Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-16.4 (Part - 2), Circles, Class 9, Maths

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q5) O is the circumference of the triangle ABC and Odis perpendicular on BC. Prove that  ∠BOD = ∠A.

Solution:

Given O is the circum centre of triangle ABC and OD⊥BC

to prove  ∠BOD = 2 ∠A

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Pro of :

In ΔOBD and ΔOCD
∠ODB = ∠ODC [Each900]
OB = OC [Radius of circle]
OD = OD [Common]

ThenΔOBD≅ΔOCD         [ by RHS Condition] .

∴ ∠BOD = ∠COD…..(i)
[PCT] .  by degree measure theorem ∠BOC = 2 ∠BAC
⇒ 2 ∠BOD = 2 ∠BAC [ by using(i)]
⇒ ∠BOD = ∠BAC.

 

Q6) In figure 16.135, O is the centre of the circle, BO is the bisector of   ∠ABC. Show that AB = AC.

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics                                                                   

Solution:

Given, BO is the bisec to r of   ∠ABC

to prove AB = BC

Pro of :

Since, BO is the bisec to r of   ∠ABC.

Then,  ∠ABO = ∠CBO…..(i)

Since, OB = OA                     [Radius of circle]

Then,  ∠ABO = ∠DAB…..(ii)             [opposite Angles to equal sides]

Since OB = OC                       [Radius of circle]

Then,  ∠OAB = ∠OCB…..(iii)  [opposite Angles to equal sides]

Compare equations (i), (ii) and (iii)

∠OAB = ∠OCB…..(iv)

InΔOAB andΔOCB

∠OAB = ∠OCB                      [From (iv)]

∠OBA = ∠OBC                      [Given]

OB = OB                                                                    [Common]

Then, ΔOAB≅ΔOCB [ by AAScondition]
∴AB = BC [CPCT]

 

Q7) In figure 16.136, O is the centre of the circle, then prove that  ∠x = ∠y+ ∠z.

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

∠3 = ∠4 [ Angles insamesegment]
∴ ∠x = 2 ∠3 [ by degree measure theorem]
⇒ ∠x = ∠3+ ∠3
⇒ ∠x = ∠3+ ∠4…..(i)             [ ∠3 = angle4]
But ∠y = ∠3+ ∠1 [ by exteriorangleproperty]
⇒ ∠3 = ∠y− ∠1…..(ii)
from(i) and(ii)
∠x = ∠y− ∠1+ ∠4
⇒ ∠x = ∠y+ ∠4− ∠1
⇒ ∠x = ∠y+ ∠z+ ∠1− ∠1 [ by exteriorangleproperty]
⇒ ∠x = ∠y+ ∠z

 

Q8) In figure 16.137, O and O’ are centers of two circles intersecting at B and C. ACD is a straight line, find x.

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

by degree measure theorem ∠AOB = 2 ∠ACB
⇒ 1300 = 2 ∠ACB
⇒ ∠ACB = Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 650
∴ ∠ACB+ ∠BCD = 1800 [Linerapair of Angles ]
⇒ 650+ ∠BCD = 1800
⇒ ∠BCD = 1800−65= 1150 by degree measure theoremreflex ∠BOD = 2 ∠BCD
⇒ reflex ∠BOD = 2×1150 = 2300
Now,reflex ∠BOD+ ∠BO′D = 3600 [Complexangle]
⇒ 2300+x = 3600
⇒ x = 3600−2300
∴x = 1300

 

Q9) In figure 16.138, O is the centre of a circle and PQ is a diameter. If  ∠ROS = 400,find ∠RTS..

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Since PQ is diameter

Then,

∠PRQ = 900 [Angleinsemicircle]
∴ ∠PRQ+ ∠TRQ = 1800 [Linearpair of angle]
900+ ∠TRQ = 1800
∠TRQ = 1800−900 = 900.

by degree measure theorem

∠ROS = 2 ∠RQS

⇒ 400 = 2 ∠RQS

⇒ ∠RQS = Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics= 200

In ΔRQT, by Angle sum property

∠RQT+ ∠QRT+ ∠RTS = 1800

⇒ 200+900+ ∠RTS = 1800

⇒ ∠RTS = 1800−200−90= 700

 

Q10) In figure 16.139, if  ∠ACB = 400, ∠DPB = 1200,find ∠CBD.

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

We have,

∠ACB = 400; ∠DPB = 1200

∴ ∠APB = ∠DCB = 400 [Angleinsamesegment]
InΔPOB, by Angles umproperty ∠PDB+ ∠PBD+ ∠BPD = 1800
⇒ 400+ ∠PBD+1200 = 1800
⇒ ∠PBD = 1800−400−1200
⇒ ∠PBD = 200
∴ ∠CBD = 200

 

Q11) A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

We have,

Radius OA = Chord AB

= >OA = OB = AB

Then triangle OAB is an equilateral triangle.

Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∴ ∠AOB = 600 [oneangle of equilateraltriangle]

by degree measure theorem

∠AOB = 2 ∠APB

⇒ 600 = 2 ∠APB

⇒ ∠APB = Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics 

Now, ∠APB+ ∠AQB = 1800 [opposite Angles of cyclicquadrilateral]

⇒ 300+ ∠AQB = 1800

⇒ ∠AQB = 1800−300 = 1500.

Therefore, Angle by chord AB at minor arc = 1500

Angle by chord AB at major arc = 300

The document Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-16.4 (Part - 2), Circles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What is the concept of a circle in mathematics?
Ans. A circle is a closed curve formed by all the points in a plane that are equidistant from a fixed point called the center. It is defined by its radius, which is the distance between the center and any point on the circle.
2. How do we calculate the circumference of a circle?
Ans. The circumference of a circle can be calculated using the formula C = 2πr, where C is the circumference and r is the radius of the circle. Alternatively, it can also be calculated as C = πd, where d is the diameter of the circle.
3. What is the relationship between the radius and diameter of a circle?
Ans. The radius of a circle is half the length of its diameter. In other words, if r is the radius and d is the diameter, then d = 2r.
4. How do we find the area of a circle?
Ans. The area of a circle can be calculated using the formula A = πr^2, where A is the area and r is the radius of the circle. This formula represents the amount of space enclosed within the circle.
5. Can the circumference of a circle be greater than its area?
Ans. No, the circumference of a circle cannot be greater than its area. The circumference represents the length of the boundary of the circle, while the area represents the space enclosed within the circle. Since the circumference is directly proportional to the radius, it cannot be greater than the square of the radius, which is the measure of the area.
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