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## AP Board Class 9 Maths Chapter 12 Circles Ex 12.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 12 Circles Ex 12.4 Book Answers

 AP Board Class 9 Maths Chapter 12 Circles Ex 12.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 12 Circles Ex 12.4 Book Answers

## Andhra Pradesh State Board Class 9th Maths Chapter 12 Circles Ex 12.4 Books Solutions

 Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 9th Subject Maths Chapters Maths Chapter 12 Circles Ex 12.4 Provider Hsslive

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## AP Board Class 9th Maths Chapter 12 Circles Ex 12.4 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 9th Maths Chapter 12 Circles Ex 12.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

## AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.4

Question 1.
In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:
’O’ is the centre
∠AOB = 100°

Thus ∠ACB = 12 ∠AOB
[∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]]
= 12 x 100° = 50°
[ ∵ Opp. angles of a cyclic quadrilateral]
∴ ∠ADB = 180°-50° = 130°
[OR]
∠ADB is the angle made by the major arc ACBˆ at D.
∴ ∠ADB = 12∠AOB [where ∠AOB is the angle; made by ACBˆ at the centre]
= 12 [360° – 100°] [from the figure]
= 12 x 260° = 130°

Question 2.
In the figure, ∠BAD = 40° then find ∠BCD.

Solution:
‘O’ is the centre of the circle.
∴ In ΔOAB; OA = OB (radii)
∴ ∠OAB = ∠OBA = 40°
(∵ angles opp. to equal sides)
Now ∠AOB = 180° – (40° + 40°)
(∵ angle sum property of ΔOAB)
= 180°-80° = 100°
But ∠AOB = ∠COD = 100°
Also ∠OCD = ∠ODC [OC = OD]
= 40° as in ΔOAB
∴ ∠BCD = 40°
(OR)
In ΔOAB and ΔOCD
∠AOB = ∠COD (vertically opp. angles)
∴ ΔOAB ≅ ΔOCD
∴ ∠BCD = ∠OBA = 40°
[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

Question 3.
In the figure, ‘O’ is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.

Solution:
‘O’ is the centre; ∠POR = 120°
∠PQR = 12∠POR [∵ angle made by an arc at the centre is, twice the angle made by it on the remaining part]
∠PSR = 12 [Angle made by PQRˆ at the centre]
∠PSR = 12 [360° – 120°] from the fig.
= 12 x 240 = 120°

Question 4.
If a parallelogram is cyclic, then it is a rectangle. Justify.
Solution:

Let □ABCD be a parallelogram such
that A, B, C and D lie on the circle.
∴∠A + ∠C = 180° and ∠B + ∠D = 180°
[Opp. angles of a cyclic quadri lateral are supplementary]
But ∠A = ∠C and ∠B = ∠D
[∵ Opp. angles of a ||gm are equal]
∴∠A = ∠C =∠B =∠D = 1802 = 90°
Hence □ABCD is a rectangle

Question 5.
In the figure, ‘O’ is the centr of the circle. 0M = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.
OM bisects AB.
∴ AM = AB2=82 = 4 cm
OA2 = OM2 + AM2 [ ∵ Pythagoras theorem]
OA =32+42‾‾‾‾‾‾‾√=9+16‾‾‾‾‾‾√=25‾‾‾√
= 5cm

Question 6.
In the figure, ‘O’ is the centre of the circle and OM, ON are the perpen-diculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6 cm. Find RS.

Solution:
‘O’ is the centre of the circle.
OM = ON and 0M ⊥ PQ; ON ⊥ RS
Thus the chords FQ and RS are equal.
[ ∵ chords which are equidistant from the centre are equal in length]
∴ RS = PQ = 6cm

Question 7.
A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the

Solution:

A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle.

Question 8.
Draw a circle with any radius and then draw two chords equidistant
from the centre.
Solution:

1. Draw a circle with centre P.
3. Mark off two points M and N oh these radii. Such that PM = PN.
4. Draw perpendicular through M and N to these radii.

Question 9.
In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD.

Solution:
‘O’ is the centre of the circle.
AB, CD are equal chords
⇒ They subtend equal angles at the centre.
∴ ∠AOB =∠COD = 70°
Now in ΔOCD
∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]
∴ ∠OCD + ∠ODC + 70° = 180°
= ∠OCD +∠ODC = 180° – 70° = 110°
∴ ∠OCD + ∠ODC = 110° = 55°

## Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.4 Textbooks for Exam Preparations

Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.4 exam preparation. The AP Board STD 9th Maths Chapter 12 Circles Ex 12.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 12 Circles Ex 12.4 Books State Board syllabus with maximum efficiency.

## FAQs Regarding Andhra Pradesh Board Class 9th Maths Chapter 12 Circles Ex 12.4 Textbook Solutions

#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?

Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

## Important Terms

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