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

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

Raman had to draw an isosceles triangle with base angles measuring 65 each and base length of 5cm. Raman said the given data is insufficient. Is he correct?