RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Created by: Abhishek Kapoor

Class 9 : RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9

Q1) In figure 16.120, O is the centre of the circle. If  ∠APB = 500,find ∠AOB and ∠OAB.

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

∠APB = 500

by degree measure theorem

∠AOB = 2 ∠APB
⇒ ∠APB = 2×500 = 1000
since OA = OB [Radius of circle]
Then ∠OAB = ∠OBA [ Angles opposite to equalsides]
Let ∠OAB = x In ΔOAB, by Angles umproperty ∠OAB+ ∠OBA+ ∠AOB = 1800

= >x + x + 1000  = 1800

= >2x = 1800 – 1000

= >2x = 800

= >x = 400

∠OAB = ∠OBA = 400

 

Q2) In figure 16.121, it is given that O is the centre of the circle and  ∠AOC = 1500.Find ∠ABC.

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

∠AOC = 1500
∴ ∠AOC+reflex ∠AOC = 3600 [Complexangle]
⇒ 1500+reflex ∠AOC = 3600
⇒ reflex ∠AOC = 3600−1500
⇒ reflex ∠AOC = 2100
⇒ 2 ∠ABC = 2100 [ by degree measure theorem]
⇒ ∠ABC = RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev = 1050

  

Q3) In figure 16.22, O is the centre of the circle. Find  ∠BAC.

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev                                                                             

 Solution:

We have  ∠AOB = 800

and  ∠AOC = 1100

Therefore,  ∠AOB+ ∠AOC+ ∠BOC = 3600 [Completeangle]

⇒ 800+1000+ ∠BOC = 3600
⇒ ∠BOC = 3600−800−1100
⇒ ∠BOC = 1700

by degree measure theorem

∠BOC = 2 ∠BAC
⇒ 1700 = 2 ∠BAC
⇒ ∠BAC = RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev = 850

 

Q4) If O is the centre of the circle, find the value of x in each of the following figures.

(i)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

∠AOC = 1350
∴ ∠AOC+ ∠BOC = 1800 [Linearpair of Angles ]
⇒ 135+ ∠BOC = 1800
⇒ ∠BOC = 1800−1350
⇒ ∠BOC = 450

by degree measure theorem ∠BOC = 2 ∠CPB
⇒ 450 = 2x
RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

 

(ii)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠ABC = 400 ∠ACB = 900 [Angle in semicircle]
In ΔABC, by Angles umproperty ∠CAB+ ∠ACB+ ∠ABC = 1800
⇒ ∠CAB+900+40= 1800
⇒ ∠CAB = 1800−900−400
⇒ ∠CAB = 500
Now, ∠CDB = ∠CAB [Angleissameinsegment]
⇒ x = 500

 

(iii)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠AOC = 120by degree measure theorem. ∠AOC = 2 ∠APC
⇒ 120= 2 ∠APC
⇒ ∠APC = RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev = 600
∠APC+ ∠ABC = 1800 [Opposite Angles of cyclicquadrilaterals]
⇒ 600+ ∠ABC = 1800
⇒ ∠ABC = 1800−600
⇒ ∠ABC = 1200
∴ ∠ABC+ ∠DBC = 1800 [Linearpair of Angles ]
⇒ 120+x = 1800
⇒ x = 1800−1200 = 600

 

(iv)  

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠CBD = 650
∴ ∠ABC+ ∠CBD = 1800 [Linearpair of Angles ]
⇒ ∠ABC = 650 = 1800
⇒ ∠ABC = 1800−65= 1150
∴reflex ∠AOC = 2 ∠ABC [ by degree measure theorem]
⇒ x = 2×1150
⇒ x = 2300

(v)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠OAB = 350
Then, ∠OBA = ∠OAB = 350 [ Angles opposite to equalradii]
InΔAOB, by Angles umproperty
⇒ ∠AOB+ ∠OAB+ ∠OBA = 1800
⇒ ∠AOB+350+350 = 1800
⇒ ∠AOB = 1800−350−350 = 1100
∴ ∠AOB+reflex ∠AOB = 3600 [Complexangle]
⇒ 1100+reflex ∠AOB = 3600
⇒ reflex ∠AOB = 3600−1100 = 2500 by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 2500 = 2x
⇒ x = RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev = 1250

 

(vi)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠AOB = 600 by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 60= 2 ∠ACB
⇒ ∠ACB = RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev= 300 [ Angles opposite to equalradii]
⇒ x = 300.

 

(vii)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave ∠BAC = 500 and ∠DBC = 700
∴ ∠BDC = ∠BAC = 500 [Angleinsamesegment]
InΔBDC, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 1800
⇒ 500+x+70= 1800
⇒ x = 1800−500−700 = 600

 

(viii)  

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave, ∠DBO = 400 and ∠DBC = 900 [Angleinasemicircle]
⇒ ∠DBO+ ∠OBC = 900
⇒ 400+ ∠OBC = 900
⇒ ∠OBC = 900−40= 500 by degree measure theorem
∠AOC = 2 ∠OBC
⇒ x = 2×500 = 1000

 

(ix)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

InΔDAB, by Angles umproperty ∠ADB+ ∠DAB+ ∠ABD = 1800
⇒ 320+ ∠DAB+500 = 1800
⇒ ∠DAB = 1800−320−500
⇒ ∠DAB = 980
Now, ∠OAB+ ∠DCB = 1800 [Opposite Angles of cyclicquadrilateral]
⇒ 98+ x = 1800
⇒ x = 1800−980 = 820

 

(x)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave, ∠BAC = 350
∠BDC = ∠BAC = 350 [Angleinsamesegment]
InΔBCD, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 1800
⇒ 350+x+650 = 1800
⇒ x = 1800−350−650 = 800

 

(xi)

RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Wehave, ∠ABD = 400
∠ACD = ∠ABD = 400 [Angleinsamesegment]
InΔPCD, by Angles umproperty
∠PCD+ ∠CPO+ ∠PDC = 1800
⇒ 400+1100+x = 1800
⇒ x = 1800−1500
⇒ x = 300

 

 (xii)

 RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev

Solution:

Giventhat, ∠BAC = 520
Then ∠BDC = ∠BAC = 520 [Angleinsamesegment]
SinceOD = OC
Then ∠ODC = ∠OCD [Oppositeangle to equalradii]
⇒ x = 520

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