Haryana State Board HBSE 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 8 Quadrilaterals Exercise 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 8.29). 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 : ABCD is a quadrilateral in which P, Q, R and S are mid points of the sides AB, BC, CD and DA respectively. 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 Î”ACD, we have

âˆ´ S and R are the mid points of AD and CD respectively.

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

(ii) In Î”ABC, we have

âˆ´ P and Q are the mid points of AB and BC respectively.

âˆ´ PQ || AC and PQ = \(\frac{1}{2}\)AC,

[By theorem 8.9] …(ii)

From (i) and (i), we get

PQ || SR and PQ = SR.

(iii) âˆµ PQ || SR and PQ = SR,

(As proved above)

âˆ´ PQRS is a parallelogram.

Hence proved

Question 2.

ABCD is a rhombus 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 rectangle.

Solution:

Given: A rhombus ABCD in which P, Q, R and S are mid points of the sides AB, BC, CD and DA respectively.

To prove: Quadrilateral PQRS is a rectangle.

Construction: Join AC and BD.

Proof: In Î”ADC, we have S and R are the mid points of AD and CD respectively.

âˆ´ SR || AC and SR = \(\frac{1}{2}\)AC ……(1)

and in Î”ABC, we have

P and Q are the mid points of AB and BC respectively.

âˆ´ PQ || AC and PQ = \(\frac{1}{2}\)AC,

[By theorem 8.9] …….. (ii)

From (i) and (ii), we get

PQ || SR and PQ = SR

âˆ´ PQRS is a parallelogram.

[By theorem 8.8]

Again in Î”DAB, we have

S and P are the mid points of AD and AB respectively.

SP || BD

â‡’ SL || MO

and SR || AC [From (i)]

â‡’ SM || LO

âˆ´ SLOM is a parallelogram.

Since, diagonals of a rhombus bisect each other at 90Â°.

âˆ´ âˆ AOD = 90Â°

âˆ MOL = 90Â°

But, âˆ LSM = âˆ MOL,

(Opposite angles of a parallelogram)

â‡’ âˆ LSM = 90Â°

â‡’ âˆ PSR = 90Â°

Since, one angle of a parallelogram PQRS is 90Â°.

Therefore, quadrilateral PQRS is a rectangle.

Hence proved

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: A rectangle ABCD in which P, Q, R and S are mid points of the sides AB, BC, CD and DA respectively.

To prove : Quadrilateral PQRS is a rhombus.

Construction: Join AC.

Proof: In Î”ADC, S and R are the mid points of AD and CD respectively.

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

[By thoerem 8.9]

In Î”ABC, P and Q are the mid points of AB and BC respectively.

âˆ´ PQ || AC and PQ = \(\frac{1}{2}\)AC …..(ii)

From (i) and (ii), we get

PQ || SR and PQ = SR

âˆ´ PQRS is a parallelogram.

[By theorem 8.8]

âˆµ ABCD is a rectangle.

âˆ A = âˆ B = 90Â° …(ii)

and AD = BC, (Opposite sides of a rectangle)

â‡’ \(\frac{1}{2}\)AD = \(\frac{1}{2}\)BC

â‡’ AS = BQ,

[âˆµ S is the mid point of AD and Q is the mid point of BC]

âˆ´ AS = \(\frac{1}{2}\)AD and BQ = \(\frac{1}{2}\)BC]

Now in Î”SAP and Î”QBP, we have

AS = BQ, [As proved above]

âˆ SAP = âˆ QBP,

[From (iii), Each = 90Â°]

and AP = BP,

[âˆµ P is the mid point of AB]

âˆ´ Î”SAP â‰… Î”QBP,

(By SAS congruence rule)

â‡’ SP = PQ, (CPCT)

Thus, adjacent sides of a parallelogram PQRS are equal.

Therefore, PQRS is a rhombus.

Hence proved

Question 4.

ABCD is a trapezium in which AB || CD, 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 8.30). Show that F is the mid point of BC.

Solution:

Given: ABCD is a trapezium in which AB || CD, EF || AB and E is the mid point of AD.

To prove: F is the mid point of BC.

Proof: Let line EF intersecting BD at G.

In Î”DAB, we have

EG || AB (âˆµ EF || AB)

and E is the mid point of AD.

âˆ´ G is the mid point of BD.

Now, AB || CD and AB || EF, (Given)

â‡’ EF || CD

â‡’ GF || CD

and G is the mid point of BD. (As proved above)

âˆ´ F is the mid point of BC.

Hence proved

Question 5.

In a parallelogram ABCD, E and Fare the mid points of sides AB and CD respectively (see figure 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Solution:

Since, ABCD is a parallelogram.

âˆ´ AB = CD,

(Opposite sides of parallelogram)

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

AE = FC,

[âˆµ E and F are mid points of AB and CD respectively

âˆ´ AE = \(\frac{1}{2}\)AB and FC = \(\frac{1}{2}\)CD]

and AB || CD

â‡’ AE || FC

âˆ´ AECF is a parallelogram.

â‡’ AF || EC …..(i)

In Î”ABP, we have

E is the mid point of AB.

and EQ || AP [âˆµ AF || EC]

âˆ´ Q is the mid point of BP

â‡’ BQ = PQ …….(ii)

In Î”CQD, we have

F is the mid point of CD.

and PF || CQ [âˆµ AF || EC]

âˆ´ P is the mid point of DQ.

â‡’ DP = PQ ……..(iii)

From (ii) and (iii), we get

DP = PQ = BQ

Hence, line segments AF and EC trisect the diagonal BD.

Hence proved

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 quadrilateral in which EG and FH are the line segments joining the mid points of opposite sides of a quadrilateral.

To prove: EG and FH bisect each other.

Construction: Join EF, FG, GH, HE and AC.

Proof: In Î”ABC, E and F are mid points of AB and BC.

âˆ´ EF || AC and EF = \(\frac{1}{2}\)AC

[By mid point theorem]

In Î”ADC, H and G are mid points of AD and CD.

âˆ´ HG || AC and HG = \(\frac{1}{2}\)AC ……(ii)

From (i), (ii), we get

EF || HG and EF = HG

âˆ´ EFGH is a parallelogram. [By theorem 8.8]

We know that diagonals of a parallelogram bisect each other.

Therefore EG and FH bisect each other.

Hence proved

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 âŠ¥ AC

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

Solution:

Given: A Î”ABC in which âˆ C = 90Â°,

M is the mid point of AB and MD || BC.

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

(ii) MD âŠ¥ AC

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

Proof: (i) In Î”ABC, M is the mid point of AB.

and MD || BC, (Given)

âˆ´ D is the mid point of AC.

(By theorem 8.10)

(ii) Since, MD || BC and âˆ ACB = 90Â°

âˆ´ âˆ ADM = 90Â°

(Corresponding angles) …(i)

(iii) But, âˆ DM + âˆ CDM = 180Â°, (Linear pair)

â‡’ 90Â° + âˆ CDM = 180Â°, [Using (i)]

â‡’ âˆ CDM = 180Â° – 90Â° = 90Â° … (ii)

Now, Î”ADM and Î”CDM, we have

AD = CD,

(As proved above, D is the mid point of AC)

âˆ ADM = âˆ CDM, [From (i) and (ii), Each = 90Â°]

and MD = MD, (Common)

âˆ´ Î”ADM â‰… Î”CDM,

(By SAS congruence rule)

â‡’ AM = CM, (CPCT)

But, AM = BM, (âˆµ M is the mid point of AB)

âˆ´ AM = CM = \(\frac{1}{2}\)AB.

Hence proved