RD Sharma Solutions Class 9 Circles Exercise 16.4

RD Sharma Solutions Class 9 Chapter 16 Exercise 16.4

RD Sharma Class 9 Solutions Chapter 16 Ex 16.4 Free Download

RD Sharma Solutions Class 9 Chapter 16 Ex 16.4

Q1) In figure 16.120, O is the centre of the circle. If \(\angle APB=50^{0},\;find\;\angle AOB\;and\;\angle OAB\).

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Solution:

\(\angle APB=50^{0}\)

By degree measure theorem

\(\angle AOB=2\angle APB\\\Rightarrow \angle APB=2\times 50^{0}=100^{0}\\since\;OA=OB\;\;\;\;\;\;\;\;[Radius\;of\;circle]\\Then\;\angle OAB=\angle OBA\;\;\;\;\;\;\;[Angles\;opposite\;to\;equal\;sides]\\Let\;\angle OAB=x\\In\;\Delta OAB,\;by\;angle\;sum\;property\\\angle OAB+\angle OBA+\angle AOB=180^{0}\)

=>x + x + 1000 = 1800

=>2x = 1800 – 1000

=>2x = 800

=>x = 400

\(\angle OAB=\angle OBA=40^{0}\)

Q2) In figure 16.121, it is given that O is the centre of the circle and \(\angle AOC=150^{0}.\;Find\;\angle ABC\).

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Solution:

\(\angle AOC=150^{0}\\\\ ∴ \angle AOC+reflex\;\angle AOC=360^{0}\;\;\;\;\;\;[Complex\;angle]\\\\ \Rightarrow 150^{0}+reflex\;\angle AOC=360^{0}\\\\ \Rightarrow reflex\;\angle AOC=360^{0}-150^{0}\\\\ \Rightarrow reflex\;\angle AOC=210^{0}\\\\ \Rightarrow 2\angle ABC=210^{0}\;\;\;\;\;\;\;[By\;degree\;measure\;theorem]\\\\ \Rightarrow \angle ABC=\frac{210^{0}}{2}=105^{0}\)

  

Q3) In figure 16.22, O is the centre of the circle. Find \(\angle BAC\).

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 Solution:

We have \(\angle AOB=80^{0}\)

And \(\angle AOC=110^{0}\)

Therefore, \(\angle AOB+\angle AOC+\angle BOC=360^{0}\;\;\;\;\;\;[Complete\;angle]\)

\(\Rightarrow 80^{0}+100^{0}+\angle BOC=360^{0}\\\Rightarrow \angle BOC=360^{0}-80^{0}-110^{0}\\\Rightarrow \angle BOC=170^{0}\)

By degree measure theorem

\(\angle BOC=2\angle BAC\\\\\Rightarrow 170^{0}=2\angle BAC\\\\\Rightarrow \angle BAC=\frac{170^{0}}{2}=85^{0}\)

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

(i)

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Solution:

\(\angle AOC=135^{0}\\\\ ∴ \angle AOC+\angle BOC=180^{0}\;\;\;\;\;\;[Linear\;pair\;of\;angles]\\\\ \Rightarrow 135^{0}+\angle BOC=180^{0}\\\\ \Rightarrow \angle BOC=180^{0}-135^{0}\\\\ \Rightarrow \angle BOC=45^{0}\\\\\)

\(By\;degree\;measure\;theorem\\\\ \angle BOC=2\angle CPB\\\\ \Rightarrow 45^{0}=2x\\\\ \Rightarrow x=\frac{45^{0}}{2}=22\frac{1}{2}^{0}\)

(ii)

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Solution:

\(We\;have\\ \angle ABC=40^{0}\\ \angle ACB=90^{0}\;\;\;\;[Angle\;in\;semicircle]\\ In\;\Delta ABC,\;by\;angle\;sum\;property\\ \angle CAB+\angle ACB+\angle ABC=180^{0}\\ \Rightarrow \angle CAB+90^{0}+40^{0}=180^{0}\\ \Rightarrow \angle CAB=180^{0}-90^{0}-40^{0}\\ \Rightarrow \angle CAB=50^{0}\\ Now,\;\\ \angle CDB=\angle CAB\;\;\;\;\;\;\;[Angle\;is\;same\;in\;segment]\\ \Rightarrow x=50^{0}\)

(iii)

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Solution:

