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BSEB Class 9 Maths Chapter 7 Triangles Ex 7.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.2 Book Answers |
Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.2 Textbooks Solutions PDF
Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.2 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 7 Triangles Ex 7.2 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 7.2 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 7 Triangles Ex 7.2 Textbooks. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.2 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 7 Triangles Ex 7.2 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 9th |
Subject | Maths Chapter 7 Triangles Ex 7.2 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 9th Maths Chapter 7 Triangles Ex 7.2 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 9th Maths Chapter 7 Triangles Ex 7.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2
Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A.
Solution:
(i) In ∆ ABC, we have
AB = AC
⇒ ∠B = ∠C [∵ Angles opposite to equal sides are equal]
12∠B = 12∠C
⇒ ∠OBC = ∠OCB
[∵ OB and OC bisect ∠s B and C respectively]
∴ ∠OBC = 12∠B and ∠OCB = 12∠c
⇒ OB = OC … (2)
[∵ Sides opp. to equal ∠s are equal]
(ii) Now, in ∆s ABO and ACO, we have
AB = AC [Given]
∠OBC = ∠CCB [From (1)]
∆OB = OC . [From (2)]
∴ By SAS criterion of congruence, we have
∆ABO ≅ ∆ACO
⇒ ∠BAO = ∠CAO
[∵ Corresponding parts of congruent triangles are equal]
⇒ AO bisects ∠BAC.
Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
Solution:
In ∆s ABD and ACD, we have
DB = DC [Given]
∠ADB = ∠ADC [∵ AD ⊥ BC]
AD = AD [Common]
∴ By SAS criterion of congruence, we have
∆ ABD ≅ ∆ ACD
⇒ AB = AC
[∵ Corresponding parts of congruent triangles are equal]
Hence, ∆ ABC is isosceles.
Question 3.
ABC is an isosceles triangle in Which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.
Solution:
In ∆s ABE and ACF, we have
∠AEB = ∠AFC [∵ Each = 90°]
∠BAE = ∠CAF [Common]
and, BE = CF [Given]
By AAS criterion of congruence, we have
∆ ABE ≅ ∆ ACF
⇒ AB = AC
[∵ Corresponding parts of congruent triangles are equal]
Hence, ∆ ABC is isosceles.
Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Solution:
Let BE ⊥ AC and CF ⊥ AB.
In ∆s ABE and ACF, we have
∠AEB = ∠AFC [∵ Each = 90°]
∠A = ∠A [Common]
and, AB = AC [Given]
∴ By AAS criterion of congruence,
∆ ABE ≅ ∆ ACF
⇒ BE = CF
[∵ Corresponding parts of congruent triangles are equal]
Question 5.
ABC and DBC are two isosceles triangles on tile same base BC (see figure). Show that ∠ABD = ∠ACD.
Solution:
In ∆ ABC, we have
AB = AC
⇒ ∠ABC = ∠ACB
[∵ Angles opposite to equal sides are equal]
In ∆ BCD, we have
BO = CD
⇒ ∠DBC = ∠DCB … (2)
[∵ Angles opposite to equal sides are equal]
Adding (1) and (2), we have
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD
Question 6.
∆ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.
Solution:
In ∆ ABC, we have
AB = AC
⇒ ∠ACB = ∠ABC … (1)
[∵ Angles opp. to equal sides are equal]
Now, AB = AD [Given]
∴ AD = AC [∵AB = AC]
Thus, in ∆ ADC, we have
AD = AC
⇒ ∠ACD = ∠ADC … (2)
[∵ Angles opp. to equal sides are equal]
Adding (1) and (2), we get
∠ACB + ∠ACD = ∠ABC + ∠ADC
⇒ ∠BCD = ∠ABC + ∠BDC [∵ ∠ADC = ∠BDC]
⇒ ∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD [Adding ∠BCD on both sides]
⇒ 2∠BCD = 180°
⇒ ∠BCD = 90°
Hence, ∠BCD is a right angle.
Question 7.
ABC is a right angled triangle in which ∠A = 90 and AB = AC. Find ∠B and ∠C.
Solution: We have, ∠A = 90°
AB = AC
⇒ ∠B = ∠C
[∵ Angles opp. to equal sides of a triangle are equal]
Also, ∠A + ∠B + ∠C = 180° [Angle-sum property]
⇒ 90° + 2∠B = 180° [∵ ∠C = ∠B]
⇒ 2∠B = 180° – 90° = 90°
⇒ ∠B = 90°2 = 45°
∴ ∠C = ∠B = 45°
Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Let ∆ ABC be an equilateral triangle so that AB = AC = BC.
Now, ∵ AB = AC
⇒ ∠B = ∠C … (1)
[∵ Angles opp. to equal sides are equal]
Also, ∵ CB = CA
⇒ ∠A = ∠B … (2)
[∵ Angles opp. to equal sides are equal]
From (1) and (2), we have
∠A = ∠B = ∠C
Also, ∠A + ∠B + ∠C = 180° [Angle-sum property]
∴ ∠A + ∠A + ∠A = 180°
⇒ 3∠A = 180° ⇒ ∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, each angle of an equilateral triangle is 60°.
BSEB Textbook Solutions PDF for Class 9th
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