NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2

In NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 takes students deeper into the study of special types of quadrilaterals, such as parallelograms, rectangles, rhombuses, and squares. In this exercise, learners explore important properties like the opposite sides and angles being equal, diagonals bisecting each other, and specific conditions that define each quadrilateral. The questions are designed to help students apply these properties to solve for unknown angles, sides, and prove certain geometrical facts.

By solving Exercise 8.2, students develop strong reasoning and proof-writing skills, which are crucial for geometry. The NCERT solutions offer clear and logical steps, making it easier for learners to understand complex concepts and approach each problem with confidence.

1.0Download NCERT Solutions of Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 Free PDF

Are you searching for full solutions and complete solutions to Exercise 8.2. You can download the NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.2 for complete reasoning and full solutions.

Key Concepts of Exercise 8.2

• Opposite angles and sides of a parallelogram
• The diagonals bisect each other in a parallelogram
• Conditions for a quadrilateral to be a parallelogram
• Using congruence triangles with geometric proofs
• Deductive reasoning based on the properties and theorems provided

2.0NCERT Exercise Solutions Class 9 Chapter 8 Quadrilaterals : All Exercises

Here is a quick overview and access to solutions for all exercises from Chapter 8, Quadrilaterals .

3.0NCERT Class 9 Maths Chapter 8 Exercise 8.2 : Detailed Solutions

1. ABCD is a quadrilateral in which P,Q,R and S are mid points of the sides AB,BC, CD and DA (fig.) and AC is a diagonal. Show that

(i) SR∥AC and SR = (1/2)AC

(ii) PQ=SR

(iii) PQRS is a parallelogram.

Sol. Given : ABCD is a quadrilateral in which P,Q,R and S are mid-points of AB, BC, CD and DA respectively. AC is a diagonal.

To prove :

(i) SR∥AC and SR = (1/2)AC

(ii) PQ=SR

(iii) PQRS is a parallelogram.

Proof:

(i) In ΔDAC:

S is the mid-point of DA (Given).

R is the mid-point of DC (Given).

By Mid-point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

Therefore, SR∥AC and SR = (1/2)AC. ... (a)

(ii) In ΔBAC:

P is the mid-point of AB (Given).

Q is the mid-point of BC (Given).

By Mid-point theorem:

Therefore, PQ∥AC and PQ = (1/2)AC. ... (b)

From (a) and (b), since both SR and PQ are equal to (1/2)AC, then PQ=SR.

(iii) From (a) SR∥AC and from (b) PQ∥AC.

Lines parallel to the same line are parallel to each other.

Therefore, PQ∥SR.

Also, from (ii), we proved PQ=SR.

A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length.

Therefore, PQRS is a parallelogram.

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

Sol. 

Given : P,Q,R and S are the mid-points of respective sides AB,BC,CD and DA of rhombus ABCD. PQ,QR,RS and SP are joined.

To prove : PQRS is a rectangle.

Construction : Join A and C, and B and D.

Proof:

From previous problem (Exercise 8.2, Q1):

In ΔABC, P and Q are mid-points of AB and BC. So, PQ∥AC and PQ = (1/2)AC. ... (1)

In ΔADC, R and S are mid-points of CD and DA. So, SR∥AC and SR = (1/2)AC. ... (2)

From (1) and (2), PQ∥SR and PQ=SR. Therefore, PQRS is a parallelogram.

Similarly, using diagonal BD:

In ΔABD, P and S are mid-points of AB and AD. So, PS∥BD and PS = (1/2)BD. ... (3)

In ΔBCD, Q and R are mid-points of BC and CD. So, QR∥BD and QR = (1/2)BD. ... (4)

From (3) and (4), PS∥QR and PS=QR. (This further confirms PQRS is a parallelogram).

Now, ABCD is a rhombus. We know that the diagonals of a rhombus bisect each other at right angles.

Therefore, AC⊥BD.

Since PQ∥AC (from (1)) and PS∥BD (from (3)).

When two lines are parallel to two other perpendicular lines, then the angle between the first two lines is also 90°.

So, PQ⊥PS. This means ∠SPQ = 90°.

Since PQRS is a parallelogram and one of its angles (∠SPQ) is 90°, it is a rectangle.

Hence, 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.

Sol. Given : A rectangle ABCD in which P,Q,R and S are the mid-points of the sides AB, BC,CD and DA respectively. PQ, QR, RS and SP are joined.

To prove : PQRS is a rhombus.

Construction : Join AC.

Proof :

In ΔABC, P and Q are the midpoints of sides AB, BC respectively.

By Mid-Point Theorem: PQ∥AC and PQ = (1/2)AC. ... (i)

In ΔADC, R and S are the mid-points of sides CD, AD respectively.

By Mid-Point Theorem: SR∥AC and SR = (1/2)AC. ... (ii)

From (i) and (ii), PQ∥SR and PQ=SR. Therefore, PQRS is a parallelogram.

Now, to prove PQRS is a rhombus, we need to show that two adjacent sides are equal (e.g., PS = PQ).

Consider ΔAPS and ΔBPQ:

AP = BP [P is the mid-point of AB]

AS = (1/2)AD

BQ = (1/2)BC

Since ABCD is a rectangle, AD = BC. So, (1/2)AD = (1/2)BC, which means AS = BQ.

∠PAS = ∠PBQ [Each = 90°, as ABCD is a rectangle]

Therefore, ΔAPS ≅ ΔBPQ [By SAS congruency rule]

Thus, PS = PQ [By C.P.C.T.]

Since PQRS is a parallelogram and its adjacent sides PS and PQ are equal, it is a rhombus.

