CHAPTER -2
UNIT -2
COMPOUND INTEREST
POINTS TO REMEMBER
1. in compound interest, principal increases for every period of time.
2. formula a= p(1+R)4 C.I. = A-P.
3.Banks charge compound interest on any type of loan.
4. In fixed deposits and cumulative deposits we get compound interest.
5. Growth rate per unit time is called growth factor.
6. (a) If the growth factor is >1, it is Appreciations.
(b) If the growth factor is < it is despreciation.
7. A ready recokner shows the compound interest calculated for Rs. 1 different
Period and different rate.
EXERCISE 2.25
1. Calcualte the amount and compound interest for the following :
(a) Rs12,000 for 2 years at 10% compounded annually.
(b) Rs 20,000 for 3 years at 8% compounded annually.
(c) Rs 5,000 for 1 year at 4% compounded semi-annually.
(d) Rs 10,000 for 1 1 years at 5% compounded half – hearly.
2
(e) Rs 500 for 1year at% compounded quarterly.
Solution:
(a)
Rs 12,000 for 2 years at 10% compounded annually.
P = Rs12,000 R =10% n-2
A = p(1+ R)2
100
= 12,000 (1+ R)2
100
=12,000H 110 H 110
100 100
Amount: Rs 14,520
Interest = A –P =14,520 -12,000
=RS 2,520
(B) Rs 20,000 for 3 years at 8% compounded annually.
P = Rs 20,000 R =8% N =3
A= P (1+ R )n
100
=20,000 (1+ 8 )3
100
= 20,000H 108 H 108 H 108
100 100 100
= 54X108X108
25
=Rs 25,194.24
A = rs 25194.24
I =A – P
25,194 – 2000
= Rs 5194
(c)Rs. 5000 for 1 year at 4% compounded semi-annually.
P = Rs 5,000 R = 4% n=1
A = (1 + R )2n
2*100
=5,000( 1 + 4 )2*1
2*100
= 5000 (1+ 1 )2
50
= 5000 X 51 X 51
50 50
Amount = RS 5202
Compound Interest = A- P
= 5202 – 5000
= Rs 202.00
(d) Rs 10000 for 1 ½ years at 5% compounded half – hearly.
P = Rs 10000
A= ( 1+
R = 5%
R )1 ½ *2
2*100
= 10000 (1 + 5 )3
200
= 10000( 1 + 1 )3
40
= 10000(41)3
40
= 10000*41*41*41
40*40*40
Rs 10768.91
Compound Interest = A – P
10768.91 – 10000.00
= Rs. 768.91
= Rs 769.00
n= 1½
(e)Rs 500 for 1 year at 2% compounded quarterly.
P= RS 500
R=2%
n=1
A= p(( 1 + R )4n
4*100
=500 (1 + 2 )4*1
4*100
= 500 ( 201 )4
200
=500*201*201*201*201
200*200*200*200
= 201*201*201*201
200*200*40*2
= 510.07
Compound interest = A – P
= 510.07 – 500
= 10.07
= RS 10.00
2. A man invest Rs 5000 for 2 years at compound interest. After one year his
money amount to Rs 5150. Find the interest for second year.
Solution:
Principal for the first year = P1 = RS 5000.00
Amount at the end of first year=A1 = Rs 5150.00
Interest = A1 –P1=5150.00 - 5000.00
Rs = 150.00
Rate of interest = R 100I = 100*150
PT 5000*1
= 3%
Interest for the scond year = PTR
100
P = 5000 + 150 = 5150
= 5150 * 1 * 3 = Rs 154.5
100
3. Find the Amount on 10000 after 2 years if the rate of interst are 3% and 4%
for successive years.
Solution:
P1 = RS 10000.00
R = 3%
T=1
I1 = PTR = 10000 * 1* 3 =RS 300.00
100
100
Principal for the second year P2= 10000.00+300.00
= 10300.00
Rate R2 = 4%
Interest = 10300*4*1 = Rs 412
100
Amount at the end of the second year A =P + I
= 10300.00 + 412.00
=Rs 10712.00
4. Pralhad invest a sum of money in a bank and gets Rs 3307.5 Rs 3472.87 in
2nd and 3rd year respectively. Find the sum he invested.
