NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 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) \[SR{\text{ }}||{\text{ }}AC\] 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.

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 \[\Delta ABC\], 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. 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

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

Conclusion

In exercise 8.2 class 9th Quadrilaterals, students delve into the properties and theorems related to parallelograms. Understanding these properties is crucial, as they form the foundation for solving problems related to angles, sides, and diagonals of quadrilaterals. Focus on mastering the conditions for a quadrilateral to be a parallelogram and the various methods to prove it.

This exercise is essential for developing problem-solving skills in geometry. In previous exams, about 3-4 questions have been asked from this chapter, emphasizing its importance. By thoroughly practicing the solutions provided by Vedantu, students can enhance their understanding and perform well in exams.

Class 9 Maths Chapter 8: Exercises Breakdown

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Chapter-Specific NCERT Solutions for Class 9 Maths

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