NCERT Solutions for Class 9 Maths 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 = 12 AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q1

Solution:

(i) In ∆ACD, We have

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

SR = 12AC and SR || AC …(1)

[By mid-point theorem]

(ii) In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ = 12AC and PQ || AC …(2)

[By mid-point theorem]

From (1) and (2), we get

PQ = 12AC = SR and PQ || AC || SR

⇒ PQ = SR and PQ || SR

(iii) In a quadrilateral PQRS,

PQ = SR and PQ || SR [Proved]

∴ 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:

We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q2

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

∴ PQ = 12AC and PQ || AC …(1)

[By mid-point theorem]

In ∆ADC, R and S are the mid-points of CD and DA respectively.

∴ SR = 12AC and SR || AC …(2)

[By mid-point theorem]

From (1) and (2), we get

PQ = 12AC = SR and PQ || AC || SR

⇒ PQ = SR and PQ || SR

∴ PQRS is a parallelogram. …….(3)

Now, in ∆ERC and ∆EQC,

∠1 = ∠2

[ ∵ The diagonals of a rhombus bisect the opposite angles]

CR = CQ [ ∵CD2 = BC2]

CE = CE [Common]

∴ ∆ERC ≅ ∆EQC [By SAS congruency]

⇒ ∠3 = ∠4 …(4) [By C.P.C.T.]

But ∠3 + ∠4 = 180° ……(5) [Linear pair]

From (4) and (5), we get

⇒ ∠3 = ∠4 = 90°

Now, ∠RQP = 180° – ∠b [ Y Co-interior angles for PQ || AC and EQ is transversal]

But ∠5 = ∠3

[ ∵ Vertically opposite angles are equal]

∴ ∠5 = 90°

So, ∠RQP = 180° – ∠5 = 90°

∴ One angle of parallelogram PQRS is 90°.

Thus, PQRS is a rectangle.

Question 3.

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

Solution:

We have,

Now, in ∆ABC, we have

PQ = 12AC and PQ || AC …(1)

[By mid-point theorem]

Similarly, in ∆ADC, we have

SR = 12AC and SR || AC …(2)

From (1) and (2), we get

PQ = SR and PQ || SR

∴ PQRS is a parallelogram.

Now, in ∆PAS and ∆PBQ, we have

∠A = ∠B [Each 90°]

AP = BP [ ∵ P is the mid-point of AB]

AS = BQ [∵ 12AD = 12BC]

∴ ∆PAS ≅ ∆PBQ [By SAS congruency]

⇒ PS = PQ [By C.P.C.T.]

Also, PS = QR and PQ = SR [∵opposite sides of a parallelogram are equal]

So, PQ = QR = RS = SP i.e., PQRS is a parallelogram having all of its sides equal.

Hence, 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.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q4

Solution:

We have,

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q4.1

In ∆DAB, we know that E is the mid-point of

AD and EG || AB [∵ EF || AB]

Using the converse of mid-point theorem, we get, G is the mid-point of BD.

Again in ABDC, we have G is the midpoint of BD and GF || DC.

[∵ AB || DC and EF || AB and GF is a part of EF]

Using the converse of the mid-point theorem, we get, 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.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q5

Solution:

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB || DC

⇒ AE || FC …(1)

and AB = DC

⇒ 12AB = 12DC

⇒ AE = FC …(2)

From (1) and (2), we have

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ

[∵ AF || CE]

⇒ DP = PQ …(3)

[By converse of mid-point theorem] Similarly, in A BAP, E is the mid-point of AB and EQ || AP [∵AF || CE]

⇒ BQ = PQ …(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, the line segments AF and EC 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:

Let ABCD be a quadrilateral, where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

Join PQ, QR, RS and SP.

Let us also join PR, SQ and AC.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q6

Now, in ∆ABC, we have P and Q are the mid-points of its sides AB and BC respectively.

∴ PQ || AC and PQ = 12 AC …(1)

[By mid-point theorem]

Similarly, RS || AC and RS = 12AC …(2)

∴ By (1) and (2), we get

PQ || RS, PQ = RS

∴ PQRS is a parallelogram.

And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other. Thus, the line segments joining the midpoints of opposite sides of a 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 = 12AB

Solution:

we have

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Q7

(i) In ∆ACB, We have

M is the mid-point of AB. [Given]

MD || BC , [Given]

∴ Using the converse of mid-point theorem,

D is the mid-point of AC.

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

∠MDA = ∠BCA

[ ∵ Corresponding angles are equal] As

∠BCA = 90° [Given]

∠MDA = 90°

⇒ MD ⊥AC.

(iii) In ∆ADM and ∆CDM, we have

∠ADM = ∠CDM [Each equal to 90°]

MD = MD [Common]

AD = CD [∵ D is the mid-point of AC]

∴ ∆ADM ≅ ∆CDM [By SAS congruency]

⇒ MA = MC [By C.P.C.T.] .. .(1)

∵ M is the mid-point of AB [Given]

MA = 12AB …(2)

From (1) and (2), we have

CM = MA = 12AB