Solution:
Amount at the end of second year A2 = RS 3307.50
Amount at the end of third year A3 = Rs 3472.87
Interest on Rs 3307.50
For a year = 3472.87 – 3307.50
= rs 165.37
Rate of interest = 100I = 100 * 165.37 = 5%
PT
3307.50*1
Let P be initial investment
Amount after 2 years =A =P( 1+ R )n
100
= P (1 + 5 )2 =3307.50
100
= P ( 21 )2 = 330750
20
P= 3307.50 *20 *20
21 * 21
= Rs.3020.00
5. On what sum of money will be difference between the simple interest and
compound interest for 2 years at 4% per annum will be equal to RS 100?
(Hint : Assuming the principal to be RS 100 first calculate the SI and CI, then
proceed)
Solution:
Let the principal be Rs 100.00
R = 4 T =2 years
S.I = PTR = 100 *2 * 4 = Rs 8.00
100
100
Amount at the end of 2 years if C.I is calculated
= 100 ( 1 + 4 )2
100
= 100 * 26 * 26 = Rs 108.16
25 25
C.I = 108.16 – 100.00
= 8.16
Difference between SI and CI
= 8.16 – 8.00 = 0.16
When the difference is RS 0.16
The amount invested = RS 100.00
The amount invested when the difference between S.I and C.I is
RS 100.00= 100*100
0.16
= Rs 62500.00
6. A sum of money is invested at compound interest payable annually. The
interest in two successive year are RS 275.00 and Rs 300.00. Find the rate of
interest.
Solution:
Interest at the end of 1 year = 275
Interest at the end of II year = 300
Interest per year = 300 -275
= RS. 25
Interest on RS 100 per year = 25
I = PTR
100
25 = 275 * 1 * R
100
= 25 *100 = 9 1
275
11
Rate of interest = 9 1%
11
7. The difference between compound interest and simple interest on a certain
sum for 2 year at 7 ½% per annum is RS. 360. Find the sum and verify
answer.
Let principal be Rs. 100 time 2 years
Rate =7 ½ %
S.F = PTR = 100 *2 * 15 =15
100
100 2
C.F = P (1 + 15 )2 -1
200
= 100
(200 + 15 )2 -1
200
= 100
( 215 * 215 ) -1
200 200
=100 46225 – 1
40000
= 100 46225 – 40000
40000
= 100 * 6225 = 6225
40000 400
=15.56
C.I – S.I =15.56 -15.00
= 0.56
When p= 100 difference in CI and SI
= 0.56 300
P = 100*360 = 100*360*100
0.56
56
= 100 * 45 * 100
7
= 64287.71
P= 64286
8. Teju invests rs 12000 At 5% interest compounded annually . If he receives an
amount RS. 13320 at the end find the period..
Solution:
P = Rs 12000
R = 5%
A= Rs 13230
A = P (1 + r )n
100
12000( 1 + 5 )n = 13230
100
(1+ 1 )n = 13230 = 1323 = 441 = (21)2
20
12000 1200 400 20
i.e (21)n = (21)2
20
20
N= 2 years
9. The present population of a village is Rs . 18000. It is estimated that the
population of the village grows by 3% per year . Find the population of the
village after 4 years .
Solution:
P= 18000
R= 3%
A = p ( 1 + R )n
100
=18000( 1 + 3 )4
100
n=4
= 18000*103*103*103*103
100*100*100*100
=20256.15858
=20259
Population after 4 years = 20259
10. Jeshu purchased a bike by paying Rs. 52000. If the value depreciates by 2%
every year. Find the value of the bike after 3 years .
Solution:
Original price of the bike =rs.52,000
Rate of depreciation =2%
Price after 3 year =a=p 1-r n
100
52000 1- 2 3
100
52,000* 98 * 98 * 98
100 100 100
=Rs.48941.98
11.
