Gujarat Board GSEB Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions 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 fig.) AC is a diagonal. Show that

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

2. PQ = SR

3. PQRS is a parallelogram.

Solution:

1. In ΔADC,

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

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

(Line segment joining the mid-points of two sides of a triangle is parallel and half of the third side.)

2. In ΔABC,

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

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

(A line segment joining the mid-points of two sides of a triangle is parallel and half of the third side)

From eqn. (1) and (2),

PQ = SR

3. From eqn. (1) and (2),

PQ II SR and PQ = SR

∴ PQRS is a parallelogram.

(If one pair of opposite sides of a quadrilateral is equal and parallel, then it 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:

Given: ABCD is a rhombus, P, Q, R, S are mid-points of AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.

To Prove: PQRS is a rectangle.

Construction: Join AC and BD.

Proof: In ΔADC,

S and R are mid-point of AD and DC

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

(A line segment joining the mid-points of two sides of a triangle is parallel to third and half of third side.)

Similarly in ΔABC,

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

From eqn. (1) and (2),

PQ = SR

and PQ || SR

Hence ABCD is a parallelogram,

∠AOD = 90°

Now, diagonals of a rhombus bisect each other at 90°

∴ ∠EOF = 90°

⇒ ∠DFS = ∠EOF = 90° (Corresponding angles)

Again, SP || BD

∴ ∠ESF = ∠DFS (Alternates interior angles)

∴ ∠ESF = 90°

Hence PQRS is a rectangle.

Question 3.

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

Solution:

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

To Prove: Quadrilateral PQRS is a rhombus.

Construction: Join AC and BD.

Proof: In ΔADC, S and R are the midpoints of AD and DC.

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

(A line segment joining the mid-points of two sides of a triangle is parallel to third and half of the third side)

Similarly, in AABC,

PQ || AC

and PQ = \(\frac {1}{2}\) AC …….(2)

From eqn. (1) and (2),

PQ || SR and PQ = SR

∴ PQRS is a parallelogram.

Now, we know that the diagonals of a rectangle are equal.

∴ AC = BD ⇒ \(\frac {1}{2}\) AC = \(\frac {1}{2}\) BD

⇒ PQ = QR [From (1) and (2)]

∴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 AB. A line is drawn through E parallel to AB intersecting BC at F (See figure). Show that F is the mid-point of BC.

Solution:

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

Proof: Let DB intersect EF at G.

In ΔDAB,

∴ E is the mid-point of DA and EG || AB.

∴ G is the mid-point of DB.

(A line drawn from mid-point of one side and parallel to another side, bisects the third)

Now in ΔDBC,

∴ G is the mid-point of DB. and GF || AB || DC

∴ F is the mid-point of BC.

(A line drawn from mid-point of one side and parallel to another side bisects the third side)

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 ALF and EC trisect the diagonal BD.

Solution:

To Prove: DP = PQ = BQ

Proof: AB || DC

and AB = DC

(Opposite sides of a parallelogram)

∴ \(\frac {1}{2}\) AB || \(\frac {1}{2}\) DC

and \(\frac {1}{2}\) AB = \(\frac {1}{2}\) DC (Dividing by 2 or both sides)

⇒ AE || FC (∴ E and Fare mid-points of AB and DC) AE = FC

Hence AECF is a parallelogram.

(If one pair of opposite sides of a quadrilateral are equal and parallel, then it is a parallelogram)

∴ AF || CE

(Opposite sides of a parallelogram)

Now in ΔDQC,

F is the mid-point and PF || QC.

∴ DP = PQ …….(1)

(A line drawn from mid-point of one side and parallel to other side bisects the third side of the triangle)

In ΔABP,

E is the mid-point of AB and EQ || AP.

∴ BQ = PQ …….(2)

(A line drawn from mid-point of one side and parallel to the other bisects the third side of the triangle)

Now from eqn. (1) and (2),

DP = PQ = BQ

Hence diagonal BD is trisected by AF and CE.

Question 6.

Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.

Solution:

Given: A quadrilateral ABCD, in which P, Q, R and S are the mid-points of the sides AB, BC, DC and AD respectively.

To Prove: PR and QS bisect each other. Construction: Join PQ, QR, RS, SP, AC and BD.

Proof: In AABD,

∴ P and S are mid-points of AB and AD respectively.

∴ PS || BD and PS = \(\frac {1}{2}\) BD ……(1)

2

(A line segment joining the mid-points of two sides parallel to the third side and half of it)

Similarly, we can show

QR || BD and QR = \(\frac {1}{2}\) BD ……(2)

From eqn. (1) and (2), we have

PS || QR

and PS = QR

If one pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram Hence PQRS is a parallelogram.

Hence PR and QS are diagonals, therefore PR and Q§ bisect each other. (Diagonals of a parallelogram bisect each other).

Question 7.

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. Show that

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

(ii) MD⊥AC.

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

Solution:

(i) In ΔABC

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

∴ D is the mid-point of AC.

(A line drawn through mid-point of one side and parallel to other side bisects the third side)

(ii) DM || BC

∠MDC + ∠BCD = 180°

(Sum of consecutive interior angles is equal to 180°)

⇒ 90° +∠BCD = 180°

⇒ ∠BCD = 90°

∴ MD ⊥ AC

(iii) In ΔADM and ΔCDM,

AD = CD (Proved above)

∠ADM = ∠CDM (Proved above)

DM = DM (Common)

Hence, ΔADM = ΔCDM

(by SAS congruency)

∴ MA = CM (by CPCT)

M is the mid-point of AB.

∴ MA = CM = – AB