## MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

Question 1.

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

- SR ∥ AC and SR = \(\frac{1}{2}\) AC
- PQ = SR
- PQRS is a parallelogram.

Solution:

Given

ABCD is in which P, Q, R and S are the mid-points of sides AB,BC, CD and DA.

To prove.

- SR ∥ AC and SR= \(\frac{1}{2}\) AC
- PQ = SR
- PQRS is a parallelogram.

Proof:

In ∆ABC. P is the mid point of AB and Q is the midpoint of BC.

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

In ∆ADC, S is the mid-point of AD and R is the mid – point of DC.

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

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

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

∴ PQ = SR

3. From (1) and (2), we get

PQ ∥ SR and PQ = SR

∴ 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 P&RS is a rectangle.

Solution:

Given

ABCD is rhombus and P, Q and R and S are the mid-points of sides AB, BC, CD and DA.

To prove

PQRS is a rectangle.

Construction:

Join AC and BD).

Proof:

In ∆ABC, P is the midpoint of AB and Q is the midpoint of BC.

∴ PQ ∥AC (By MPT) …(1)

In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.

SR ∥ AC (By MPT) …(2)

From (1) and (2), we get

PQ ∥ SR …(3)

In ∆ABD, P is the midpoint of AB and S is the midpoint of AD.

PS ∥ BD …(4)

In ABCD, Q is the midpoint of BC and R is the midpoint of CD

∴ QR ∥ BD (ByMPT) …(5)

From (4) and (5), we get

PS ∥ QR …(6)

In quadrilateral PQRS, PQ ∥ SR and PS ∥ QR

∴ PQRS is a parallelogram.

In quadrilateral OESF

SF ∥ EO (∴ SR ∥ AO …(7)

SE ∥ FO (∴ SP ∥ BD) …(8)

SEOF is a parallelogram.

We know that, that in a parallelogram opposite angles are equal.

∴ ∠ESF = ∠EOF= 90°

(∴ In a rhombus, diagonals intersect each other at right angles)

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.

Given

ABCD is a rectangle P,Q,R and S are the mid-points of AB, BC, CD and DA.

To prove.

PQRS is a rhombus.

Proof:

In ∆ABC, P is the midpoint of AB and Q is the midpoint of BC.

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

In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.

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

From (1) and (2),we get

PQ ∥ SB and PQ = SR

PQRS is a parallelogram

AD = BC (∴ ABCD is a rectangle)

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

AS = BQ

In ∆PAS and ∆PBQ, AS = BQ (proved)

AP = BP P is the mid-point of AB

∠A = ∠B (each 90°)

and so PS = PQ (ByCPCT)

In a ∥^{gm}, if adjacent sides are equal, then it a rhombus.

∴ 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 and F (see Fig. below). Show that F is the mid-point of BC.

Solution:

Given

AB ∥ DC, DE = AE and EF ∥ AS.

To prove

F is the mid-point of BC

Proof:

AB ∥ DC (given) …(1)

AB ∥ EF (given) …(2)

From (1) and (2) we get,

DC ∥ EF

In ∆ABD, E is the mid – point of AD

EP ∥ AB (∴ EF∥AB)

∴ P is the mid – point of BD (By CMPT)

In ∆BCD, P is the midpoint of BD.

PF ∥ DC (∴ EF ∥ DC)

F is the midpoint of BC (By CMPT).

Question 5.

In a parallelogram ABCD, E and F are the mid-point of sides AB and CD respectively (See Fig.). Show that the line segments AD and EC trisect the diagonal BD.

Solution:

Given

ABCD is a ∥^{gm}, E and F are the mid-point of AB and CD.

To prove:

BQ = PQ = DP

Proof:

AB ∥ DC and AB = DC (∴ ABCD is a ∥gm)

⇒ \(\frac{1}{2}\) ∥ AB \(\frac{1}{2}\) DC and \(\frac{1}{2}\) AB = \(\frac{1}{2}\)DC

⇒ AE ∥ FC and AE = FC (∴ \(\frac{1}{2}\) AB = AE and \(\frac{1}{2}\) DC = FC)

AECF is a parallelogram (∴ AECF is a ∥^{gm})

In ∆ABP, E is the mid – point of AB,

EQ ∥AP (∴ AECF is a ∥<sup<gm)

∴ Q is the midpoint of BP (∴ by CMPT) …(1)

BQ = PQ

In ∆DQC, F is the midpoint of DC

EF ∥ CQ AECF is a ∥<sup<gm)

P is the midpoint of DQ

i.e., DP = PQ

From (1) and (2), we get BQ = PQ = DP.

Question 6.

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

Solution:

Given

ABCD is a in which P, Q.R and S are the mid-points of AB, BC, CD and DA respectively.

To prove

PR and SQ bisect each other.

Construction:

Join AC. Join PQ, QR, RS and SP.

Proof:

In ∆ABC, P is the midpoint of AB and Q is the midpointof BC.

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

In ∆ADC, S is the midpoint of AD and R is the midpoint of DC.

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

Form (1) and (2), we get

PQ ∥ SR and PQ = SR

∴ PQRS is a parallelogram in which PR and SQ are diagonals^ and so PR and SQ bisect each other (In a ∥^{gm}, diagnaols 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

- D is the mid – point of AC
- MD ⊥ AC
- CM = MA = \(\frac{1}{2}\) AB.

Solution:

Given

ABC is a right ∆. ∠C = 90°

AM = BM and MD ∥ BC.

To prove:

- D is the mid-point of AC, i.e., AD – CD
- MD ⊥ AC
- CM = MA = \(\frac{1}{2}\) AB

Proof:

1. In ∆ACB, M is the mid-point of AB and MD ∥ BC

∴ D is the mid – point of AC (By CMPT)

i.e., AD = CD

2. MD ∥ BC and AC is the transversal

∴ ∠ADM = ∠ACB (CA’s)

⇒ ∠ADM = 90° (∠ACB = 90°)

∴ MD ⊥ AC

3. In ∆ADM and ∆CDM,

AD = CD (proved)

∠ADM = ∠CDM (each 90°)

[∴ ∠ADM + ∠CDM = 180° (LPA’s); 90° + ∠CDM= 180°; ∠CDM = 90°]

MD = MD (common)

∴ ∆ADM = ∆CDM (By SAS)

and so MA = MC (By cpct) …(1)

MA = \(\frac{1}{2}\) AB (Given) …(2)

Form (1) and (2), we get MA = MC = \(\frac{1}{2}\) AB.