Bihar Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th 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.

Solution:

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof :

(i) In Δ ACD, we have

S is the mid-point of AD and R is the mid-point of CD. Then SR || AC and SR = \(\frac { 1 }{ 2 }\)AC [Mid-point theorem]

(ii) In Δ ABC, we have

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

Then PQ || AC

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

Thus, we have proved that:

(iii) Since PQ = SR and PQ || SR

⇒ 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 respectively the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Solution:

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 prove : PQRS is a rectangle.

Construction : Join AC.

Proof:

In Δ ABC, P and Q are the mid-points of AB and

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\)AC … (1)

Similarly, in A ADC, R and S are the mid-points of CD and PD.

∴ SR || AC and SR = \(\frac { 1 }{ 2 }\)AC

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

∴ AB = BC [Sides of a rhombus]

⇒ \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 2 }\)BC

⇒ PB = BQ

⇒ ∠3 = ∠4 [∠s opp. to equal sides of a A]

Now, in ∆ APS and ∆ CQR, we have

AP = CQ [Halves of equal sides AB, BC]

AS = CR [Halves of equal sides AD, CD]

PS = QR [Opp. sides of parallelogram PQRS]

∴ ∆ APS = ∆ CQR [SSS Cong. Theorem]

⇒ ∠1 = ∠2 [C.P.C.T.]

Now, ∠1 + ∠SPQ + ∠3 = 180° [Linear pair axiom]

∴ ∠1 + ∠PQ + ∠3 = ∠2 + ∠PQR + ∠4

But ∠1 = ∠2 and ∠3 = ∠4[Proved above]

∴ ∠SPQ = ∠PQR … (3)

∴ SP || RQ intersects them,

∴ ∠SPQ + ∠PQR = 180° … (4)

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.

Solution:

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 prove : PQRS is a rhombus.

Construction : Join AC.

Proof:

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

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

Similarly, in ∆ ADC, R and S are 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 …(4)

Now, AD = BC [Opp. sides ofrect. ABCD]

⇒ \(\frac { 1 }{ 2 }\) AD = \(\frac { 1 }{ 2 }\) BC ⇒ AS = BQ

In ∆s APS and BPQ, we have

AP = BP [∵ P is the mid-point of AB]

∠PAS = ∠PBQ [Each = 90°]

AS = BQ [Proved above]

∴ ∆ APS ≅ ∆ BPQ [SAS Cong. axiom]

⇒ PS = PQ [C.P.C.T] … (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.

Solution:

Given : In trapezium ABCD, AB || DC

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

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

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

Proof:

In A DAB, E is the mid-point of AD [Given]

EG || AB [∵ EF || AB]

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

In ∆ BCD, G is the mid-point of BD [Proved]

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 mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Solution:

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

To prove : BQ = QP = PD

Proof:

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 … (1)

F is the mid-point of CD

∴ CF = \(\frac { 1 }{ 2 }\)CD

⇒ CF = \(\frac { 1 }{ 2 }\) AB [∵ CD = AB] … (2)

From (1) and (2),

AE = CF

Also AE || CF [∵ AB || DC]

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

Quad. AECF is a parallelogram

⇒ EC || AF

⇒ EQ || AP and QC || MF

In ∆ BMA, E is the mid-point of BA [Gtuen]

EQ || AP [Proved]

∴ BQ = LP

[Converse of mid-point theorem] … (3)

Similarly by taking ∆ CLD, we can prove that

DP = QP … (4)

From (3) and (4), we get

BQ = QP = PD

Hence, AF and CE trisect the diagonal AC.

Question 6.

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

Solution:

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 prove : OP = OR,

OQ = OS.

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

Proof:

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

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC

Similarly, we can prove 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.

∴ Diagonals PR and QS of a ||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

Solution:

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

To prove that :

(i) D is the mid-points of AC

(ii) MD ⊥ AC

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

Proof :

(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 …(1)

(ii) Since MD || BC. Therefore,

∠ADM = ∠ACB [Corresponding angles]

⇒ ∠ADM = 90° [∵ ∠ACB = 90° (given)]

But, ∠ADM + ∠CDM = 180°

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

∴ 90° + ∠CDM = 180° ⇒ ∠CDM = 90°

Thus, ∠ADM = ∠CDM = 90° … (2)

⇒ MD ⊥ AC

(iii) In ∆s AMD and CMD, we have

AD = CD [From (1)]

∠ADM = ∠CDM [From (2)]

and, MD = MD [Common]

∴ By SAS criterion of congruence

∆ AMD ≅ ∆ CMD

⇒ MA = MC

[∵ 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.