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BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Book Answers

BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Book Answers
BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Book Answers


BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks. These Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Books Solutions

Board BSEB
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Class 9th
Subject Maths Chapter 8 Quadrilaterals Ex 8.2
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BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

BSEB Bihar Board Class 9th 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 = 12 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
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 prove : (i) SR || AC and SR = 12 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Proof :
(i) In Δ ACD, we have
S is the mid-point of AD and R is the mid-point of CD. Then SR || AC and SR = 12AC [Mid-point theorem]

(ii) In Δ ABC, we have
P is the mid-point of the side AB and Q is the mid-point of the side BC.
Then PQ || AC
and PQ = 12AC
Thus, we have proved that:

(iii) Since PQ = SR and PQ || SR
⇒ 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 respectively 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 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 prove : PQRS is a rectangle.
Construction : Join AC.
Proof:
In Δ ABC, P and Q are the mid-points of AB and
∴ PQ || AC and PQ = 12AC … (1)
Similarly, in A ADC, R and S are the mid-points of CD and PD.
∴ SR || AC and SR = 12AC
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
∴ AB = BC [Sides of a rhombus]
⇒ 12AB = 12BC
⇒ PB = BQ
⇒ ∠3 = ∠4 [∠s opp. to equal sides of a A]
Now, in ∆ APS and ∆ CQR, we have
AP = CQ [Halves of equal sides AB, BC]
AS = CR [Halves of equal sides AD, CD]
PS = QR [Opp. sides of parallelogram PQRS]
∴ ∆ APS = ∆ CQR [SSS Cong. Theorem]
⇒ ∠1 = ∠2 [C.P.C.T.]
Now, ∠1 + ∠SPQ + ∠3 = 180° [Linear pair axiom]
∴ ∠1 + ∠PQ + ∠3 = ∠2 + ∠PQR + ∠4
But ∠1 = ∠2 and ∠3 = ∠4[Proved above]
∴ ∠SPQ = ∠PQR … (3)
∴ SP || RQ intersects them,
∴ ∠SPQ + ∠PQR = 180° … (4)
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.
Solution:
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 prove : PQRS is a rhombus.

Construction : Join AC.
Proof:
In ∆ ABC, P and Q; are the mid-points of sides AB and BC.
∴ PQ = AC and PQ = 12AC …(1)
Similarly, in ∆ ADC, R and S are the mid-points of sides CD and AD. .
∴ SR || AC and SR = 12AC …(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 …(4)
Now, AD = BC [Opp. sides ofrect. ABCD]
⇒ 12 AD = 12 BC ⇒ AS = BQ
In ∆s APS and BPQ, we have
AP = BP [∵ P is the mid-point of AB]
∠PAS = ∠PBQ [Each = 90°]
AS = BQ [Proved above]
∴ ∆ APS ≅ ∆ BPQ [SAS Cong. axiom]
⇒ PS = PQ [C.P.C.T] … (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.

Solution:
Given : In trapezium ABCD, AB || DC
E is the mid-point of AD, EF || AB.
To prove : F is the mid-point of BC.

Construction : Join DB. Let it intersect EF in G.
Proof:
In A DAB, E is the mid-point of AD [Given]
EG || AB [∵ EF || AB]
∴ By converse of mid-point theorem G is the mid-point of DB.
In ∆ BCD, G is the mid-point of BD [Proved]
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 mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.

Solution:
Given : E and F are the mid-points of sides AB and CD of the parallelogram ABCD whose diagonal is BD.
To prove : BQ = QP = PD
Proof:
ABCD is parallelogram [Given]
∴ AB || DC and AB = DC
[Opp. sides of parallelogram]
E is the mid-point of AB [Given]
∴ AE = 12 AB … (1)
F is the mid-point of CD
∴ CF = 12CD
⇒ CF = 12 AB [∵ CD = AB] … (2)
From (1) and (2),
AE = CF
Also AE || CF [∵ AB || DC]
Thus, a pair of opposite sides of a quadrilateral AECF are parallel and equal.
Quad. AECF is a parallelogram
⇒ EC || AF
⇒ EQ || AP and QC || MF
In ∆ BMA, E is the mid-point of BA [Gtuen]
EQ || AP [Proved]
∴ BQ = LP
[Converse of mid-point theorem] … (3)
Similarly by taking ∆ CLD, we can prove that
DP = QP … (4)
From (3) and (4), we get
BQ = QP = PD
Hence, AF and CE trisect the diagonal AC.

Question 6.
Show that the line segments joining the mid¬points of the opposite sides of a quadrilateral bisect each other.
Solution:
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 prove : OP = OR,
OQ = OS.
Construction : Join PQ, QR, RS, SP, AC and BD.
Proof:
In ∆ ABC, P and Q are mid-points of AB and BC respectively.
∴ PQ || AC and PQ = 12 AC
Similarly, we can prove that
RS || AC and RS = 12AC
∴ 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.
∴ Diagonals PR and QS of a ||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 = 12 AB
Solution:
Given : ∆ ABC is right angled at C, M is the mid-point of hypotenuse AB. Also MD || BC.

To prove that :
(i) D is the mid-points of AC
(ii) MD ⊥ AC
(iii) CM = MA = 12AB.
Proof :
(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 …(1)

(ii) Since MD || BC. Therefore,
∠ADM = ∠ACB [Corresponding angles]
⇒ ∠ADM = 90° [∵ ∠ACB = 90° (given)]
But, ∠ADM + ∠CDM = 180°
[∵∠ADM and ∠CDM are angles of a linear pair]
∴ 90° + ∠CDM = 180° ⇒ ∠CDM = 90°
Thus, ∠ADM = ∠CDM = 90° … (2)
⇒ MD ⊥ AC

(iii) In ∆s AMD and CMD, we have
AD = CD [From (1)]
∠ADM = ∠CDM [From (2)]
and, MD = MD [Common]
∴ By SAS criterion of congruence
∆ AMD ≅ ∆ CMD
⇒ MA = MC
[∵ Corresponding parts of congruent triangles are equal]
Also, MA = 12AB, since M is the mid-point of AC
Hence, CM = MA = 12AB.


BSEB Textbook Solutions PDF for Class 9th


Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 exam preparation. The BSEB STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Books State Board syllabus with maximum efficiency.

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