The document RD Sharma Solutions -Ex-16.4 (Part - 2), 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

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

**Solution:**

Given O is the circum centre of triangle ABC and ODâŠ¥BC

to prove âˆ BOD = 2 âˆ A

Pro of :

In Î”OBD and Î”OCD

âˆ ODB = âˆ ODC [Each900]

OB = OC [Radius of circle]

OD = OD [Common]

ThenÎ”OBDâ‰…Î”OCD [ by RHS Condition] .

âˆ´ âˆ BOD = âˆ CODâ€¦..(i)

[PCT] . by degree measure theorem âˆ BOC = 2 âˆ BAC

â‡’ 2 âˆ BOD = 2 âˆ BAC [ by using(i)]

â‡’ âˆ BOD = âˆ BAC.

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

**Solution:**

Given, BO is the bisec to r of âˆ ABC

to prove AB = BC

Pro of :

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

Then, âˆ ABO = âˆ CBOâ€¦..(i)

Since, OB = OA [Radius of circle]

Then, âˆ ABO = âˆ DABâ€¦..(ii) [opposite Angles to equal sides]

Since OB = OC [Radius of circle]

Then, âˆ OAB = âˆ OCBâ€¦..(iii) [opposite Angles to equal sides]

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

âˆ OAB = âˆ OCBâ€¦..(iv)

InÎ”OAB andÎ”OCB

âˆ OAB = âˆ OCB [From (iv)]

âˆ OBA = âˆ OBC [Given]

OB = OB [Common]

Then, Î”OABâ‰…Î”OCB [ by AAScondition]

âˆ´AB = BC [CPCT]

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

**Solution:**

We have,

âˆ 3 = âˆ 4 [ Angles insamesegment]

âˆ´ âˆ x = 2 âˆ 3 [ by degree measure theorem]

â‡’ âˆ x = âˆ 3+ âˆ 3

â‡’ âˆ x = âˆ 3+ âˆ 4â€¦..(i) [ âˆ 3 = angle4]

But âˆ y = âˆ 3+ âˆ 1 [ by exteriorangleproperty]

â‡’ âˆ 3 = âˆ yâˆ’ âˆ 1â€¦..(ii)

from(i) and(ii)

âˆ x = âˆ yâˆ’ âˆ 1+ âˆ 4

â‡’ âˆ x = âˆ y+ âˆ 4âˆ’ âˆ 1

â‡’ âˆ x = âˆ y+ âˆ z+ âˆ 1âˆ’ âˆ 1 [ by exteriorangleproperty]

â‡’ âˆ x = âˆ y+ âˆ z

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

**Solution:**

by degree measure theorem âˆ AOB = 2 âˆ ACB

â‡’ 130^{0} = 2 âˆ ACB

â‡’ âˆ ACB = = 65^{0}

âˆ´ âˆ ACB+ âˆ BCD = 180^{0} [Linerapair of Angles ]

â‡’ 65^{0}+ âˆ BCD = 180^{0}

â‡’ âˆ BCD = 180^{0}âˆ’65^{0 }= 115^{0} by degree measure theoremreflex âˆ BOD = 2 âˆ BCD

â‡’ reflex âˆ BOD = 2Ã—115^{0} = 230^{0}

Now,reflex âˆ BOD+ âˆ BOâ€²D = 360^{0} [Complexangle]

â‡’ 230^{0}+x = 360^{0}

â‡’ x = 360^{0}âˆ’230^{0}

âˆ´x = 130^{0}

**Q9) In figure 16.138, O is the centre of a circle and PQ is a diameter. If âˆ ROS = 40 ^{0},find âˆ RTS..**

**Solution:**

Since PQ is diameter

Then,

âˆ PRQ = 90^{0} [Angleinsemicircle]

âˆ´ âˆ PRQ+ âˆ TRQ = 180^{0} [Linearpair of angle]

90^{0}+ âˆ TRQ = 180^{0}

âˆ TRQ = 180^{0}âˆ’90^{0} = 90^{0}.

by degree measure theorem

âˆ ROS = 2 âˆ RQS

â‡’ 40^{0} = 2 âˆ RQS

â‡’ âˆ RQS = = 20^{0}

In Î”RQT, by Angle sum property

âˆ RQT+ âˆ QRT+ âˆ RTS = 180^{0}

â‡’ 20^{0}+90^{0}+ âˆ RTS = 180^{0}

â‡’ âˆ RTS = 180^{0}âˆ’20^{0}âˆ’90^{0 }= 70^{0}

**Q10) In figure 16.139, if âˆ ACB = 40 ^{0}, âˆ DPB = 120^{0},find âˆ CBD.**

**Solution:**

We have,

âˆ ACB = 40^{0}; âˆ DPB = 120^{0}

âˆ´ âˆ APB = âˆ DCB = 40^{0} [Angleinsamesegment]

InÎ”POB, by Angles umproperty âˆ PDB+ âˆ PBD+ âˆ BPD = 180^{0}

â‡’ 40^{0}+ âˆ PBD+120^{0} = 180^{0}

â‡’ âˆ PBD = 180^{0}âˆ’40^{0}âˆ’120^{0}

â‡’ âˆ PBD = 20^{0}

âˆ´ âˆ CBD = 20^{0}

**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.

âˆ´ âˆ AOB = 60^{0} [oneangle of equilateraltriangle]

by degree measure theorem

âˆ AOB = 2 âˆ APB

â‡’ 60^{0} = 2 âˆ APB

â‡’ âˆ APB =

Now, âˆ APB+ âˆ AQB = 180^{0} [opposite Angles of cyclicquadrilateral]

â‡’ 30^{0}+ âˆ AQB = 180^{0}

â‡’ âˆ AQB = 180^{0}âˆ’30^{0} = 150^{0}.

Therefore, Angle by chord AB at minor arc = 150^{0}

Angle by chord AB at major arc = 30^{0}