BSEB Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

Tuesday, June 28, 2022

BSEB Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Book Answers

BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 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 9 Areas of Parallelograms and Triangles Ex 9.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 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 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks. These Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 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 9 Areas of Parallelograms and Triangles Ex 9.4 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4
Chapters	All
Provider	Hsslive

How to download Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions Answers PDF Online?

Visit our website - Hsslive
Click on the Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Answers.
Look for your Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks PDF.
Now download or read the Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions for PDF Free.

BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

BSEB Bihar Board Class 9th Maths Solutions Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

Question 1.
Parallelogram ABCD and rectangle ABEF arc on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Solution:
Given : A ||gm ABCD and a rectangle ABEF with same base AB and equal areas.

To prove : Perimeter of ||gm ABCD > Perimeter of rectangle ABEF.
Proof : Since opposite sides of a ||gm and rectangle are equal.
∴ AB = DC [∵ ABCD is a ||gm]
AB = EF [∵ ABEF is a rectangle]
∴ DC = EF … (1)
⇒ AB + DC = AB + EF … (2)
Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.
∴ BE < BC and AF < AD ⇒ BC > BE and AD > AF
⇒ BC + AD > BE + AF … (3)
Adding (2) and (3), we get
AB + DC + BC + AD > AB + EF + BE + AF
⇒ AB + BC + CD + DA > AB + BE + EF + FA
Hence, perimeter of ||gm ABCD > perimeter of rectangle ABEF.

Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Can you now answer the question that you have left in the ‘Introduction’, of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
Solution:
Let AL be perpendicular to BC. So, AL is the height of ∆s ABD, ADE and AEC.

∴ ar(ABD) = 12 x BD x AL
ar(ADE) = 12 x DE x AL
and, ar(AEC) = 12 x EC x AL
Since BD = DE = EC
∴ ar(ABD) = ar(ADE) = ar(AEC).
Yes, altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.

Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).

Solution:
Since opposite sides of a ||gm are equal.
∴ AD = BC [∵ ABCD is a ||gm]
DE = CF [∵ DCFE is a ||gm]
and, AE = BF [∵ ABFE is a ||gm]
Consider As ADE and BCF, in which AE = BF, AD = BC and DE = CF
∴ By SSS criterion Of congruence
∆ ADE ≅ ∆ BCF
⇒ ar(ADE) = ar(BCF).

Question 4.
In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(BPC) = ar(DPQ).

Solution:
Join AC.
Since ∆s APC and BPC are on the same base PC and between the same parallels PC and AB.

∴ ar(APC) = ar(BPC) … (1)
Since AD = CQ
and AD || CQ [Given]
∴ In the quadrilateral ADQC, one pair of opposite sides is equal and parallel.
∴ ADQC is a parallelogram.
⇒ AP = PQ and CP = DP [∵ Diagonals of a l|gm bisect each other]
In ∆s APC and DPQ, we have
AP = PQ [Proved above]
∠APC = ∠DPQ [Vertically opp.∠s]
and, PC = PD [Proved above]
∴ By SAS criterion of congruence,
∆ APC ≅ ∆ DPQ
⇒ ar(APC) = ar(DPQ) [∵ Congruent As have equal area]
∴ ar(BPC) = ar(DPQ).

Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i) ar(BDE) = 14ar(ABC)
(ii) ar(BDE) = 12 ar(BAE)
(iii) ar(ABC) = 2 ar(BEC)
(iv) ar(BFE) = ar(AFD)
(v) ar(BFE) = 2 ar(FED)
(vi) ar(FED) = 18 ar(AFC).
Solution:
Join EC and AD. Let a be the side of ∆ ABC. Then,

(ii) We have, ar(BDE) = 12 ar(BEC) … (1) [∵ DE is a median of ∆ BEC and each median divides a triangle in two other As of equal area]
Now, ∠EBC = 60°
and ∠BCA = 60°
⇒ ∠EBC = ∠BCA.
But these are alternate angles with respect to the line- segments BE and CA and their transversal BC.
Hence, BE || AC.
Now, As BEC and BAE stand on the same base BE and lie between the same parallels BE and AC.
∴ ar(BEC) = ar(BAE)
∴ From (1), ar(∆ BDE) = 12 ar(BAE).

(iii) Since ED is a median of A BEC and we know that each median divides a triangle in two other As of equal area.
∴ ar(BDE) = 12 ar(BEC)
From part (i),ar(BDE) = 14 ar(ABC)
Combining these results, we get
14 ar(ABC) = 14 ar(BEC)
⇒ ar(ABC) = 2ar(BEC).

