ABC=\mathrm{\angle }DCB$$

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.

2. 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 $$\mathrm{\Delta }ABC$$and $$\mathrm{\Delta }DCB$$,

$$AB=DC$$ (Sides of the square are equal)

$$\mathrm{\angle }ABC=\mathrm{\angle }DCB$$ (All the interior angles are of the value $${90}^o$$)

$$BC=CB$$(Common side)

$$\therefore \mathrm{\Delta }ABC\cong \mathrm{\Delta }DCB$$(By SAS congruency)

$$\therefore AC=DB$$(By CPCT)

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

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

$$\mathrm{\angle }AOB=\mathrm{\angle }COD$$(Vertically opposite angles)

$$\mathrm{\angle }ABO=\mathrm{\angle }CDO$$(Alternate interior angles)

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

$$\therefore \mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$$(By AAS congruence rule)

$$\therefore AO=CO$$ and $$OB=OD$$(By CPCT)

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

In $$\mathrm{\Delta }AOB$$and $$\mathrm{\Delta }COB$$,

Because we already established that diagonals intersect each other,

$$AO=CO$$

$$AB=CB$$(Sides of a square are equal)

$$BO=BO$$(Common)

$$\therefore \mathrm{\Delta }AOB\cong \mathrm{\Delta }COB$$(By SSS congruency)

$$\therefore \mathrm{\angle }AOB=\mathrm{\angle }COB$$(By CPCT)

However,  (Linear pair)

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

3. Diagonal AC of a parallelogram ABCD is bisecting $$\mathrm{\angle }A$$(see the given figure). Show that

(i) It is bisecting $$\mathrm{\angle }C$$also,

(ii) ABCD is a rhombus

Answer: 

Given: Diagonal AC of a parallelogram ABCD is bisecting $$\mathrm{\angle }A$$

To find: (i) It is bisecting $$\mathrm{\angle }C$$ also,

(ii) ABCD is a rhombus

 (i) ABCD is a parallelogram.

$$\mathrm{\angle }DAC=\mathrm{\angle }BCA$$ (Alternate interior angles) ... (1)

And, $$\mathrm{\angle }BAC=\mathrm{\angle }DCA$$(Alternate interior angles) ... (2)

However, it is given that AC is bisecting  $$\mathrm{\angle }A$$.

$$\mathrm{\angle }DAC=\mathrm{\angle }BAC$$... (3)

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

$$\mathrm{\angle }DAC=\mathrm{\angle }BCA=\mathrm{\angle }BAC=\mathrm{\angle }DCA$$... (4)

$$\mathrm{\angle }DCA=\mathrm{\angle }BCA$$

Hence, AC is bisecting $$\mathrm{\angle }C$$.

(ii) From Equation (4), we obtain

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

$$DA=DC$$ (Side opposite to equal angles are equal)

However, $$DA=BC$$ and $$AB=CD$$ (Opposite sides of a parallelogram)

$$AB=BC=CD=DA$$

As a result, ABCD is a rhombus.

4. ABCD is a rectangle in which diagonal AC bisects $$\mathrm{\angle }A$$ as well as $$\mathrm{\angle }C$$. Show that:

(i) ABCD is a square

(ii) Diagonal BD bisects $$\mathrm{\angle }B$$ as well as $$\mathrm{\angle }D$$.

Answer:

Given: ABCD is a rectangle where the diagonal AC bisects $$\mathrm{\angle }A$$ as well as $$\mathrm{\angle }C$$.

To find: (i) ABCD is a square

              (ii) Diagonal BD bisects $$\mathrm{\angle }B$$ as well as $$\mathrm{\angle }D$$.

It is given that ABCD is a rectangle.$$\mathrm{\angle }A=\mathrm{\angle }C$$$$$$

$\Rightarrow {\displaystyle \frac{1}{2}}\mathrm{\angle }A={\displaystyle \frac{1}{2}}\mathrm{\angle }C$ (AC bisects $$\mathrm{\angle }A$$ and $$\mathrm{\angle }C$$)

$\Rightarrow \mathrm{\angle }DAC={\displaystyle \frac{1}{2}}\mathrm{\angle }DCA$

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

Also, $$DA=BC$$ and $$AB=CD$$ (Opposite sides of the rectangle are same)

$$AB=BC=CD=DA$$

ABCD is a rectangle with equal sides on all sides.

Hence, ABCD is a square.

(ii) Let us now join BD.

