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

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.

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.

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.

Solution:

by degree measure theorem ∠AOB = 2 ∠ACB
⇒ 130⁰ = 2 ∠ACB
⇒ ∠ACB =  = 65⁰
∴ ∠ACB+ ∠BCD = 180⁰ [Linerapair of Angles ]
⇒ 65⁰+ ∠BCD = 180⁰
⇒ ∠BCD = 180⁰−65⁰ = 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..

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 = = 20⁰

In ΔRQT, by Angle sum property

∠RQT+ ∠QRT+ ∠RTS = 180⁰

⇒ 20⁰+90⁰+ ∠RTS = 180⁰

⇒ ∠RTS = 180⁰−20⁰−90⁰ = 70⁰

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

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.

∴ ∠AOB = 60⁰ [oneangle of equilateraltriangle]

by degree measure theorem

∠AOB = 2 ∠APB

⇒ 60⁰ = 2 ∠APB

⇒ ∠APB =  

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