(iv) Now, ∠ABD = ∠BDE = 60° (given)
But ∠ABD and ∠BDE are alternate angles with respect to the line-segment BA and DE and their transversal BD. Hence BA || ED.
Now, As BDE and AED stand on the same base ED and lie between the same parallels BA and DE
∴ ar (BDE) = ar(AED)
⇒ ar(BDE) – ar (FED) = ar(AED) – ar(FED)
⇒ ar(BEF) = ar (AFD)

From (1) and (2), we have ar (AFD) = 2 ar (EFD)
Combining this result with part (iv).
We bave ar (BFE) = ar (AFD) = 2ar (EFD)
(vi) From part (i)

Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
Solution:
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

Draw AM ⊥ BD and CN ⊥ BD.
Now, ar(APB) x ar(CPD)
= (12 x BP x AM) x (12 x DP x CN)
= 14 x BP x DP x AM x CN … (1)
and, ar(APD) x ar(BPC)
= ( 12 x DP x AM) x (12 x BP x CN)
= 14 x BP x DP x AM x CN … (2)
From (1) and (2), we have
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).

Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 12 ar(ARC)
(ii) ar(RQC) = 38 ar(ABC)
(iii) ar(PBQ) = ar(ARC).
Solution:
P and Q are respectively the npd-points of sides AB and BC of Δ ABC and R is the mid-point of AP.
Join AQ and PC.

(i) We have,
ar(PQR) = 12 ar(APQ) [∵ QR is a median of ∆ APQ and it divides the ∆ into two other ∆s of equal area]

From (1) and (2), we get .
ar(PQR) = 18 ar(ABC) = 12 x 14 ar(ABC)
= 12 ar(ARC).

(ii) We have,

From (3), (4), (5) and (6), we have

(iii) We have,

Question 8.
In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that :

(i) ∆ MBC ≅ ∆ ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆ FCB ≅ ∆ ACE
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
Solution:
(i) In As MBC and ABD, we have
BC = BD [Sides of the square BCED]
MB = AB [Sides of the square ABMN]
∠MBC = ∠ABD [∵ Each = 90° + ∠ABC]
By SAS criterion of congruence, we have
∆ MBC ≅ ∆ ABD.

(ii) ∆ ABD and square BYXD have the same base BD and are between the same parallels BD and AX.
∴ ar(ABD) = 12ar(BYXD)
But ∆ MBC ≅ ∆ ABD [Proved in part (i)]
⇒ ar(MBC) = ar(ABD)
∴ ar(MBC) = ar(ABD) = 12 ar(BYXD)
⇒ ar(BYXD) = 2 ar(MBC).

(iii) Square ABMN and ∆ MBC have the same base MB and are between same parallels MB and NAC.
ar(MBC) = 12 ar(ABMN)
⇒ ar(ABMN) = 2ar(MBC)
= ar(BYXD) [Using part (ii)]

(iv) In ∆s ACE and BCF, we have
CE = BC [Sides of the square BCED]
AC = CF [Sides of the square ACFG]
and, ∠ACE = ∠BCF [∵ Each = 90° + ∠BCA]
∴ By SAS criterion of congruence,
∆ ACE ≅ ∆ BCF.

(v) ∆ ACE and square CYXE have the same base CE and are between same parallels CE and AYX.
∴ ar(ACE) = 12 ar(CYXE)
⇒ ar(FCB) = 12 ar(CYXE)[∵ ∆ ACE ≅ ∆ BCF, part (iv)]
⇒ ar(CYXE) = 2 ar(FCB).

(vi) Square ACFG and A BCF have the same base CF and are between same parallels CF and BAG.
∴ ar(BCF) = 12ar(ACFG)
⇒ 12 ar(CYXE) = 12ar(ACFG) [Using part (v)]
ar(CyXE) = ar(ACFG)

(vii) From part (Hi) and (vi), we have ar(BYXD) = ar(ABMN)
and,ar( CYXE) = ar(ACFG)
On adding, we get
ar(BYXD) + ar(CYXE) = ar(ABMN) + ar(ACFG) ar(BCED) = ar(ABMN) + ar(ACFG)

BSEB Textbook Solutions PDF for Class 9th

Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 exam preparation. The BSEB STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Books State Board syllabus with maximum efficiency.

FAQs Regarding Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Solutions

How to get BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Answers??

Students can download the Bihar Board Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Answers PDF from the links provided above.

Can we get a Bihar Board Book PDF for all Classes?

Yes you can get Bihar Board Text Book PDF for all classes using the links provided in the above article.

Important Terms

Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4, BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks, Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4, Bihar Board Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook solutions, BSEB Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks Solutions, Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4, BSEB STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks, Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4, Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook solutions, BSEB STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbooks Solutions,

HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

Hsslive.co.in: Kerala Higher Secondary News, Plus Two Notes, Plus One Notes, Plus two study material, Higher Secondary Question Paper.

Tuesday, June 28, 2022