\(We\;have\\ \angle AOC=120^{0}\\ By\;degree\;measure\;theorem.\\ \angle AOC=2\angle APC\\ \Rightarrow 120^{0}=2\angle APC\\ \Rightarrow \angle APC=\frac{120^{0}}{2}=60^{0}\\ \angle APC+\angle ABC=180^{0}\;\;\;\;\;[Opposite\;angles\;of\;cyclic\;quadrilaterals]\\ \Rightarrow 60^{0}+\angle ABC=180^{0}\\ \Rightarrow \angle ABC=180^{0}-60^{0}\\ \Rightarrow \angle ABC=120^{0}\\ ∴ \angle ABC+\angle DBC=180^{0}\;\;\;\;\;\;[Linear\;pair\;of\;angles]\\ \Rightarrow 120+x=180^{0}\\ \Rightarrow x=180^{0}-120^{0}=60^{0}\)

(iv)  

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Solution:

\(We\;have\\ \angle CBD=65^{0}\\ ∴ \angle ABC+\angle CBD=180^{0}\;\;\;\;\;\;[Linear\;pair\;of\;angles]\\ \Rightarrow \angle ABC=65^{0}=180^{0} \Rightarrow \angle ABC=180^{0}-65^{0}=115^{0}\\ ∴  reflex\;\angle AOC=2\angle ABC\;\;\;\;\;\;[By\;degree\;measure\;theorem]\\ \Rightarrow x=2\times 115^{0}\\ \Rightarrow x=230^{0}\)

(v)

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Solution:

\(We\;have\\ \angle OAB=35^{0}\\ Then,\;\angle OBA=\angle OAB=35^{0}\;\;\;\;\;\;[Angles\;opposite\;to\;equal\;radii]\\\\ In\;\Delta AOB,\;by\;angle\;sum\;property\\ \Rightarrow \angle AOB+\angle OAB+\angle OBA=180^{0}\\ \Rightarrow \angle AOB+35^{0}+35^{0}=180^{0}\\ \Rightarrow \angle AOB=180^{0}-35^{0}-35^{0}=110^{0}\\\\∴ \angle AOB+reflex\angle AOB=360^{0}\;\;\;\;\;\;[Complex\;angle]\\ \Rightarrow 110^{0}+reflex\angle AOB=360^{0}\\ \Rightarrow reflex\angle AOB=360^{0}-110^{0}=250^{0}\\ By\;degree\;measure\;theorem\;reflex\angle AOB=2\angle ACB\\ \Rightarrow 250^{0}=2x\\ \Rightarrow x=\frac{250^{0}}{2}=125^{0}\)

(vi)

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Solution:

\(We\;have\\ \angle AOB=60^{0}\\ By\;degree\;measure\;theorem\;reflex\\ \angle AOB=2\angle ACB\\ \Rightarrow 60^{0}=2\angle ACB\\ \Rightarrow \angle ACB=\frac{60^{0}}{2}=30^{0}\;\;\;\;\;\;[Angles\;opposite\;to\;equal\;radii]\\ \Rightarrow x = 30^{0}.\)

(vii)

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Solution:

\(We\;have\\ \angle BAC=50^{0}\;and\;\angle DBC=70^{0}\\ ∴ \angle BDC=\angle BAC=50^{0}\;\;\;\;\;[Angle\;in\;same\;segment]\\ In\;\Delta BDC,\;by\;angle\;sum\;property\;\\ \angle BDC+\angle BCD+ \angle DBC=180^{0}\\ \Rightarrow 50^{0}+x+70^{0}=180^{0}\\ \Rightarrow x=180^{0}-50^{0}-70^{0}=60^{0}\)

(viii)  

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Solution:

\(We\;have,\\ \angle DBO=40^{0}\;and\;\angle DBC=90^{0}\;\;\;\;\;\;[Angle\;in\;a\;semi\;circle]\\ \Rightarrow \angle DBO+\angle OBC=90^{0}\\ \Rightarrow 40^{0}+\angle OBC=90^{0}\\ \Rightarrow \angle OBC=90^{0}-40^{0}=50^{0}\\\\ By\;degree\;measure\;theorem\;\\ \angle AOC=2\angle OBC\\ \Rightarrow x=2\times 50^{0}=100^{0}\)

(ix)

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Solution:

\(In\;\Delta DAB,\;by\;angle\;sum\;property\\ \angle ADB+\angle DAB+\angle ABD=180^{0}\;\\ \Rightarrow 32^{0}+\angle DAB+50^{0}=180^{0}\\ \Rightarrow \angle DAB=180^{0}-32^{0}-50^{0}\\ \Rightarrow \angle DAB =98^{0}\\ Now,\\ \angle OAB+\angle DCB=180^{0}\;\;\;\;\;\;\;[Opposite\;angles\;of\;cyclic\;quadrilateral]\\ \Rightarrow 98^{0}+x=180^{0}\\ \Rightarrow x=180^{0}-98^{0}=82^{0}\)

(x)

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Solution:

\(We\;have,\\ \angle BAC=35^{0}\;\\ \angle BDC=\angle BAC=35^{0}\;\;\;\;\;\;[Angle\;in\;same\;segment]\\ In\;\Delta BCD,\;by\;angle\;sum\;property\\ \angle BDC+\angle BCD+\angle DBC=180^{0}\\ \Rightarrow 35^{0}+x+65^{0}=180^{0}\\ \Rightarrow x=180^{0}-35^{0}-65^{0}=80^{0}\)

(xi)

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Solution:

\(We\;have,\\ \angle ABD=40^{0}\;\\ \angle ACD=\angle ABD=40^{0}\;\;\;\;\;\;[Angle\;in\;same\;segment]\\ In\;\Delta PCD,\;by\;angle\;sum\;property\\ \angle PCD+\angle CPO+\angle PDC=180^{0}\\ \Rightarrow 40^{0}+110^{0}+x=180^{0}\\ \Rightarrow x=180^{0}-150^{0}\\ \Rightarrow x=30^{0}\)

 (xii)

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Solution:

\(Given\;that,\\ \angle BAC=52^{0}\;\\ Then\;\angle BDC=\angle BAC=52^{0}\;\;\;\;\;\;[Angle\;in\;same\;segment]\\ Since\;OD=OC\\ Then\;\angle ODC=\angle OCD\;\;\;\;\;\;[Opposite\;angle\;to\;equal\;radii]\\ \Rightarrow x=52^{0}\)

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

Solution:

Given O is the circum centre of triangle ABC and \(OD\perp BC\)

To prove \(\angle BOD=2\angle A\)

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Proof:

In \(\Delta OBD\;and\;\Delta OCD\\ \angle ODB=\angle ODC\;\;\;\;\;\;\;\;[Each\;90^{0}]\\ OB=OC\;\;\;\;\;\;\;\;[Radius\;of\;circle]\\ OD=OD\;\;\;\;\;\;\;\;[Common]\)

\(Then\;\Delta OBD\cong \Delta OCD\)         [By RHS Condition].

\(∴ \angle BOD=\angle COD…..(i)\;\;\;\;[PCT]\). \(By\;degree\;measure\;theorem\\ \angle BOC=2\angle BAC\\ \Rightarrow 2\angle BOD=2\angle BAC\;\;\;\;\;[By\;using\;(i)]\\ \Rightarrow \angle BOD=\angle BAC\).

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

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Solution:

Given, BO is the bisector of \(\angle ABC\)

To prove AB = BC

Proof:

Since, BO is the bisector of \(\angle ABC\).

Then, \(\angle ABO=\angle CBO…..(i)\)

Since, OB = OA                     [Radius of circle]

Then, \(\angle ABO=\angle DAB…..(ii)\)             [opposite angles to equal sides]

Since OB = OC                       [Radius of circle]

Then, \(\angle OAB=\angle OCB…..(iii)\) [opposite angles to equal sides]

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

\(\angle OAB=\angle OCB…..(iv)\)

\(In\;\Delta OAB\;and\;\Delta OCB\)

\(\angle OAB=\angle OCB\)                      [From (iv)]

\(\angle OBA=\angle OBC\)                      [Given]

OB = OB                                                                    [Common]

Then, \(\Delta OAB\cong \Delta OCB\;\;\;\;\;\;\;[By\;AAS\;condition]\\ ∴AB=BC\;\;\;\;\;\;[CPCT]\)

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

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Solution:

We have,

\(\angle 3=\angle 4\;\;\;\;\;\;[Angles\;in\;same\;segment]\\ ∴ \angle x=2\angle 3\;\;\;\;\;[By\;degree\;measure\;theorem]\\ \Rightarrow \angle x=\angle 3+\angle 3 \Rightarrow \angle x=\angle 3+\angle 4…..(i)\;\;\;\;\;[\angle 3=\;angle 4]\\ But\;\angle y=\angle 3+\angle 1\;\;\;\;\;\;[By\;exterior\;angle\;property]\\ \Rightarrow \angle 3=\angle y-\angle 1…..(ii)\\ from\;(i)\;and\;(ii)\\ \angle x=\angle y-\angle 1+\angle 4\\ \Rightarrow \angle x=\angle y+\angle 4-\angle 1\\ \Rightarrow \angle x=\angle y+\angle z+\angle 1-\angle 1\;\;\;\;\;\;[By\;exterior\;angle\;property]\\ \Rightarrow \angle x=\angle y+\angle z\)

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.

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Solution:

\(By\;degree\;measure\;theorem\\ \angle AOB=2\angle ACB\\ \Rightarrow 130^{0}=2\angle ACB \Rightarrow \angle ACB=\frac{130^{0}}{2}=65^{0}\\ ∴\angle ACB+\angle BCD=180^{0}\;\;\;\;\;\;\;[Linera\;pair\;of\;angles]\\ \Rightarrow 65^{0}+\angle BCD=180^{0}\\ \Rightarrow \angle BCD=180^{0}-65^{0}=115^{0}\\ By\;degree\;measure\;theorem\\ reflex\angle BOD=2\angle BCD\\ \Rightarrow reflex\angle BOD=2\times 115^{0}=230^{0}\\ Now,\;reflex\angle BOD+\angle BO’D=360^{0}\;\;\;\;\;\;[Complex\;angle]\\ \Rightarrow 230^{0}+x=360^{0}\\ \Rightarrow x=360^{0}-230^{0}\\ ∴ x=130^{0}\)

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

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Solution:

Since PQ is diameter

Then,

\(\angle PRQ=90^{0}\;\;\;\;\;\;[Angle\;in\;semi\;circle]\\ ∴ \angle PRQ+\angle TRQ=180^{0}\;\;\;\;\;[Linear\;pair\;of\;angle]\\ 90^{0}+\angle TRQ=180^{0}\\ \angle TRQ=180^{0}-90^{0}=90^{0}.\\\)

By degree measure theorem

\(\angle ROS=2\angle RQS\\ \Rightarrow 40^{0}=2\angle RQS\\ \Rightarrow \angle RQS = \frac{40^{0}}{2}=20^{0}\)

In \(\Delta RQT\), by Angle sum property

\(\angle RQT+\angle QRT+\angle RTS=180^{0}\\ \Rightarrow 20^{0}+90^{0}+\angle RTS=180^{0}\\ \Rightarrow \angle RTS=180^{0}-20^{0}-90^{0}=70^{0}\)

Q10) In figure 16.139, if \(\angle ACB=40^{0},\;\angle DPB=120^{0},\;find\;\angle CBD.\)

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Solution:

We have,

\(\angle ACB=40^{0};\;\angle DPB=120^{0}\)

\(∴ \angle APB=\angle DCB=40^{0}\;\;\;\;\;\;[Angle\;in\;same\;segment]\\ In\;\Delta POB,\;by\;angle\;sum\;property\\ \angle PDB+\angle PBD+\angle BPD=180^{0}\\ \Rightarrow 40^{0}+\angle PBD+120^{0}=180^{0}\\ \Rightarrow \angle PBD=180^{0}-40^{0}-120^{0}\\ \Rightarrow \angle PBD=20^{0}\\ ∴ \angle CBD=20^{0}\)

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.

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\(∴ \angle AOB=60^{0}\;\;\;\;\;[one\;angle\;of\;equilateral\;triangle]\)

By degree measure theorem

\(\angle AOB=2\angle APB\\ \Rightarrow 60^{0}=2\angle APB\\ \Rightarrow \angle APB=\frac{60^{0}}{2}=30^{0}\)

\(Now,\;\angle APB+\angle AQB=180^{0}\;\;\;\;\;\;[opposite\;angles\;of\;cyclic\;quadrilateral]\\ \Rightarrow 30^{0}+\angle AQB=180^{0}\\ \Rightarrow \angle AQB=180^{0}-30^{0}=150^{0}.\)

Therefore, Angle by chord AB at minor arc = 1500

Angle by chord AB at major arc = 300


Practise This Question

Say true or false:
Any two diameters of a circle will necessarily intersect.