In $$\mathrm{\Delta }BCD$$,

$$BC=CD$$ (Sides of a square are equal to each other)

$$\mathrm{\angle }CDB=\mathrm{\angle }CBD$$ (Angles opposite to equal sides are equal)

However, $$\mathrm{\angle }CDB=\mathrm{\angle }ABD$$ (Alternate interior angles for $$AB||CD$$)

$$\mathrm{\angle }CBD=\mathrm{\angle }ABD$$

BD bisects $$\mathrm{\angle }B.$$

Also, $$\mathrm{\angle }CBD=\mathrm{\angle }ADB$$(Alternate interior angles for $$BC||AD$$)

$$\mathrm{\angle }CDB=\mathrm{\angle }ABD$$

BD bisects $$\mathrm{\angle }D$$ and $$\mathrm{\angle }B$$.

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

(i) $$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$$

(ii) $$AP=CQ$$

(iii) $$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$$

(iv) $$AQ=CP$$

(v) APCQ is a parallelogram

Answer:

Given: A parallelogram is given.

To prove: (i) $$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$$

                  (ii) $$AP=CQ$$

      (iii) $$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$$

      (iv) $$AQ=CP$$

      (v) APCQ is a parallelogram

(i) In $$\mathrm{\Delta }APD$$ and $$\mathrm{\Delta }CQB$$,

$$\mathrm{\angle }ADP=\mathrm{\angle }CBQ$$ (Alternate interior angles for $$BC||AD$$)

$$AD=CB$$ (Opposite sides of the parallelogram ABCD)

$$DP=BQ$$ (Given)

$$\therefore \mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$$ (Using SAS congruence rule)

(ii) As we had observed that $$\mathrm{\Delta }APD\cong \mathrm{\Delta }CQB$$,

$$\therefore AP=CQ$$ (CPCT)

(iii) In $$\mathrm{\Delta }AQB$$ and $$\mathrm{\Delta }CPD$$,

$$\mathrm{\angle }ABQ=\mathrm{\angle }CDP$$ (Alternate interior angles for $$AB||CD$$)

$$AB=CD$$ (Opposite sides of parallelogram ABCD)

$$BQ=DP$$ (Given)

$$\therefore \mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$$ (Using SAS congruence rule)

(iv) Since we had observed that $$\mathrm{\Delta }AQB\cong \mathrm{\Delta }CPD$$,

$$\therefore AQ=CP$$ (CPCT)

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

$$AQ=CP$$ and

$$AP=CQ$$

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

6. 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) $$\mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$$

(ii) $$AP=CQ$$

Answer:

(i) In $$\mathrm{\Delta }APB$$and $$\mathrm{\Delta }CQD$$,

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

$$AB=CD$$ (The opposite sides of a parallelogram ABCD)

$$\mathrm{\angle }ABP=\mathrm{\angle }CDQ$$ (Alternate interior angles for $$AB||CD$$)

$$\therefore \mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$$ (By AAS congruency)

(ii) By using

$$\therefore \mathrm{\Delta }APB\cong \mathrm{\Delta }CQD$$, we obtain

$$AP=CQ$$ (By CPCT)

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

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$$

(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) $$\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$$

(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) $$AD=CE$$ (Opposite sides of parallelogram AECD)

However, $$AD=BC$$ (Given)

Therefore, $$BC=CE$$

$$\mathrm{\angle }CEB=\mathrm{\angle }CBE$$ (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

$$\mathrm{\angle }A=\mathrm{\angle }B$$

 (ii) $$AB||CD$$

 (Angles on a same side of the transversal)

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

$$\therefore \mathrm{\angle }A+\mathrm{\angle }D=\mathrm{\angle }C+\mathrm{\angle }B$$

However, $$\mathrm{\angle }A=\mathrm{\angle }B$$ (Using the result obtained in (i))

$$\therefore \mathrm{\angle }C=\mathrm{\angle }D$$

 (iii) In $$\mathrm{\Delta }ABC$$ and $$\mathrm{\Delta }BAD$$,

$$AB=BA$$ (Common side)

$$BC=AD$$ (Given)

$$\mathrm{\angle }B=\mathrm{\angle }A$$ (Proved before)

$$\therefore \mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$$ (SAS congruence rule)

(iv) We had seen that,

$$\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$$

$$\therefore AC=BD$$ (By CPCT)

Exercise 8.2

1. 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) $$SR||AC$$ and $$SR={\displaystyle \frac{1}{2}}AC$$

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Answer

Given: ABCD is a quadrilateral

To prove: (i) $$SR||AC$$ and $$SR={\displaystyle \frac{1}{2}}AC$$

     (ii) PQ = SR

     (iii) PQRS is a parallelogram.

