# HBSE 9th Class Maths Solutions Chapter 8 Quadrilaterals Ex 8.2

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
âˆ´ AM = CM = $$\frac{1}{2}$$AB.