AP Board Class 9 Maths Chapter 12 Circles Ex 12.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 12 Circles Ex 12.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

AP Board Class 9 Maths Chapter 12 Circles Ex 12.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 12 Circles Ex 12.3 Book Answers

AP Board Class 9th Maths Chapter 12 Circles Ex 12.3 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 9th Maths Chapter 12 Circles Ex 12.3 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 9th Maths Chapter 12 Circles Ex 12.3 book solutions pdf online from this page.

Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 9th Maths Chapter 12 Circles Ex 12.3 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 12 Circles Ex 12.3
Provider	Hsslive

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AP Board Class 9th Maths Chapter 12 Circles Ex 12.3 Textbooks Solutions with Answer PDF Download

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AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.3

Question 1.
Draw the following triangles and construct circumcircles for them.
(OR)
In ΔABC, AB = 6 cm, BC = 7 cm and ∠A = 60°.
Construct a circumcircle to the triangle XYZ given XY = 6cm, YZ = 7cm and ∠Y = 60°. Also, write steps of construction.
Solution:

Steps of construction :

Draw the triangle with given mea-sures.
Draw perpendicular bisectors to the sides.
The point of concurrence of per-pendicular bisectors be S’.
With centre S; SA as radius, draw a circle which also passes through B and C.
This is the required circumcircle.

ii) In ΔPQR; PQ = 5 cm, QR = 6 cm and RP = 8.2 cm.

Steps of construction:

Draw ΔPQR with given measures.
Draw perpendicular to PQ, QR and RS; let they meet at ‘S’.
With S as centre and SP as radius draw a circle.
This is the required circumcircle.

iii) In ΔXYZ, XY = 4.8 cm, ∠X = 60°and ∠Y = 70°.

Steps of construction:

Draw ΔXYZ with given measures.
Draw perpendicular bisectors to the sides of ΔXYZ, let the point of con-currence be S’.
Draw the circle (S, SX⎯⎯⎯⎯⎯⎯⎯⎯).
This is the required circumcircle.

Question 2.
Draw two circles passing A, B where AB = 5.4 cm.
(OR)
Draw a line segment AB with 5.4 cm. length and draw two different circles that passes through both A and B.
Solution:

Steps of construction:

Draw a line segment AB = 5.4 cm.
Draw the perpendicular bisector 𝑋𝑌↔ of AB.
Take any point P on 𝑋𝑌↔.
With P as centre and PA as radius draw a circle.
Let Q be another point on XY.
Draw the circle (Q, QA⎯⎯⎯⎯⎯⎯⎯⎯⎯).

Question 3.
If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.
Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.
Join A, B to form the common chord
AB⎯⎯⎯⎯⎯⎯⎯⎯. Let ‘O’ be the midpoint of AB.
Join ‘O’ with P and Q.
Now in ΔAPO and ΔBPO
AP = BP (radii)
PO = PO (common)
AO = BO (∵ O is the midpoint)
∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)
Also ∠AOP = ∠BOP (CPCT)
But these are linear pair of angles.
∴ ∠AOP = ∠BOP = 90°
Similarly in ΔAOQ and ΔBOQ
AQ = BQ (radii)
AO = BO (∵ O is the midpoint of AB)
OQ = OQ (common)
∴ AAOQ ≅ ABOQ
Also ∠AOQ = ∠BOQ (CPCT)
Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)
∴ ∠AOQ = ∠BOQ = 180∘2 = 90°
Now ∠AOP + ∠AOQ = 180°
∴ PQ is a line.
Hence the proof.

Question 4.
If two intersecting chords of a circle make equal angles with diameter pass¬ing through their point of intersection, prove that the chords are equal.

Solution:
Let ‘O’ be the centre of the circle.
PQ is a drametre.
AB⎯⎯⎯⎯⎯⎯⎯⎯ and CD⎯⎯⎯⎯⎯⎯⎯⎯ are two chords meeting at E, a point on the diameter.
∠AEO = ∠DEO
Drop two perpendiculars OL and OM from ‘O’ to AB and CD;
Now in ΔLEO and ΔMEO
∠LEO = ∠MEO [given]
EO = EO [Common]
∠ELO = ∠EMO [construction 90°]
∴ ΔLEO ≅ ΔMEO
[ ∵ A.A.S. congruence]
∴ OL = OM [CPCT]
i.e., The two chords AB⎯⎯⎯⎯⎯⎯⎯⎯ and CD⎯⎯⎯⎯⎯⎯⎯⎯ are at equidistant from the centre ‘O’.
∴ AB = CD
[∵ Chords which are equi-distant from the centre are equal]
Hence proved.

Question 5.
In the given figure, AB is a chord of circle with centre ‘O’. CD is the diam-eter perpendicular to AB. Show that AD = BD.

Solution:
CD is diameter, O is the centre.
CD ⊥ AB; Let M be the point of inter-section.
Now in ΔAMD and ΔBMD
AM = BM [ ∵ radius perpendicular to a chord bisects it]
∠AMD =∠BMD [given]
DM = DM (common)
∴ ΔAMD ≅ ΔBMD
⇒ AD = BD [C.P.C.T]

AP Board Textbook Solutions PDF for Class 9th Telugu

Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 Textbooks for Exam Preparations

Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.3 exam preparation. The AP Board STD 9th Maths Chapter 12 Circles Ex 12.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 12 Circles Ex 12.3 Books State Board syllabus with maximum efficiency.

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