NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2

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

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 8 Quadrilaterals Ex 8.2.

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

Ex 8.2 Class 9 Maths 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 = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:
(i)
In ∆ACD, we have
S is the mid-point of AD and R is the mid-point of CD.
SR = ½ AC 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 = ½ AC and PQ || AC        ….…(2)              [By mid-point theorem]
From (1) and (2), we get
PQ = SR and PQ || SR
PQ = SR and PQ || SR               [Proved in (ii)]
PQRS is a parallelogram.

Ex 8.2 Class 9 Maths 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 and P, Q, R and S are the mid-points of the sides AB, BC, CD, and DA, respectively.
To prove: Quadrilateral PQRS is a rectangle.
Proof: By mid-point theorem.
In Î”ADC, we have S and R are the mid-points of AD and CD, respectively.
SR || AC and SR = ½ AC                        …………..(i)
In Î”ABC, we have P and Q are the mid-points of AB and BC, respectively.
PQ || AC and PQ = ½ AC                      ……..…..(ii)
From eqs. (i) and (ii), we get PQ || SR and PQ = SR = ½ AC
Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.

So, PQRS is a parallelogram.
We know that diagonals of a rhombus bisect each other at right angles.
COD = EOF = 90°
Now, in ABCD, R and Q are the mid-points of CD and BC, respectively.

RQ || DB                           [by mid-point theorem]
RE || OF
Also, SR || AC                   [from eq. (i)]
FR || OE
So, OERF is a parallelogram.
ERF = EOF = 90°       [Opposite angles of a parallelogram are equal]
Thus, PQRS is a parallelogram with
R = 90°
Hence, PQRS is a rectangle.
Hence proved.

Ex 8.2 Class 9 Maths 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, ABCD is a rectangle.
In Î”ASP and Î”BQP,

AP = BP                (Given)
AS = BQ                (Given)
A = B               (Each 90°)
Î”ASP Î”BQP  (By SAS congruency rule)
SP = PQ             (By CPCT)                ……(i)
In Î”PBQ and Î”RCQ,

PB = RC                 (Given)

BQ = CQ                (Given)
B = C                (Each 90°)
Î”PBQ Î”RCQ  (By SAS congruency rule)
PQ = RQ             (By CPCT)              ……(ii)
Similarly, we can prove that RQ = SR and SR = SP        (iii)

From eqs. (i), (ii) and (iii), we get SP = PQ = QR = RS.

Therefore, PQRS is a rhombus.

Ex 8.2 Class 9 Maths 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.

Solution.
Given:
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD and a line is drawn through E parallel to AB intersecting BC at F such that EF || AB.

To prove: F is the mid-point of BC.
Proof: Let EF intersects BD at P.
In Î”ABD, we have EP || AB [
EF || AB] and E is the mid-point of AD.
So, by the converse of mid-point theorem, P is the mid-point of BD.
Similarly, in Î”BCD, we have PF || CD [
EF || AB and AB || CD]
and P is the mid-point of BD.

So, by the converse of mid-point theorem, F is the mid-point of BC.

Ex 8.2 Class 9 Maths 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.

Solution.

Given: ABCD is a parallelogram, and E and F are the mid-points of sides AB and CD, respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since ABCD is a parallelogram.
AB || DC and AB = DC              [opposite sides of a parallelogram]
AE || FC and ½ AB = ½ DC
AE || FC and AE = FC            [ E and F are the mid-points of AB and CD]

Since a pair of opposite sides of a quadrilaterals AECF is equal and parallel.
So, AECF is a parallelogram.
Then, AF || EC
AP || EQ and FP || CQ     [Since opposite sides of a parallelogram are parallel]
In Î”BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.
[by the converse of mid-point theorem]
BQ = PQ                     ……(i)
Again, in Î”DQC, F is the mid-point of DC and FP || CQ.
So, P is the mid-point of DQ.         [by the converse of mid-point theorem]
QP = DP                    ……(ii)
From eqs. (i) and (ii), we get BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.
Hence proved.

Ex 8.2 Class 9 Maths Question 6.
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 = ½ AB

Solution.

Given, ABC is a right-angled triangle.
C = 90° and M is the mid-point of AB. Also, DM || BC

(i) In Î”ABC, BC || MD and M is the mid-point of AB.
D is the mid-point of AC.               (By converse of mid-point theorem)
(ii) Since MD || BC and CD is transversal.              .
But
ACB = 90°

MD AC
(iii) Now, in Î”ADM and Î”CDM, we have
DM = MD                                             (Common)
AD = CD                                                (D is the mid-point of AC)
Î”CDM
CM = AM                           …….(i)

Also, M is the mid-point of AB.
AM = BM = ½ AB                …….(ii)
From eqs. (i) and (ii), we get
CM = AM = ½ AB
Hence proved.