Using the ready reckoner find the compound interest in the following:
a.Principal Rs 15000 for 4 years at 6.5% p.a.
b.Principal Rs 22000 for 5 years at 12% p.a.
Solution:
a.P=Rs 15000
R = 6.5%
n = 4 years
From the table the interest on Rs 1.00 at 6.5% for 4 years is Rs 0.2865
C.I on Rs 15000 = 0.2865 * 15000
= Rs 4297.5
b. P = RS 22000
R=12%
n = 5 years
From the table the interest on RS 1 at 12% for 5 years is Rs 0.7623
C.I on Rs 22000 for 5 years
= 0.7623 H 22000
= Rs 16770.60
12. Using the ready reckoner find the period of interest in the following :
a. Compound interest Rs 4347 at 7% p.a. principal Rs 30000
b. Amount RS 16939.2 principal RS 12000 at 9% p.a.
Solution:
a. P=30000
R= 7%
C.I Rs 4347
The C.i on RS 30000 at 7% for a
Certain period = RS 4347.00
The C.I on RS 1 4347
30000
= 0.1449
From the ready reckorner find the C.I on Rs 1 is 0.1449
If n =2
Thw term period = 2 years
b. Amount RS 16939.2 principal Rs 12000 at 9% p.a.
A=RS 16939.2
p =RS 12000
R= 9%
C.I .A.P = 16939.2 -12000
= 4939.2
Interest on Re 1= 4939.2
12000
= 0.4116
From the Mady welcome we consu that the C.I and Re 1 is 0.4116
If n =4
Period =4 years
13. Using the ready reckorner ,Find the rate of interest in the following :
a. Compound interest RS 1733.6 for 3 years on principal of RS 11000
b. Amount RS 35246 principal SR 20000 for 5 years .
Solution :
a. P=RS 11000
C.I Rs 1733.6
n=3 years
Inersest on RS 11000.00
For three years RS
=1733.6
Interset on RS 1
=1733.6
11000
=0.1576
From ready reckoner we find ,
The interest on re1.00 is 0.1576
When the rate is 5%
Rate of interest = 5%
b. Amount RS 35,246 principal Rs 20000 for 5 years
C.I =A – P
=35246.00 – 20000
= 15246.00
Interest on Re 1.00 = 15246
20000
= 0.7623
From the ready reckoner we find the intersest on Re 1.00 for 5
years at 12% = 0.7623.
Rate of interest R = 12%
ADDITIONAL PROBLEMS
 The difference between the simple interest and the compound interest on a
certain principal for 2 years in Rs 160. If the Simple interest is Rs 2880 find
the rate of interest.
Solution:
Number of years n =2
Simple Interest = RS 2880.00
Difference between S.I and CI = RS 160.00
Compound Interest = 2880.00 + 160.00
= 3040.00
Simple Interest for one year = 2880
2
=1440
The difference between the SI and CI is the interest on
RS 1440.00 for 1 year
R= 100I = 100 * 160
PT
1440*1
=100
9
=11 1/9%
 A sum amount to RS 2916 in 2 years and RS 3149.28 in 3 years when
compounded annually , find the principal.
Solution:
Amount at the end of 2 nd year = principal for 3 rd year
Rs 3149.28
P= rs 2916
A= P (1 + R )n
100
3149.28 = 2916 (1 + R )1
100
3194.28 = (1 + R )
2916
100
1 + R =1.09
100
R =?
R = 1.09 – 10.09
100
R = 9%
Amount at the end of 2 years = Rs 2916
P( 1 + R )2 = 2916
100
P( 1 +0.09)2 = 2.916
P(1.1881) = 2.916
P = 2916
1.1881
= 2.454
The principal = Rs 2454
 A sum of Rs 5500 was taken a loan. This is to be paid back in two equal
instalments . If the rate of interest is 20% per annum compounded annually
find the value of each instalment.
Solution:
P= Rs 5500
R= 20%
A= p( 1 + R )n
100
A= 5500 ( 1 + 20 )2
100
A = 5500( 1 + 0.02)2
n = 2 years A=?
