RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions 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 = \(\frac{1}{2}\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Answer:
Given: A quadrilateral ABCD in which P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA. Also AC is its diagonal.
To show :
(i) SR ∥ AC and SR = \(\frac{1}{2}\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

(i) In ∆ACD, we have :
S is the mid-point of AD and R is the mid-point of CD.
Therefore, SR ∥ AC and SR = \(\frac{1}{2}\)AC (Mid-point theorem)

(ii) In ∆ABC, we have :
P is the mid-point of the side AB and Q is the mid-point of the side BC.
Therefore, PQ ∥ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem)

Thus, we have shown that:

(iii) Since PQ = SR and PQ ∥ SR, therefore one pair-of opposite sides are equal and parallel.
∴ 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.
Answer:
Given : ABCD is a rhombus in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quadrilateral PQRS.

To show: PQRS is a rectangle.

Construction: Join AC.
In ∆ABC, P and Q are respectively the mid-points of AS and BC.
PQ ∥ AC and PQ = \(\frac{1}{2}\)AC

Similarly, in ∆ADC, R and S are respectively the mid-points of CD and AD.
SR ∥ AC and SR = \(\frac{1}{2}\)AC ……………(2)
From (1) and (2), we get PQ ∥ RS and PQ = SR

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is equal and parallel.
∴ PQRS is a parallelogram.
Also, AS = BC (Sides of a rhombus)

Now, SP ∥ RQ and PQ intersects them.
∴ ∠SPQ + ∠PQR = 180°
From (3) and (4), we get
∠SPQ = ∠PQR = 90°
Thus, PQRS is a parallelogram whose one angle ∠SPQ = 90°.
Hence, 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
Answer:
Given : ABCD is a rectangle in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined to obtain a quad. PQRS.

To show : PQRS is a rhombus.
Construction: Join AC.
In ∆ABC, P and Q are respectively the mid-points of sides AB and BC.
PQ ∥ AC and PQ = \(\frac{1}{2}\)AC ………..(1)

Similarly, in AADC, R and S are respectively the mid-points of sides CD and AD.
∴ SR ∥ AC and SR = \(\frac{1}{2}\)AC …………….(2)
From (1) and (2), we get PQ ∥ SR and PQ = SR …………….. (3)

Now, in quad. PQRS, its one pair of opposite sides PQ and SR is parallel and equal. [From (3)]
∴ PQRS is a parallelogram.
Now, AD = BC (Opp. sides of rect. ABCD) ….(4)
AP = BP ⇒ AS = BQ (∵ P is the mid-point of AB)
∠PAS = ∠PBQ (Each = 90°)
AS = BQ (Shown above)
∴ ∆APS = ∆BPQ (SAS Cong, axiom)
PS = PQ (CPCT) ………….(5)
From (4) and (5), we get 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.

Answer:
Given : In trapezium ABCD, AB ∥ DC,
E is the mid-point of AD and EF ∥ AB.

To show: F is the mid-point of BC.
Construction: Join DB. Let it intersect EF in G.
In ADAB, E is the mid-point of AD (Given) and EG ∥ AB (∵ EF ∥AB)
∴ By converse of mid-point theorem, G is the midpoint of DB.
In ABCD, G is the mid-point of BD and GF ∥ DC (AB ∥ DC, EF ∥ AB ⇒ DC ∥ EF)

∴ By converse of mid-point theorem,
F is the mid-point of BC.

Question 5.
In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Answer:
Given : E and F are the mid-points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
To-show : BQ = QP = PD.
ABCD is parallelogram. (Given)
∴ AB ∥ DC and AB = DC (Opp. sides of parallelogram)
E is the mid-point of AB. (Given)
AE = \(\frac{1}{2}\)AB
F is the mid-point of CD.
CF = \(\frac{1}{2}\)CD (∵ CD = AB)
So, AE = CF
From (1) and (2) AE ∥ CF (∵ AB ∥ DC)
Thus, a pair of opposite sides of a quadrilateral AECF are parallel and equal.
Quad. AECF is a parallelogram.
So, EC ∥ AF
∴ EQ ∥ AP and QC ∥ PF

In ABPA, E is the mid-point of BA (Given)
and EQ ∥ AP (Shown)
BQ = QP (Converse of mid-point theorem) ……….. (3)
Similarly by taking ACQD, we can prove that
DP = QP …………… (4)
From (3) and (4), we get
BQ = QP = PD
Hence, AF and CE 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.
Answer:
Given : A quad. ABCD, P, Q, R and S are respectively the mid-points of AB, BC, CD and DA. PR and QS intersect each other at O.
To show: OP = OR, OQ = OS.

Construction: Join PQ, QR, RS, SP, AC and BD
In ∆ABC, P and Q are mid-points of AB and BC respectively
∴ PQ ∥ AC and PQ = \(\frac{1}{2}\)AC
Similarly, we can show that
RS ∥ AC and RS = \(\frac{1}{2}\)AC
∴ PQ ∥SR and PQ = SR
Thus, a pair of opposite sides of a quadrilateral PQRS are parallel and equal.
∴ Quadrilateral PQRS is a parallelogram.
Since, the diagonals of a parallelogram bisect each other, therefore diagonals PR and QS of ∥gm PQRS, i.e., the line segments joining the mid-points of opposite sides of 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 = \(\frac{1}{2}\)AB.
Answer:
Given : ∆ABC is right angled at C, M is the mid-point of hypotenuse AB. Also, MD ∥ BC.

To show that:
(i) D is the mid-point of AC.
(ii) MD ⊥ AC.
(iii) CM = MA = \(\frac{1}{2}\)AB.

(i) In ∆ABC, M is the mid-point of AB and MD ∥ BC. Therefore, D is the mid-point of AC.
i.e., AD = DC

(ii) Since MD ∥ BC, therefore
∠ADM = ∠ACB (Corresponding angles)
So, ∠ADM = 90° [v ∠ACB = 90° (given)]
But, ∠ADM + ∠CDM = 180°
[ v ∠ADM and ∠CDM are angles of a linear pair]
∴ 90° + ∠CDM = 180° or ∠CDM =90°
Thus, ∠ADM = ∠CDM =90° ……………….. (2)
So, MD ⊥ AC

(iii) In As AMD and CMD, we have :
AD = CD [From (1)]
∠ADM = ∠CDM [From (2)]
and MD = MD (common)
∴ By SAS criterion of congurence
∆AMD ≅ ∆CMD
So, MA = MC
(v Corresponding parts of congruent triangles are equal)
Also, MA = \(\frac{1}{2}\)AB, since M is the mid-point of AC.
Hence, CM = MA = \(\frac{1}{2}\)AB.