NCERT Solutions Class 9th Maths Chapter – 8 Quadrilaterals Exercise 8.2

NCERT Solutions Class 9th Maths Chapter – 8 Quadrilaterals

TextbookNCERT
Class 9th
Subject Mathematics
Chapter8th
Chapter NameQuadrilaterals
CategoryClass 9th Math Solutions 
Medium English
SourceLast Doubt

NCERT Solutions Class 9th Maths Chapter – 8 Quadrilaterals

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). AC is a diagonal. Show that
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
ch 8 8.2 vk
Solution
(i) In ∆ACD, We have
∴ S is the mid-point of AD and R is the mid-point of CD.
SR = 1/2AC 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 = 12AC and PQ || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 1/2AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
(iii) In a quadrilateral PQRS,
PQ = SR and PQ || SR [Proved]
∴ 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.
Solution
We have a rhombus ABCD and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Join AC.
ch 8 8.2 vk
In ∆ABC, P and Q are the mid-points of AB and BC respectively.
∴ PQ = 1/2AC and PQ || AC …(1)
[By mid-point theorem]
In ∆ADC, R and S are the mid-points of CD and DA respectively.
∴ SR = 1/2AC and SR || AC …(2)
[By mid-point theorem]
From (1) and (2), we get
PQ = 1/2AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
∴ PQRS is a parallelogram. …….(3)
Now, in ∆ERC and ∆EQC,
∠1 = ∠2
[ ∵ The diagonals of a rhombus bisect the opposite angles]
CR = CQ [ ∵CD2 = BC2]
CE = CE [Common]
∴ ∆ERC ≅ ∆EQC [By SAS congruency]
⇒ ∠3 = ∠4 …(4) [By C.P.C.T.]
But ∠3 + ∠4 = 180° ……(5) [Linear pair]
From (4) and (5), we get
⇒ ∠3 = ∠4 = 90°
Now, ∠RQP = 180° – ∠b [ Y Co-interior angles for PQ || AC and EQ is transversal]
But ∠5 = ∠3
[ ∵ Vertically opposite angles are equal]
∴ ∠5 = 90°
So, ∠RQP = 180° – ∠5 = 90°
∴ One angle of parallelogram PQRS is 90°.
Thus, 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.
Solution
We have,
Now, in ∆ABC, we have
PQ = 1/2AC and PQ || AC …(1)
[By mid-point theorem]
Similarly, in ∆ADC, we have
SR = 1/2AC and SR || AC …(2)
From (1) and (2), we get
PQ = SR and PQ || SR
∴ PQRS is a parallelogram.
Now, in ∆PAS and ∆PBQ, we have
∠A = ∠B [Each 90°]
AP = BP [ ∵ P is the mid-point of AB]
AS = BQ [∵ 1/2AD = 12BC]
∴ ∆PAS ≅ ∆PBQ [By SAS congruency]
⇒ PS = PQ [By C.P.C.T.]
Also, PS = QR and PQ = SR [∵opposite sides of a parallelogram are equal]
So, PQ = QR = RS = SP i.e., PQRS is a parallelogram having all of its sides equal.
Hence, 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.
ch 8 8.2 vk
Solution:
We have,
ch 8 8.2 vk
In ∆DAB, we know that E is the mid-point of
AD and EG || AB [∵ EF || AB]
Using the converse of mid-point theorem, we get, G is the mid-point of BD.
Again in ABDC, we have G is the midpoint of BD and GF || DC.
[∵ AB || DC and EF || AB and GF is a part of EF]
Using the converse of the mid-point theorem, we get, F is the mid-point of BC.
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.
ch 8 8.2 vk
Solution – Since, the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC
⇒ AE || FC …(1)
and AB = DC
⇒ 1/2AB = 12DC
⇒ AE = FC …(2)
From (1) and (2), we have
AE || PC and AE = PC
∴ ∆ECF is a parallelogram.
Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ
[∵ AF || CE]
⇒ DP = PQ …(3)
[By converse of mid-point theorem] Similarly, in A BAP, E is the mid-point of AB and EQ || AP [∵AF || CE]
⇒ BQ = PQ …(4)
[By converse of mid-point theorem]
∴ From (3) and (4), we have
DP = PQ = BQ
So, the line segments AF and EC 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.
Solution
Let ABCD be a quadrilateral, where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
Join PQ, QR, RS and SP.
Let us also join PR, SQ and AC.
ch 8 8.2 vk
Now, in ∆ABC, we have P and Q are the mid-points of its sides AB and BC respectively.
∴ PQ || AC and PQ = 1/2 AC …(1)
[By mid-point theorem]
Similarly, RS || AC and RS = 1/2AC …(2)
∴ By (1) and (2), we get
PQ || RS, PQ = RS
∴ PQRS is a parallelogram.
And the diagonals of a parallelogram bisect each other, i.e., PR and SQ bisect each other. Thus, the line segments joining the midpoints of opposite sides of a 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 = 1/2AB
Solution
we have

ch 8 8.2 vk

(i) In ∆ACB, We have
M is the mid-point of AB. [Given]
MD || BC , [Given]
∴ Using the converse of mid-point theorem,
D is the mid-point of AC.

(ii) Since, MD || BC and AC is a transversal.
∠MDA = ∠BCA
[ ∵ Corresponding angles are equal] As
∠BCA = 90° [Given]
∠MDA = 90°
⇒ MD ⊥AC.

(iii) In ∆ADM and ∆CDM, we have
∠ADM = ∠CDM [Each equal to 90°]
MD = MD [Common]
AD = CD [∵ D is the mid-point of AC]
∴ ∆ADM ≅ ∆CDM [By SAS congruency]
⇒ MA = MC [By C.P.C.T.] .. .(1)
∵ M is the mid-point of AB [Given]
MA = 12AB …(2)
From (1) and (2), we have
CM = MA = 1/2AB

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