A = 5500(1.2)2
=5500 *1.44
= 7920
This is to be paid in two equal installments
Each installment =7920
2
= Rs 3960
4. What is the compound interest on rs. 5,000 for 3 year at 8% for The firswt year
10% for the second year and 12% for the third year?
Solution:
P1 = 5,000,
R1=8% ,
I=PTR = 5000X1X8= 4000.00
100
100
P1 =5,000+400 =Rs.5,400.00
P2 =10%
T=1YEAR
I2 = 5,400X1X10 =Rs.540.00
100
P3 =5,400+540=Rs. 5,940.00
I3= 5,940x1x12 =Rs . 712.8
100
T =1YEAR
Compound interest for 3 years =400+
540
712.8
1,652.8
5. find the minimum number of complete year in whish a sum of money put at
20% compound interest will be more than doubled.
Solution:
Let the principal be Rs 100
Interest in 1 year at 20% = 200 x100 = Rs.20
100
Interest in 2 years at 20% =120 x 20 =Rs.24
100
Interest in 3 years at 20% =144x20 =Rs 288
100
10
Interest in 4 years at 20% = 172.8x20
100
= Rs.34.56
For Rs100, the interest in 4 years = Rs = .107.36
Compound interest more than double in 4 years.
CHAPTER -4
UNIT – 4
CIRCLES
EXERCISE 4.4.2
1.
In a circle whose radious is 8 cm.a chord is drawn at a point 3cm. from the
centre of the circle. The chord is divided into two segments by a point on it. If one
segment of the chord is 9cm .what is the length of the other segment?
Solution:
O
A
Pp
8cm
Q 3cm
B
Given : Let O is the center of the circle AB chord OQ = 3cm , OB = 8cm, AP
and BP are segments BP = 9cm .
To find : AP
Proof: IN
OQB; ∟OQB = 900
( OQ┴ AB)
BQ2+OQ2= OB2
BQ2 = OB2- OQ2
Or BQ = √ OB2 – OQ2
= √ 82 – 32
=
√ 64 – 9
BQ = √ 55
AB = 2√ 55
AP =AB – BP
AP = ( 2 √ 55 - 9)cm
2. Suppose two chords of a circle areequidistant from the center of the circle .
Prove that th chords have equal length.
Solution:
P
A
B
O
D
Q
C
Given: In the fig O is the center of the circle. AB and CD are the chords of a
circle which are equidistant from the center i.e OP =OQ
To prove: AB = CD
Proof: IN
PBO and
QDO
∟BPO = ∟DQO (900)
PO = QO(data)
OB = OD (radil)
PBO = QDO(RHS)
PB = DQ
Hence AB = CD
3. Suppose two chords of a circle are unequal in length . Prove that the chord
of larger length is nearer to the centre than the chord of smaller length.
Solution:
A
2
P
2
C
O
Q
B
D
Given : In the fir O is the center of the circle AB and CD Are chords. AB is
larger and CD is smaller chord . OP and OQ the distance from the chord AB
and CD respectively.
To prove : OP< OQ
Proof : In
APO;
In
CQO;
AO = OC
AO2 = AP2 + PO2……(1)
OC2 = CQ2 + OQ2……(2)
radius of the circle.
From (1) and (2)
AP2 + PO2 = CQ2 + OQ2
Let AP = CQ + x….
( AB>CQ ½ AB > ½ CD. AP> CQ)
(CO+X)2 + PO2 = CQ2 + OQ2
CQ2 + 2.CQ.X + PO2 = CQ2 + OQ2
OQ2 = OP2 + 2.CQ.X
OR OQ2 > OP2
OQ > OP
OP <OQ
4. Let AB and CD be parallel chords of a circle with center O. M is the
midpoint of AB and N is the midpoint of cd . prove that O, N ,M are
collinerar. If MN = 3cm , AB= 4cm, CD = 10 cm find the radious of the
circle.
Solution:
O
C
D
N
A
M
B
Given : AB and CD are to parallel chords . M and N Are midpoint Of AB and CD
respectively.
