BSEB Class 9 Maths Chapter 10 Circles Ex 10.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 10 Circles Ex 10.3 Book Answers |
Bihar Board Class 9th Maths Chapter 10 Circles Ex 10.3 Textbooks Solutions PDF
Bihar Board STD 9th Maths Chapter 10 Circles Ex 10.3 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 10 Circles Ex 10.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 10 Circles Ex 10.3 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 10 Circles Ex 10.3 Textbooks. These Bihar Board Class 9th Maths Chapter 10 Circles Ex 10.3 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 10 Circles Ex 10.3 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 9th |
Subject | Maths Chapter 10 Circles Ex 10.3 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 9th Maths Chapter 10 Circles Ex 10.3 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 9th Maths Chapter 10 Circles Ex 10.3 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.3
Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
On drawing a different pairs of circles, we find that they have two points in common. Now, we shall prove that a pair of circle cannot intersect each other at more than two points. Let the two circles cut each other at three points. But from the three points only one circle passes.
Hence, two circles if intersect each other then they will intersect at two points. These two points are the ends of their common chords. This chord can be a common chord of a number of circles which will pass through the end points of this common chord.
Question 2.
Suppose you are given a construction to find its centre.
Solution:
Steps of Construction :
1. Take 3 points A, B and C on the circumference of the circle.
2. Join AB and BC.
3. Draw PQ and RS, the perpendicular bisectors of AB and BC, which intersect each other at O. Then, O is the centre of the circle.
Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
Given : Two circles, with centres O and O’ intersect at two points A and B so that AB is the common chord of the two circles and 00′ is the line segment joining the centres of the two circles. Let OO’ intersect AB at M.
To prove : OO’ is the perpendicular bisector of AB.
Construction : Draw line segments OA, OB, O’A and O’B.
Proof:
In As OAO’ and OBO
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
and; OO’ = OO’
By SSS criterion of congruence, we have
∆ OAO’ ≅ ∆ OBO’
⇒ ∠AO O’ = ∠BOO’
⇒ ∠AOM = ∠BOM …(1) [∵ ∠AOO’ = ∠AOM and ∠BOM = ∠BOO’]
In ∆s AOM and BOM, we have
OA = OB [Radii of the same circle]
∠AOM = ∠BOM [From (1)]
and, OM = OM [Common]
∴ By SAS criterion of congruence, we have
∆ AOM ≅ ∆ BOM
⇒ AM = BM and ∠AMO = ∠BMO
But ∠AMO + ∠BMO = 180°
∴ ∠AM0 = 180° ⇒ ∠AMO = 90°
Thus, AM = BM and ∠AMO = ∠BMO = 90°
Hence, OO’ is the perpendicular bisector of AB.
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