BSEB Class 9 Maths Chapter 7 Triangles Ex 7.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.1 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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BSEB Class 9 Maths Chapter 7 Triangles Ex 7.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.1 Book Answers

BSEB Class 9th Maths Chapter 7 Triangles Ex 7.1 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 7 Triangles Ex 7.1 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.1 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 7 Triangles Ex 7.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 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 7 Triangles Ex 7.1 Textbooks. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 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 7 Triangles Ex 7.1 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths Chapter 7 Triangles Ex 7.1
Chapters	All
Provider	Hsslive

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BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Solution:
Now, in As ABC and ABD, we have
AC = AD [Given]
∠CAB = ∠BAD (∵ AB bisects ∠A]
and, AB = AB [Common]
∴ By SAS congruence criterion, we have
∆ ABC ≅ ∆ ABD
⇒ BC = BD
[∵ Corresponding parts of congruent triangles are equal]

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) ∆ ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

Solution:
In ∆s ABD and BAC, we have
AD = BC
∠DAB = ∠CBA lGiven]
AB = AB [Common]
By SAS criterion of congruence, we have A ABD = A BAC, which proves (i)
⇒ BD = AC
and, ∠ABD = ∠BAC, which proves (ii) and (iii)
[∵ Corresponding parts of congruent triangles are equal]

Question 3.
AD and BC are equal „ perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:
Since AB and CD intersect at O. Therefore,
∠AOD =∠BOC … (1) [Vertically opp. angles]
In ∆s AOD and BOC, we have
∠AOD = ∠BOC [From(1)]
∠D AO = ∠OBG [Each = 90°]
and, AD = BC [Given]
∴ By AAS congruence criterion, we have ∆ AOD ≅ ∆ BOC
OA = OB
[∵ Corresponding parts of congruent triangles are equal] i.e., O is the mid-point AB.
Hence, CD bisects AB.

Question 4.
1 and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ ABC ≅ ∆ CDA.

Solution:
Since l and m are two parallel lines intersected by another pair of parallel lines p and q. Therefore, AD || BC and AB || CD ⇒ ABCD is a parallelogram.
i.e., AB = CD
and BC = AD
Now, in ∆s ABC and CDA, we have
AB = CD [Prove above]
BC = AD [Prove above]
and AC = AC [Common]
∴ By SSS criterion of congruence ∆ ABC ≅ ∆ CDA.

Question 5.
Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that :
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Solution:
In ∆s APB and AQB, we have
∠APB = ∠AQB [∵ Each = 90°]
∠PAB = ∠QAB [∵ AB bisects ∠PAQ]
AB = AB [Common]
By AAS congruence criterion, we have
∆ APB ≅ ∆ AQB, which proves (i)
⇒ BP = PQ
[∵Corresponding parts of congruent triangles are equal]
i.e., B is equidistant from the arms of ∠A, which proves

Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
AB = AD
∠BAC = ∠DAE
[∵ ∠BAD = ∠EAC ⇒ ∠BAD + ∠DAC = ∠EAC + ∠DAC ⇒ ∠BAC =∠DAE]
and, AC = AE [Given]
∴ By SAS criterion of congruence, we have
∆ ABC ≅ ∆ADE
⇒ BC = DE
[∵ Corresponding parts of congruent triangles are equal]

Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that

(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE.
Solution:
We have, ∠EPA = ∠DPB
⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE’
⇒ ∠DPA = ∠EPB … (1)
Now, in ∆s EBP and DAP, we have
∠EPB = ∠DPA [From (1)]
BP = AP [Given]
and, ∠EBP = ∠DAP [Given]
So, by ASA criterion of congruence, we have
∆ EBP ≅ ∆ DAP
⇒ BE = AD i.e., AD = BE
[∵ Corresponding parts of congruent triangles are equal]

Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = 12 AB.

Solution:
(i) In ∆s AMC and BMD, we have
AM = BM [∵ M is the mid-point of AB]
∠AMC = ∠BMD [Vertically opp. ∠s]
and, CM = MD [Given]
∴ By SAS criterion of congruence, we have
∴ ∆ AMC ≅ ∆ BMD

(ii) Now, ∆ AMC ≅ ∆ BMD
⇒ BD = CA and ∠BDM = ∠ACM … (1)
[∵Corresponding parts of congruent triangles are equal]
Thus, transversal CD cuts CA and BD at C and D respectively such’that the alternate angles ∠BDM and ∠ACM are equal. Therefore, BD || CA.
⇒ ∠CBD + ∠BCA = 180°
[∵ Sum of the interior angles on the same side of ” transversal = 180°]
⇒ ∠CBD + 90° = 180° [∵ ∠BCA = 90°]
⇒ ∠DBC = 90°.

(iii) Now, in ∆s DBC and ACB, we have
BD = CA [From (1)]
∠DBC = ∠ACB [∵ Each = 90°)
BC = BC [Common]
By SAS criterion of congruence, we have ∆ DBC ≅ ∆ ACB.

(iv) CD = AB
[∵ Corresponding parts of congruent triangles are equal]
⇒ 12CD = 12AB ⇒ CM = 12AB.

BSEB Textbook Solutions PDF for Class 9th

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.1 exam preparation. The BSEB STD 9th Maths Chapter 7 Triangles Ex 7.1 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 7 Triangles Ex 7.1 Books State Board syllabus with maximum efficiency.

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Tuesday, June 28, 2022