To prove : O,N,M are collinear.
To find : RAdious OF the circle i.e OD
Proof: AB ! CD
∟OMB =900 … M mid point .
∟OND =900 and ∟MND = 900
∟OND + ∟MND 1800
ON and M are collinear
Let ON =x, in
OMB
OB2 = OM2 + MB2
= (3+X)2 + 22
(AB=4cm
MB=2cm)
= 9+6X+X2+4
OB2 = X2+6X+13….(1)
In
OND;
OD2 = ON2 + ND2
= X2 + 52
( CD = 10cm ND = 5cm )
OD2 = X2 + 25…..
OB2 = X2 + 25……(2) (OB=OD)
(CD= 10cm ND = 5cm)
From (1) and (2)
X2+6X+13 =X2 + 252
6X= 25 -13
6X= 12
X= 12/6
X= 2Cm
OD2= X2 + 52
OD2= 22+25
OD2=4+25
OD2= 29
OD=√29
5. In the fig A,B,Care collinear circles and Ac =CB : prove that PQ = RS .
where the line PQRS passes through the point of intersection Q and R and
cuts other two circles at P and Q .( Hint: draw perpendiculars from A,C,B to
PQ)
Solution:
P
Q
DD
E
R
S
F
C
B
Given: AC = CB
AD | | CE | | BF…. Corresponding angles are egual ( i.e 90 0 )
DE = EF……. Are intercepts .
DQ + QE = ER + RF
½ PQ +QE = ER+ ½ RS
( QE = ER)
½ PQ = ½ RS
PQ = RS
6. Suppose ab and cd are two parallel chords in a circle .prove that the line
joining this mid point pass through the center of the circle.
Solution:
M
A
B
N
C
D
Given : O center of the circle and AB ||CD
Const : Join OM and ON
Proof: M is the mid point of CD
∟OMB =900
Similarly N is mid point of CD
∟OND =900
∟OMB + ∟BMN = 1800
(∟BMN =900)
i.e OMN is a straight line
MN passes through O.
7. Prove that a parelellogram inscribed in a circle is a rectangle
Solution:
A
D
O
B
C
Solution:
gm ABCD is cyclic.
∟A +∟C = 1800
(opposite angle of cyclic quadrilateral)
But ∟A +∟C = 900 …….(1)
(opposite angle of parallelogram are equal)
lly ∟B+∟D = 1800
But ∟B +∟C = 900 …….(2)
(opposite
gm angles)
AB = CD and AD = BC ……(3)
(oppite side of
gm)
Opposite sides are equal and each angle is right angle
ABCD is rectangle
EXERCISE 4.4.3
1.
In the fig ∟ACB =420 find ∟x.
Solution:
X
42
In the fig ∟AOB angle at the center and ∟ACB angle at the
circumference.
∟AOB = 2 ∟ACB
∟AOB = 2 H 42 0
∟AOB = ∟x = 840
2. In the fig ∟ABC is a circle with centre O and reflex ∟AOB=2500 find ∟X.
Solution: In the fig ∟ACB angle at the circumference ∟AOB reflex
angle at the centre.
C
A
X
B
O 2500
∟AOB = ½ reflex ∟AOB
∟ACB = ½ H 2500
∟X = ∟ACB = 1250
3. In the fig, AOB is diameter , Find ∟C .
C
O
A
B
In the fig AOB is diameter.
∟AOB =1800
∟ACB = ½ ∟AOB
= ½ H 1800
∟ACB= 900
Angle at the circumference reflex angle at the centre.
4. In the figure ABCDE is a circle withcentre O, ∟X= 1800 find
i)∟d
ii) ∟y
vi) ∟b + ∟e
iii) ∟a
iv) ∟b
v) ∟b + ∟d
D
d
y
E
x
A
b
C
B
A
Solution:
In the fig, ∟ADC angle at the circumference ∟AOC angle at the centre .
