Exercise 16.1 Page No: 16.5: RD Sharma Solutions For Class 9 Maths Chapter 16 Circles
Exercise 16.1 Page No: 16.5: RD Sharma Solutions For Class 9 Maths Chapter 16 Circles
Exercise 16.1 Page No: 16.5: RD Sharma Solutions For Class 9 Maths Chapter 16 Circles
(i) All points lying inside/outside a circle are called ______ points/_______ points.
(ii) Circles having the same centre and different radii are called _____ circles.
(iii) A point whose distance from the center of a circle is greater than its radius lies in _________ of
the circle.
(vi) An arc is a __________ when its ends are the ends of a diameter.
(vii) Segment of a circle is a region between an arc and _______ of the circle.
Solution:
(i) Interior/Exterior
(ii) Concentric
(iv) Arc
(v) Diameter
(vi) Semi-circle
(vii) Center
(viii) Three
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Question 2: Write the truth value (T/F) of the following with suitable reasons:
(ii) Line segment joining the center to any point on the circle is a radius of the circle,
(iii) If a circle is divided into three equal arcs each is a major arc.
(v) A chord of a circle, which is twice as long as its radius is the diameter of the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure of an arc is the complement of the central angle containing the arc.
Solution:
(i) T
(ii) T
(iii) T
(iv) F
(v) T
(vi) T
(vii) F
(viii) T
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
In right ΔOCA:
64 = 36 + OC2
OC2 = 64 – 36 = 28
Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of
radius 10 cm.
Solution:
In ΔOCA:
Using Pythagoras theorem,
100 = AC2 + 25
AC2 = 100 – 25 = 75
AC = √75 = 8.66
Therefore, AC = BC = 8.66 cm
Answer: AB = 17.32 cm
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of
radius 6 cm.
Solution:
In ΔOCA:
Using Pythagoras theorem,
36 = AC2 + 16
AC2 = 36 – 16 = 20
AC = √20 = 4.47
Or AC = 4.47cm
Therefore, AC = BC = 4.47 cm
Answer: AB = 8.94 cm
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Question 4: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the
distance between AB and CD is 3 cm, find the radius of the circle.
Solution:
Let OP = x cm and OC = OA = r cm
And OQ⊥AB
AQ = BQ = 5/2 cm
In ΔOCP:
By Pythagoras theorem,
r2 = x2 + (11/2) 2 …..(1)
In ΔOQA:
By Pythagoras theorem,
OA2=OQ2+AQ2
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
x2 + 6x + 9 + 25/4 = x2 + 121/4
(using identity, (a+b) 2 = a2 + b2 + 2ab )
6x = 121/4 – 25/4 − 9
6x = 15
or x = 15/6 = 5/2
Substitute the value of x in equation (1), and find the length of radius,
r2 = (5/2)2 + (11/2) 2
= 25/4 + 121/4
= 146/4
or r = √146/4 cm
Solution:
Steps of Construction:
Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.
Step 4: This point O is a centre of the circle, because we know that, the Perpendicular bisectors of
chord always pass through the centre.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes
through the mid-point of the corresponding minor arc.
Solution:
Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle
subtended by the chord at the centre of the circle.
Solution:
Now,
OA = OB [Radius]
OC = OC [Common side]
AC = BC [Given]
Solution:
In ΔOAR:
By Pythagoras theorem,
OA2+AR2=OR2
OA2+122=202
Or OA = 16 m …(1)
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
From figure, OABC is a kite since OA = OC and AB = BC. We know that, diagonals of a kite are
perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.
=>1/2×OA×RS = 1/2 x RC x OS
=> OA ×RS = RC x OS
=> 16 x 24 = RC x 20
=> RC = 19.2
RM = 2(19.2) = 38.4
Question 2: A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand
are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each
other. Find the length of the string of each phone.
Solution:
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Radius = OA = 40 m (Given)
We know, medians of equilateral triangle pass through the circumcentre and intersect each other at
the ratio 2 : 1.
OA/OD = 2/1
or 40/OD = 2/1
or OD = 20 m
Now, In ΔADC:
By Pythagoras theorem,
AC2 = 4800
or AC = 40√3 m
Solution:
Let ∠OAB = m
In ΔOAB,
By angle sum property: ∠OAB+∠OBA+∠AOB=1800
Question 2: In figure, it is given that O is the centre of the circle and ∠AOC = 1500. Find ∠ABC.
Solution:
Solution:
or ∠BOC = 1700
∠BOC = 2∠BAC
1700 = 2∠BAC
Question 4: If O is the centre of the circle, find the value of x in each of the following figures.