(i) In $$\mathrm{\Delta }ADC$$, 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.

$$\therefore SR||AC$$ and $$SR={\displaystyle \frac{1}{2}}AC$$... (1)

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

$$PQ||AC$$and $$PQ={\displaystyle \frac{1}{2}}AC$$... (2)

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

$$PQ||SR$$ and $$PQ={\displaystyle \frac{1}{2}}SR$$... (3)

$$\therefore PQ=SR$$

 (iii) From Equation (3), we obtained

$$PQ||SR$$ and  $$PQ=SR$$

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

PQRS is thus a parallelogram.

2. 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 $$\mathrm{\Delta }ABC$$, P and Q are the mid-points of sides AB and BC respectively.

$$PQ||AC\text{ , }PQ={\displaystyle \frac{1}{2}}AC$$ (Using mid-point theorem) ... (1)

In $$\mathrm{\Delta }ADC$$,

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

$$RS||AC\text{ , }RS={\displaystyle \frac{1}{2}}AC$$ (Using mid-point theorem) ... (2)

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

$$PQ||RS$$ and $$PQ=RS$$

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,

$$MQ\left|\left|ON(PQ\right|\right|AC)$$

$$QN\left|\left|OM(QR\right|\right|BD)$$

Hence , OMQN is a parallelogram.

$$\begin{array}{l}\therefore \mathrm{\angle }MQN=\mathrm{\angle }NOM\\ \therefore \mathrm{\angle }PQR=\mathrm{\angle }NOM\end{array}$$

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

$$\therefore \mathrm{\angle }PQR={90}^{\circ }$$

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 $$\mathrm{\Delta }ABC$$,

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

$$\therefore PQ||AC$$ and $$PQ={\displaystyle \frac{1}{2}}AC$$(Mid-point theorem) ... (1)

Similarly in $$\mathrm{\Delta }ADC$$,

$$SR||AC\text{ , }SR={\displaystyle \frac{1}{2}}AC$$ (Mid-point theorem) ... (2)

Clearly, $$PQ||SR$$ and $$PQ=SR$$

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

$$\therefore PS||QR,PS=QR$$ (Opposite sides of parallelogram) ... (3)

In $$\mathrm{\Delta }BCD$$, Q and R are the mid-points of side BC and CD respectively.

$$\therefore QR||BD\text{ , }QR={\displaystyle \frac{1}{2}}BD$$ (Mid-point theorem) ... (4)

Also, the diagonals of a rectangle are equal.

$$\therefore AC=BD$$…(5)

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

$$PQ=QR=SR=PS$$

So , PQRS is a rhombus

4. ABCD is a trapezium in which $$AB||DC$$, 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 $$AB||DC$$, 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 $$\mathrm{\Delta }ABD$$,

$$EF||AB$$ and E is the mid-point of AD.

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

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

$$\therefore EF||CD$$ (Two lines parallel to the same line are parallel)

In $$\mathrm{\Delta }BCD$$, $$GF||CD$$ 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.

$$AB||CD$$

And hence, $$AE||FC$$

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

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

$$AE=FC$$ (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.

$$\therefore AF||EC$$ (Opposite sides of a parallelogram)

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

$$\therefore DP=PQ$$... (1)

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

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

$$\therefore PQ=QB$$... (2)

From Equations (1) and (2),

$$DP=PQ=BQ$$

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

6. 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 $\mathrm{\perp }$ AC

(iii) $$CM=MA={\displaystyle \frac{1}{2}}AB$$

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 $\mathrm{\perp }$ AC

(iii) $$CM=MA={\displaystyle \frac{1}{2}}AB$$

(i) In $$\mathrm{\Delta }ABC$$,

It is given that M is the mid-point of AB and $$MD||BC$$.

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

(ii) As $$DM||CB$$ and AC is a transversal line for them, therefore,

 (Co-interior angles)

(iii) Join MC.

In $$\mathrm{\Delta }AMD$$ and $$\mathrm{\Delta }CMD$$,

$$AD=CD$$ (D is the mid-point of side AC)

$$\mathrm{\angle }ADM=\mathrm{\angle }CDM$$ (Each )

DM = DM (Common)

$$\therefore \mathrm{\Delta }AMD\cong \mathrm{\Delta }CMD$$ (By SAS congruence rule)

Therefore, $$AM=CM$$(By CPCT)

However, $$AM={\displaystyle \frac{1}{2}}AB$$ (M is mid-point of AB)

Therefore, it is said that

$$CM=AM={\displaystyle \frac{1}{2}}AB$$

Overview of Deleted Syllabus for CBSE Class 9 Maths  Quadrilaterals

Class 9 Maths Chapter 8: Exercises Breakdown

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 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. Questions from Chapter 8 - Quadrilaterals in Class 9 Maths might range from 2-4 questions, including both short and long answer types.

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

CBSE Class 9 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9  Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Study Materials for Class 9 Maths