∟ADC = ½ ∟AOC
∟d = ∟ADC = ½ H 1080
∟d =540
ii)reflex ∟AOC =3600 - ∟AOB
= 3600 - 1080
∟y = reflex ∟AOC = 2520
∟y= 2520
iii) ∟b= ∟ABC = ½ reflex ∟AOC
= ½ H 2520
∟b= 1260
iv)
∟a + ∟b =1800
∟a + 1260 = 1800
∟a = 1800 – 1260
∟a = 540
v)
∟b + ∟d=1260 – 540
∟b + ∟d = 1800
vi)
∟e = ½ ∟AOC
∟b + ∟c = 1800
(opposite angle of cycle is quadrilateral)
5.
In the fig diameter AB and chord QR are produced to meet at P. if
∟QPA=260, ∟QAR = 360 find ∟x and ∟y.
Solution:
A
360
Q
O
X
B
y
R
P
∟AOB =1800
∟AOB = ½ H1080
∟AOB = 900
∟PAR = ∟BQP =X0
In the
le AQP
∟PAQ + ∟AQP+ ∟APQ =1800
(360+x) + ( 900+x + 260 =1800
2∟x =1520 =1800
2∟x =1800-1520
2∟x =28
∟x =140
In the
le APQ
∟AQR + ∟QAR + ARQ = 1800
(900+x) + 360 + ∟y =1800
900+14+360 +∟y =1800
1400 + ∟y = 1800
∟y = 1800 – 1400
∟y = 400
6. In the fig , ∟CBD = 1100 find ∟AOC
Solution:
C
O
1100
A
B
∟ABC + ∟DBC = 1800
∟ABC + 1100 =1800
∟ABC = 1800 – 1100
∟ABC = 700
∟AOC = 2 ∟ABC
= 2H700
∟AOC = 1400
7. In the fig ∟ADC = 840 and AB = BC Find ∟BDC.
Solution:
D
D
A
840
C
B
ABCD is a cyclic quadrilateral
∟ABC + ∟ADC = 1800
∟ABC + 840 = 1800
∟ABC = 1800- 840
= 960
In the ABC , AB = BC
∟BAC = ∟BCA = 1800 – ABC
2
= 1800 – 960
2
∟BAC = 420
But ∟BAC = ∟BDC
∟BDC = 420
8. In the fig AB is diameter ∟BAC = 380 find ∟ADC.
Solution:
D
A
C
B
380
O
AOB is diameter
∟ACB = ½ AOB
= ½ H1800
∟ACB = 900
∟ABC + ∟ACB +∟BAC = 1800
∟ABC+ 900 + 380 = 1800
∟ABC = 1800-900-380
∟ABC = 520
∟ABC + ∟ADC = 1800
( ABCD is cyclic quadrilateral)
520 + ∟ADC = 1800
∟ADC = 1800 – 520
= 1280
9. In the fig ∟QXR = 250, ∟QRX = 330 find ∟XYZ and ∟PZQ.
P
Q
Z
330
1250
X
Y
Solution:
In
le QXR: RQ is produced to P
Exterior ∟PQX = ∟Q H ∟R +∟QRX
= 250+330
∟PQX = 580
∟PQX = ∟PYX……. Angles in a same segment
But ∟PQX = 580
∟PYX = 580
∟XYZ =580
In the
le XYZ
∟X +∟Y + ∟Z =1800
250 + 580 + ∟Z = 1800
R
830 + ∟Z = 1800
∟Z = 1800-830
= 970
∟ABC = 970
10.
In the fig AFD and BFE are straight lines find ∟ACF.
C
C
B
230
D
1150
1100
F
A
Solution:
In the fig ABCD is Cyclic quadrilateral
∟BCF = 1800 - ∟BAF
= 1800 – 1100
= 700
E
But ∟BCA = 230
∟ACF = ∟BCF - ∟BCA
=70 – 23
= 470 ……(1)
∟FCD = ∟FED = 1800
∟FCD = 1800 – 1150
∟FCD = 650
∟ACB = ∟AFB but ∟ACB = 230
But ∟AFB = ∟EFB
∟EFB = 230
∟AFB = 230
But ∟EFB = ∟∟ECD…angles in the same segments
∟ECD = 230
∟FCE
= ∟FCD - ∟ECD
= 650 – 230
∟FCE
∟ACD
= 420
=∟ACF + ∟FCE
= 470 + 420
∟ACD
= 890
EXERCISE 4.4.4
1. Construct the circum circle of a triangle ABC in which AB =3cm ∟B= 780
and BC= 4.6cm.