Hence, 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 (fig.). Show that F is the mid-point of BC.

Sol. Given : ABCD is a trapezium with AB∥DC. 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.

Proof :

Let the line through E parallel to AB be denoted by 'l'. This line 'l' intersects diagonal BD at point G.

In ΔABD:

E is the mid-point of AD (Given).

EG∥AB (Part of line 'l' which is parallel to AB).

By the Converse of Mid-Point Theorem, a line through the mid-point of one side of a triangle parallel to another side bisects the third side.

Therefore, G is the mid-point of BD.

Now, consider the trapezium ABCD again.

We know that l∥AB and also AB∥CD (given).

Since lines parallel to the same line are parallel to each other, l∥CD.

Now consider ΔBCD:

G is the mid-point of BD (Proved above).

GF∥CD (Part of line 'l' which is parallel to CD).

By the Converse of Mid-Point Theorem:

Therefore, 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 (fig.). Show that the line segments AF and EC trisect the diagonal BD.

Sol. Given : ABCD is a parallelogram. E and F are mid-points of AB and DC respectively.

To prove: Diagonal BD is trisected by AF and EC, i.e., DP=PQ=QB.

Proof:

Since E and F are the mid-points of AB and CD respectively.

AE = (1/2)AB and CF = (1/2)CD.

Since ABCD is a parallelogram, AB=CD and AB∥DC.

Therefore, (1/2)AB = (1/2)CD, which means AE = FC.

Also, since AB∥DC, then AE∥FC.

Since one pair of opposite sides (AE and FC) are equal and parallel, AECF is a parallelogram.

Since AECF is a parallelogram, its opposite sides are parallel.

Therefore, AF∥EC.

This implies FP∥QC (as FP is part of AF and QC is part of EC). ... (i)

Now, consider ΔDCQ:

F is the mid-point of CD (Given).

FP∥CQ (from (i)).

By the Converse of Mid-Point Theorem (a line through the mid-point of one side of a triangle parallel to another side bisects the third side):

Since FP is drawn from F (mid-point of CD) and is parallel to QC, it must bisect DQ.

Therefore, P is the mid-point of DQ, which means DP = PQ. ... (ii)

Similarly, consider ΔABP:

E is the mid-point of AB (Given).

EQ∥AP (since EQ is part of EC and AP is part of AF, and AF∥EC).

By the Converse of Mid-Point Theorem:

Since EQ is drawn from E (mid-point of AB) and is parallel to AP, it must bisect BP.

Therefore, Q is the mid-point of BP, which means BQ = PQ. ... (iii)

From (ii) and (iii), we have DP = PQ and BQ = PQ.

Combining these, we get DP = PQ = BQ.

Since the diagonal BD is divided into three equal segments by points P and Q, the line segments AF and EC trisect the diagonal BD.

Hence, AF and CE 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⊥AC

(iii) CM=MA=(1/2)AB

Sol. Given : ΔABC is right-angled at C. M is the mid-point of hypotenuse AB. A line is drawn through M parallel to BC, intersecting AC at D.

To prove :

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

(ii) MD⊥AC.

(iii) CM=MA=(1/2)AB.

Proof:

(i) In ΔABC:

M is the mid-point of AB (Given).

MD∥BC (Given).

By the Converse of Mid-Point Theorem, a line through the mid-point of one side of a triangle parallel to another side bisects the third side.

Therefore, D is the mid-point of AC.

(ii) Since MD∥BC and AC is a transversal.

∠ADM = ∠ACB (Corresponding angles).

Given that ΔABC is right-angled at C, so ∠ACB = 90°.

Therefore, ∠ADM = 90°.

This means MD⊥AC.

(iii) In ΔCMD and ΔAMD:

CD = AD [Proved in (i), D is the mid-point of AC]

MD = MD [Common side]

∠CDM = ∠ADM [Each = 90°, Proved in (ii), MD⊥AC]

Therefore, ΔCMD ≅ ΔAMD [By SAS congruence rule]

Thus, CM = AM [By C.P.C.T.]

We are given that M is the mid-point of AB, so AM = (1/2)AB.

Combining these, we get CM = AM = (1/2)AB.

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

Sol. Given : A quadrilateral ABCD. Let P,Q,R and S be the mid-points of the sides AB, BC, CD and DA respectively.

To prove that PR and QS bisect each other.

Proof :

Join P, Q, R, S to form quadrilateral PQRS.

Consider ΔABC: P is mid-point of AB, Q is mid-point of BC.

By Mid-point theorem, PQ∥AC and PQ = (1/2)AC. ... (1)

Consider ΔADC: R is mid-point of CD, S is mid-point of DA.

By Mid-point theorem, SR∥AC and SR = (1/2)AC. ... (2)

From (1) and (2):

PQ∥SR (Since both are parallel to AC).

PQ = SR (Since both are equal to (1/2)AC).

Since one pair of opposite sides of quadrilateral PQRS (PQ and SR) are parallel and equal, PQRS is a parallelogram.

Now, PR and QS are the diagonals of the parallelogram PQRS.

We know that the diagonals of a parallelogram bisect each other.

Therefore, PR and QS bisect each other.

4.0Key Features and Benefits: Class 9 Maths Chapter 8 Exercise 8.2

• Well Organized Solutions: Every solution is well organized with logical reasoning for each step.
• CBSE Curriculum: Completion of the CBSE Chapter is fully aligned to the most recent NCERT Syllabus.
• Builds Proofing Skills: Assist in writing proof with triangles and properties of triangles.
• Visualize Learning: To develop understanding sample images of laminate and labeled figures are included.
• Exam Focused Practice: Aid in developing speed and accuracy in applying concepts in exam settings.