(i)
Solution:
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
(ii)
Solution:
∠ABC=400 (given)
∠ACB = 900 [Angle in semicircle]
In ΔABC,
∠CAB+∠ACB+∠ABC=1800 [angle sum property]
∠CAB+900+400=1800
∠CAB=1800−900−400
∠CAB=500
(iii)
Solution:
Again, ∠APC + ∠ABC = 1800 [Sum of opposite angles of cyclic quadrilaterals = 180 o ]
600 + ∠ABC=1800
∠ABC=1800−600
∠ABC = 1200
(iv)
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
(v)
Solution:
InΔAOB:
2500 = 2x
x = 2500/2=1250
(vi)
Solution:
60 o = 2∠ OAC
Or x = 300
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
(vii)
Solution:
From figure:
∠BDC = ∠BAC = 500 [Angle on same segment]
Now,
In ΔBDC:
Using angle sum property, we have
∠BDC+∠BCD+∠DBC=1800
Substituting given values, we get
500 + x0 + 700 = 1800
x0 = 1800−500−700=600
or x = 60o .
(viii)
Solution:
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Form figure:
or x = 2×500=1000
(ix)
Solution:
∠CAD = 28, ∠ADB = 32 and ∠ABC = 50 (Given)
From figure:
In ΔDAB:
Angle sum property: ∠ADB + ∠DAB + ∠ABD = 1800
∠DAB=1800−320−500
∠DAB = 980
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Now,
980+x=1800
or x = 1800−980=820
(x)
Solution:
∠BAC = 350 and ∠DBC = 650
From figure:
∠BDC = ∠BAC = 350 [Angle in same segment]
In ΔBCD:
(xi)
Solution:
Form figure:
∠ACD = ∠ABD = 400 [Angle in same segment]
In ΔPCD,
Angle sum property: ∠PCD+∠CPO+∠PDC=1800
400 + 1100 + x = 1800
x=1800−1500 =300
(xii)
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
From figure:
∠BDC = ∠BAC = 520 [Angle in same segment]
So, x = 520
Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that
∠BOD = ∠A.
Solution:
Again,
By degree measure theorem: ∠BOC = 2∠BAC
∠BOD = ∠BAC
Hence proved.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Question 6: In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Solution:
From figure:
Radius of circle = OB = OA = OB = OC
∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]
∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]
Question 7: In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
From the figure:
∠x = ∠y − ∠1 + ∠z + ∠1
or ∠x = ∠y + ∠z + ∠1 − ∠1
or x = ∠y + ∠z
Hence proved.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
ΔABC is an equilateral triangle. (Given)
In quadrilateral ABEC:
∠BEC = 180 o - 60 o
∠BEC = 120 o
Question 2: In figure, Δ PQR is an isosceles triangle with PQ = PR and m∠PQR=35°. Find m∠QSR and
m∠QTR.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
In ΔPQR:
∠PQR = ∠PRQ = 35o (Angle opposite to equal sides)
Question 3: In figure, O is the centre of the circle. If ∠BOD = 160o, find the values of x and y.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
From figure: ∠BOD = 160 o
Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100o and ∠ABD = 70o, find ∠ADB.
Solution:
From figure:
In quadrilateral ABCD,
∠DCB + ∠BAD = 180o (Opposite angles of Cyclic quadrilateral)
100 o + ∠BAD = 180o
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
∠BAD = 800
In Δ BAD:
By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180 o
∠ADB + 80o + 70 o = 180 o
∠ADB = 30o
Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that ∠B = ∠C.
Solution:
[Co-interior angles]
∠B = ∠C
Hence proved.
Solution:
Given: ∠BOC = 100o
By degree measure theorem: ∠AOC = 2 ∠APC
100 o = 2 ∠APC
or ∠APC = 50 o
Again,
∠APC + ∠ABC = 180 o (Opposite angles of a cyclic quadrilateral)
50o + ∠ABC = 180 o
or ∠ABC = 130 o
Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 500, find ∠AOC.
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 300. Find
m(∠ACB) and m(∠ABC).
Solution:
Given: m(∠CAB)= 300
Question 9: In a cyclic quadrilateral ABCD if AB||CD and ∠B = 70o , find the remaining angles.
Solution:
Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.
Solution:
Given : ∠DAB = 50o
By degree measure theorem: ∠BOD = 2 ∠BAD
so, x = 2( 500) = 1000
∠A + ∠C = 1800
500 + y = 1800
y = 1300
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
Solution:
By degree measure theorem: ∠AOB = 2 ∠APB
so, ∠AOB = 2 × 70° = 140°
Question 2: In figure, two congruent circles with centres O and O’ intersect at A and B. If ∠AO’B =
50°, then find ∠APB.
Solution:
RD Sharma Solutions for Class 9 Maths Chapter 16 Circles
As we are given that, both the triangle are congruent which means their corresponding angles are
equal.
Therefore, ∠AOB = AO'B = 50°
Question 3: In figure, ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC=77°, AC
and BD intersect at P. Then, find ∠DPC.
Solution:
∠DBA = ∠DCA = 580 ...(1)
[Angles in same segment]
Solution:
Join OC.
In triangle OCA,
OC = OA
[same radii]
∠OCA = ∠CAO ...(2)
[Angle opposite to equal sides]
In triangle COA,
∠CAO = 600