Solution:
Steps
A
3cm
S
780
S
B
C
4.6cm
4.
2. Construct a triangle ABC ∟B = 550 ∟C = 480 BC = 3.8cm . Construct a
sircumcircle.
A
S
550
B
480
C
3. Construct A cyclic Quadrilateral ABCD in which AB = 4cm BC= 5cm CD=
2.8cm and ∟B=600
E
A
D
S
B
C
4. Construct a cycle quadrilateral ABCD in which AB = 2.8cm BC = 4cm CD =
3cm and ∟B 1050
E
A
S
D
105
B
3
4cm
C
4. Construct cyclic quadrilateral ABCD in which AB= 3.4cm BC=3cm
∟B=960 and the vertex D lies on the perpendicular bisector of BC.
A
D
3.4 cm
960
B
C
3cm
6.Construct a cycle Quadrilateral XYZW in which WX= 3.5 cm ∟x =750 XY
=2.8 cm and the vertex Z lie the angle bisector of ∟WXY.
P
W
Z
S
X
Y
7.Construct a square side length 5cm and inseribe it in a circle.
A
D
S
B
C
8.Construct a square in a circle of radious 3cm.
D
A
O
B
C
9.Construct a regular pentagon in a circle of radious 3.6 cm
C
B
D
O
360
720
A
E
10.Construct a rectangular hexagon in a circle of 5cm (Hint the interior angle of a
regular hexagon is 1200 ).
E
D
F
C
A
B
ADDITIONAL PROBLEMS ON CIRCLES AND CYCLIC
QUADRILATERALS
1. Prove that a trapezium inscribed in a circle is an isosceles trapezium
Solution:
A
B
D
C
Data: ABCD is a trapezium AD || BC
Construction: Produce BA and CD to meet a E
To prove AB = CD
Proof : AB | | CD, BE transversal
∟EAD = ∟EBC = X0 ………(1)
(corresponding angles)
∟EAD = ∟ECB =x ………..(2)
ABCD is a trapezium
AD | | BC
∟A + ∟B = 180 (interior angle are supplementary AB| | CD)
∟A + ∟C = 180 (opp.angle of cyclic quadrilateral)
∟A + ∟B = ∟A + ∟C
∟B = ∟C
Hence ABCD is an isosceles trapezium.
2. Prove that a rhombus inscribed in a circle is square .
Solution:
D
A
C
B
Data : ABCD is a cyclic rhombus
AB = CD = BC = AD
To prove: ABCD in a square .
Proof : ∟DAB + ∟BCD = 1800
But ∟DAB = ∟BCD ……opposite angle of rhombus
∟DAB = ∟BCD = 900 ………(1)
|||ly ∟ADC = ∟ABC = 900 ……(2)
From (1) and (2)
Each angle of ABCD is equal to 900
All sides are equal from data
ABCD is a square.
3. In a adjoining figure C1 and C2 are Concentric circle and AB in a chord of
C2 intersecting C2 at D and E . prove that AD=EB.
M
A
D
C2
E
O
B
C1
Solution:
Data: C1 and C2 concentric circle with center O. AB is a chord of C2
cutting C1 at D and E.
To prove AD = EB
Constn : Draw OM ┴ AB
Proof : OM ┴ AB
OM ┴ DE
AM = MB …..(1)
DM = ME ….(2)
AM = MB
AD + DM = ME + EB ( DM = ME from (2))
AD = EB
4. Two circles with centre P and Q intersect at M and N . prove that PQ bisects
MN perpendicular.
