Students can Download Class 9 Maths Chapter 7 Quadrilaterals Ex 7.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 7 Quadrilaterals Ex 7.2

Question 1.

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

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Solution:

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

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

(ii) PQ = SR

(iii) PQRS is a parallelogram.

Proof: (i) In ∆ADC, S, and R are midpoints of AD and DC sides.

As per mid-point theorem,

SR || AC

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

(ii) In ∆ABC, P and Q are mid-points of AB and BC.

As per mid-point theorem,

PQ || AC

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

But, SR = \(\frac{1}{2}\) AC (Proved)

∴ PQ = SR

(iii) PQ = SR (Proved)

SR || AC and PQ || AC

∴ SR || PQ

Opposite sides of a quadrilateral PQRS are equal and parallel, hence 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.

Solution:

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

To Prove: PQRS is a rectangle.

Construction: Diagonals AC and BD are drawn.

Proof: To prove PQRS is a rectangle, one of its angles should be the right angle.

In ∆ADC, S and R are the midpoints of AD and DC.

∴ SR || AC

SR = \(\frac{1}{2}\) AC (mid-point formula)

In ∆ABC, P and Q are the midpoints AB and BC.

∴ PQ || AC

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

∴ SR || PQ and SR = PQ

∴ PQRS is a parallelogram.

But diagonals of a rhombus bisect at right angles. 90° angle is formed at ’O’.

∴ ∠P = 90°

∴ PQRS is a parallelogram, each of its angles is the right angle.

This is the property of the rectangle.

∴ 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:

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

To Prove: PQRS is a rhombus.

Construction: Diagonals AC and BD are drawn.

Proof: In ∆ABC, P and Q are the mid-points of AD and BC.

∴ PQ || AC (Mid-point theorem)

PQ = \(\frac{1}{2}\) AC ………….. (i)

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

∴ SR || AC

SR = \(\frac{1}{2}\) AC …………… (ii)

Similarly, in ∆ABD,

SP || BD

SP = \(\frac{1}{2}\) BD ……………….. (iii)

Similarly, in ∆BCD,

QR || BD

QR = \(\frac{1}{2}\) BD ……………… (iv)

From (i), (ii), (iii) and (iv),

PQ = QR = SR = PS and Opposite sides are parallel.

∴ 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. Show that F is the mid-point of BC.

Solution:

Data: 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.

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

Proof: Straight line EF cuts at G.

In ∆ABD, E is the mid-point of AD (Data)

EF || AB

EG || AB

∴ G is the mid-point of BD.

∵ Converse of mid-point formula.

DC || AB and EF || AB

⇒ DC || EF

In ∆BDC,

GF || DC

G is the mid-point of BD (proved)

∴ 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. Show that the line segments AF and EC trisect the diagonal BD.

Solution:

Data: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.

To Prove : Line segments AF and EC trisect the diagonal BD.

Proof: ABCD is a parallelogram.

AB || DC and AB = DC

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

AE = CF

and AE || CF (∵ AB || CD)

In AECF quadrilateral,

∴ AE || CF and AE = CF.

∴ AECF is a parallelogram.

∴ AF || EC

In ∆DQC, PF || QC (∵ AF || EC)

∴ P is the mid-point of DQ.

∴ DP = PQ ………….. (i)

In ∆APB, EQ || AP.

But E is the mid-point of AB (Data)

∴ Q is the mid-point of PB.

∴ PQ = QB …………… (ii)

From (i) and (ii),

DP = PQ = QB

∴ AF and EC line segments 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.

Solution:

Data : P, Q, R, and S are mid-points of AB, BC, CD, and DA respectively in quadrilateral ABCD.

To Prove: PR and SQ line segments bisect mutually.

Construction: Join the diagonal AC.

Proof: In ∆ADC,

S and R are the mid-points of AD and DC.

∴ SR || AC

SR = \(\frac{1}{2}\) AC ………… (i) (Mid-point Theorem)

Similarly, in ∆ABC,

PQ || AC

PQ = \(\frac{1}{2}\) AC …………. (ii)

From (i) and (ii),

SR = PQ and SR || PQ

∴ PQRS is a parallelogram.

PR and SQ are the diagonals of parallelogram PQRS.

∴ PR and SQ meet at O.

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:

Data: ABC is a triangle right angled at C.

A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.

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

(ii) MD ⊥ AC

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

Proof: (i) In ∆ABC,

M is the mid point of AB and

MB || BC Mid-point of BC (Data)

∴ D is the mid-point of AC (Mid-point formula)

(ii) MD || BC

∠BCD + ∠MDC = 180° (Sum of interior angles)

90 + ∠MDC = 180°

∠MDC = 180 – 90

∠MDC = 90°

∴ MD ⊥ AC

(iii) In this fig. CM is joined.

M is the mid-point of AB.

∴ MA = \(\frac{1}{2}\) AB ……….. (i)

Now, in ∆AMD and ∆CMD.

AD = DC (proved)

∠MDC = ∠MDA = 90°

MD is common.

∴∆AMD ≅ ∆CMD (SAS Postulate)

∴ CM = MA …………… (ii)

Comparing (i) and (ii),

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