Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 1.

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

(i) SR ∥ AC and SR = \(\frac{1}{2}\)AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Answer:

Given: A quadrilateral ABCD in which P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA. Also AC is its diagonal.

To show :

(i) SR ∥ AC and SR = \(\frac{1}{2}\)AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

(i) In ∆ACD, we have :

S is the mid-point of AD and R is the mid-point of CD.

Therefore, SR ∥ AC and SR = \(\frac{1}{2}\)AC (Mid-point theorem)

(ii) In ∆ABC, we have :

P is the mid-point of the side AB and Q is the mid-point of the side BC.

Therefore, PQ ∥ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem)

Thus, we have shown that:

(iii) Since PQ = SR and PQ ∥ SR, therefore one pair-of opposite sides are equal and parallel.

∴ PQRS is a parallelogram.

Question 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 in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

To show: PQRS is a rectangle.

Construction: Join AC.

In ∆ABC, P and Q are respectively the mid-points of AS and BC.

PQ ∥ AC and PQ = \(\frac{1}{2}\)AC

Similarly, in ∆ADC, R and S are respectively the mid-points of CD and AD.

SR ∥ AC and SR = \(\frac{1}{2}\)AC ……………(2)

From (1) and (2), we get PQ ∥ RS and PQ = SR

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is equal and parallel.

∴ PQRS is a parallelogram.

Also, AS = BC (Sides of a rhombus)

Now, SP ∥ RQ and PQ intersects them.

∴ ∠SPQ + ∠PQR = 180°

From (3) and (4), we get

∠SPQ = ∠PQR = 90°

Thus, PQRS is a parallelogram whose one angle ∠SPQ = 90°.

Hence, PQRS is a rectangle.

Question 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 in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quad. PQRS.

To show : PQRS is a rhombus.

Construction: Join AC.

In ∆ABC, P and Q are respectively the mid-points of sides AB and BC.

PQ ∥ AC and PQ = \(\frac{1}{2}\)AC ………..(1)

Similarly, in AADC, R and S are respectively the mid-points of sides CD and AD.

∴ SR ∥ AC and SR = \(\frac{1}{2}\)AC …………….(2)

From (1) and (2), we get PQ ∥ SR and PQ = SR …………….. (3)

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is parallel and equal. [From (3)]

∴ PQRS is a parallelogram.

Now, AD = BC (Opp. sides of rect. ABCD) ….(4)

AP = BP ⇒ AS = BQ (∵ P is the mid-point of AB)

∠PAS = ∠PBQ (Each = 90°)

AS = BQ (Shown above)

∴ ∆APS = ∆BPQ (SAS Cong, axiom)

PS = PQ (CPCT) ………….(5)

From (4) and (5), we get PQRS is a rhombus.

Question 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 figure). Show that F is the mid-point of BC.

Answer:

Given : In trapezium ABCD, AB ∥ DC,

E is the mid-point of AD and EF ∥ AB.

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

Construction: Join DB. Let it intersect EF in G.

In ADAB, E is the mid-point of AD (Given) and EG ∥ AB (∵ EF ∥AB)

∴ By converse of mid-point theorem, G is the midpoint of DB.

In ABCD, G is the mid-point of BD and GF ∥ DC (AB ∥ DC, EF ∥ AB ⇒ DC ∥ EF)

∴ By converse of mid-point theorem,

F is the mid-point of BC.

Question 5.

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

Answer:

Given : E and F are the mid-points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.

To-show : BQ = QP = PD.

ABCD is parallelogram. (Given)

∴ AB ∥ DC and AB = DC (Opp. sides of parallelogram)

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

AE = \(\frac{1}{2}\)AB

F is the mid-point of CD.

CF = \(\frac{1}{2}\)CD (∵ CD = AB)

So, AE = CF

From (1) and (2) AE ∥ CF (∵ AB ∥ DC)

Thus, a pair of opposite sides of a quadrilateral AECF are parallel and equal.

Quad. AECF is a parallelogram.

So, EC ∥ AF

∴ EQ ∥ AP and QC ∥ PF

In ABPA, E is the mid-point of BA (Given)

and EQ ∥ AP (Shown)

BQ = QP (Converse of mid-point theorem) ……….. (3)

Similarly by taking ACQD, we can prove that

DP = QP …………… (4)

From (3) and (4), we get

BQ = QP = PD

Hence, AF and CE trisect the diagonal BD.

Question 6.

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

Answer:

Given : A quad. ABCD, P, Q, R and S are respectively the mid-points of AB, BC, CD and DA. PR and QS intersect each other at O.

To show: OP = OR, OQ = OS.

Construction: Join PQ, QR, RS, SP, AC and BD

In ∆ABC, P and Q are mid-points of AB and BC respectively

∴ PQ ∥ AC and PQ = \(\frac{1}{2}\)AC

Similarly, we can show that

RS ∥ AC and RS = \(\frac{1}{2}\)AC

∴ PQ ∥SR and PQ = SR

Thus, a pair of opposite sides of a quadrilateral PQRS are parallel and equal.

∴ Quadrilateral PQRS is a parallelogram.

Since, the diagonals of a parallelogram bisect each other, therefore diagonals PR and QS of ∥gm PQRS, i.e., the line segments joining the mid-points of opposite sides of quadrilateral ABCD bisect each other.

Question 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 ⊥ AC

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

Answer:

Given : ∆ABC is right angled at C, M is the mid-point of hypotenuse AB. Also, MD ∥ BC.

To show that:

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

(ii) MD ⊥ AC.

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

(i) In ∆ABC, M is the mid-point of AB and MD ∥ BC. Therefore, D is the mid-point of AC.

i.e., AD = DC

(ii) Since MD ∥ BC, therefore

∠ADM = ∠ACB (Corresponding angles)

So, ∠ADM = 90° [v ∠ACB = 90° (given)]

But, ∠ADM + ∠CDM = 180°

[ v ∠ADM and ∠CDM are angles of a linear pair]

∴ 90° + ∠CDM = 180° or ∠CDM =90°

Thus, ∠ADM = ∠CDM =90° ……………….. (2)

So, MD ⊥ AC

(iii) In As AMD and CMD, we have :

AD = CD [From (1)]

∠ADM = ∠CDM [From (2)]

and MD = MD (common)

∴ By SAS criterion of congurence

∆AMD ≅ ∆CMD

So, MA = MC

(v Corresponding parts of congruent triangles are equal)

Also, MA = \(\frac{1}{2}\)AB, since M is the mid-point of AC.

Hence, CM = MA = \(\frac{1}{2}\)AB.

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