Solution:
M
P
R
R
N
Data: P and Q are two intersecting circles at M and N
To prove : PQ is a perpendicular bisects of MN OR
PQ ┴ MN and RM = RN
Construction : Join PM , MQ , PN and NQ
Proof: In
les PMQ and PNQ
PM = PN … radii
QM = QN … radii
PQ is common
PQM =
PNQ
∟MPR =∟NPR = and ∟MQR = ∟NQR
PR bisects ∟MPN and QR bisects ∟MQN
Q
In a
le PMN
PM= PN …. Radii
PMN is isosceles
∟MPR = ∟NPR
PR ┴ MN …… PR is bisector of angle at
MR = RN
the vertex of isosceles
le
|||ly QR ┴ MN
PQ ┴MN and MR = RN
5.Suppose AB and AC are equal chords in a circle . prove that the bisector of
∟BAC passes through the centre .
Solution:
A
C
B
C
D
Data: AB and AC are equal chords . AD is bisects ∟BAC.
To prove : AD passes through C.
Construct : Join BD and DC
Proof : In
le ABD and
ACD
AB = AC
∟BAD = ∟CAD … AD bisector of ∟BAC
AD in commo
ABD n =
ACD
B= C
But ∟B + ∟C =1800
(opposite angles of cyclic quadrilateral)
∟B + ∟C = 900
AD in diameter
AD passes trhough center C.
6.Suppose P is point inside a circle such that there are two points . A and b on the
circle with PA = PB . prove that P is the centre of circle .
Solution:
P
A
B
Data: A and B are on circles and P is inside the circle PA = PB
To prove: mP is the centre of the circle .
Proof: PA = PB
This is possible only when PA and PB are radious.
i.e P Is the centre of the circle .
7.Suppose two equal chords in a circle intersects show that the segments of one
chords are respectively equal to the segments of the order.
Solution:
B
C
x
X
O
A
Data: AB and CD are equal chords intersects at O.
To prove: OA = OD and BO = OC
Construction : Join AC , BD and AD
Proof: in
ABD and ∟ACD
AB = CD …. Given
∟ABD = ∟ACD = x
D
AD is common
AC = BD
In
le ACO and BDO
∟AOC = ∟BOD ……V.O A
∟CAO = ∟CDB ….Angles in the same segment
AC = BD…. From (1)
ACO =~
BDO
AO = OD
BO = OC
8.In the adjoining fig AB and CD are two chords intersecting in E . Suppose EO is
the bisector of ∟BED . prove that AB = CD.
Solution:
D
Q
A
O
C
Given : O is the centre of circle .
To proov : chord AB = chord CD.
O
P
B
Construction : Draw ┴ OP and OQ to AB and CD
Proof: EO is the bisector of ∟BED
In
POE and
OQE
∟PEO = ∟QEO
Given construction
∟OPE = ∟OQE
900
FO = EO
OPE =
OP = OQ
common side
OQE
SAS
CP CT
The chords are equideritant from the centre . They are equal
AB = CD
9.In the adjoining figure ∟BDC =740 find ∟BAC , ∟BOC and ∟OBC .
A
D
740
O
B
Solution:
C
∟BAC = ∟BDC …… angles in same segments
1) ∟BAC =740
∟BOC = 2 ∟BDC
….(centre angle and angle at the circuferural)
∟BOC = 2 H740
∟BOC = 1480
2) In
le BOC
BO = OC
∟OBC = ∟OCB = 1480 – BOC
2
=1800 – 1480
2
= 320
2
∟OBC = 160
10.In the adjoining fig BD andb CA intersects in X . Suppose ∟DXC = 40 0 and
∟XCD = 250 find ∟BAC and the reffex angle BOC.
A
D
X
B
250
C
O
Solution: In
DCX ; ∟X = 400; ∟C = 250
∟BDC = 1800 – (∟X + ∟C)
= 1800 – (400 + 250)
= 1800 – 650
= 1150
∟BAC = ∟BDC … ( ngles in the same segment)
∟BAC = 1150
Reflex ∟BOC = 2BAC
= 2H1150
∟BOC = 2300