BSEB Class 9 Maths Chapter 7 Triangles Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.4 Book Answers |
Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.4 Textbooks Solutions PDF
Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.4 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 7 Triangles Ex 7.4 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 7.4 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 7 Triangles Ex 7.4 Textbooks. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.4 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 7 Triangles Ex 7.4 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 9th |
Subject | Maths Chapter 7 Triangles Ex 7.4 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 9th Maths Chapter 7 Triangles Ex 7.4 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 9th Maths Chapter 7 Triangles Ex 7.4 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 7.4
Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let ABC be a right angled A in which ∠ABC = 90°.
But ∠ABC + ∠BCA + ∠CAB = 180°
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90°
⇒ ∠BCA and ∠CAB are acute angles
⇒ ∠BCA < 90° and ∠CAB < 90°
⇒ ∠BCA < ∠ABC and ∠CAB < ∠ABC ⇒ AC > AB and AC > BC
[∵ Side opp. to greater angle is larger]
⇒ In a right triangle, the hypotenuse is the longest side.
Question 2.
In figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
Since ∠PBC < ∠QCB ⇒ – ∠PBC > – ∠QCB
⇒ 180° – ∠PBC > 180° – ∠QCB
∠ABC > ∠ACB
⇒ AC > AB [∵ Side opp, to greater angle is larger]
Question 3.
In figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Solution:
Since ∠B < ∠A and ∠C < ∠D
∴ AO < BO and OD < OC
[∵Side opp. to greater angle is larger]
Adding these results, we have
AO + OD < BO + OC
⇒ AD < BC.
Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
ABCD is a quadrilateral such that AB is its smallest side and CD is its largest side.
Join AC and BD.
Since AB is the smallest side of quadrilateral ABCD.
∴ In ∆ ABC, we have BC > AB ⇒ ∠8 > ∠3
[∵ Angle opp. to longer side is greater] Since CD is the longest side of quadrilateral ABCD.
∴ In A ACD, we have
CD > AD
⇒ ∠7 > ∠4 … (2)
[∵ Angle opp. to longer side is greater]
Adding (1) and (2),
we get ∠8 + ∠7> ∠3 + ∠4
⇒ ∠A > ∠C
Again, in ∆ ABD. we have
AD > AB [∵AB is the shortest side]
⇒ ∠1 > ∠6 … (3)
In ∆ BCD, we have
CD > BC [∵ CD is the longest side]
⇒ ∠2 > ∠5 … (4)
Adding (3) and (4), we get
∠1 + ∠2 > ∠5 + ∠6 ⇒ ∠B > ∠D
Thus, ∠A > ∠C and ∠B > ∠D.
Question 5.
In figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.
Solution:
In ∆ PQR, we have
PR > PQ [Given]
⇒ ∠PQR > ∠PRQ [∵ Angle opp. to larger side is greater]
⇒ ∠PQR + ∠1 > ∠PRQ + ∠1 [Adding ∠1 on both sides]
⇒ ∠PQR + ∠1 > ∠PRQ + ∠2 [∵ PS is the bisector of ∠P ∴∠1 = ∠2 ]
Now, in ∆s PQS and PSR,
we have ∠PQR + ∠1 + ∠PSQ = 180°
and ∠PRQ + ∠2 + ∠PSR = 180°
⇒ ∠PQR + ∠1 = 180° – ∠PSQ
and ∠PRQ + ∠2 = 180° – ∠PSR
∴ 180° – ∠PSQ > 180° – ∠PSR [From (1)]
⇒ – ∠PSQ > – ∠PSR
⇒ ∠PSQ < ∠PSR i.e., ∠PSR > ∠PSQ.
Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Lot P be any point not on the straight line l. PM ⊥ l and N is any point on l other than M.
In ∆ PMN, we have
∠M = 90°
⇒ ∠N < 90°
[ ∵ ∠M = 90° ⇒ ∠MPN + ∠PNM = 90°
⇒ ∠P + ∠N – 90° ⇒ ∠N < 90°]
⇒ ∠N < ∠M
⇒ PM < PN [∵ Side opp. to greater angle is larger]
Hence, PM is the shortest of all line segments from P to AB.
BSEB Textbook Solutions PDF for Class 9th
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- BSEB Class 9 Maths Chapter 1 Number Systems Ex 1.6 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 1 Number Systems Ex 1.6 Book Answers
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- BSEB Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 3 Coordinate Geometry Ex 3.1 Book Answers
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- BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Book Answers
- BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Book Answers
- BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Book Answers
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- BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.1 Book Answers
- BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.2 Book Answers
- BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Book Answers
- 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 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
- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Book Answers
- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.4 Book Answers
- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.5 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.5 Book Answers
- BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.1 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 9 Areas of Parallelograms and Triangles Ex 9.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.1 Book Answers
- BSEB Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.2 Book Answers
- BSEB Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Book Answers
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- BSEB Class 9 Maths Chapter 10 Circles Ex 10.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 10 Circles Ex 10.2 Book Answers
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- BSEB Class 9 Maths Chapter 11 Constructions Ex 11.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 11 Constructions Ex 11.2 Book Answers
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