Download 4th - chapter - solutions - e-ABHYAS Academy . Perfection Through

File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
ANDHRA PRADESH - TELANGANA
2014-2015 PROGRAM M E
4TH - CHAPTER - SOLUTIONS
4. GEOMETRY
1.
DE II BC  ADE II ABC
AD AE
4x  3 8x  7

or

DB EC
3x  1 5x  3
or20x 2  15x  12x  9  24x 2  8x  21x  7
or 4x 2  2x  2  0
or2x 2  2x  x  1  0
2x  x  1  1 x  1  0
 x  1 2x  1  0
 x  1or
2.
1
2
AOC  BOC  1800 [linear pairs]
AOC  750  1800
AOC  1800  750  1050
3.
a  b  1800.......... 1linearpairs 
2014 - 2015
0
a  b  80 ..........  2 given
Adding (1) and (2), we get
2a  1800  80 0  2600
a
2600
 1300
2
Subtracting (2) from (1), we get
2b  180 0  800  100 0
4.
 b
100
 500
2
Given AD  8x  9,CD  x  3
BE  3x  4,CE  x
To find value of x that makes, DE II AB :
www.eabhyasacademy.com
If DE II AB , then by Thales Theorem, we have

AD BE

DC EC
8x  9 3x  4

x3
x
 8x 2  9x  3x 2  9x  4x  12
 8x 2  9x  3x 2  13x  12  0
PATHFINDER
1
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
 5x 2  4x  12  0
 5x 2  10x  6x  12  0
 5x  x  2  6  x  2  0
 x  2 5x  6  0
 x  2or  6 / 5
5.
POQ  POC  COQ

1
1
BOC  AOC  OP bisects BOC
2
2

1
1
 BOC  AOC  POC  BOC
2
2
1
 1800
2

6.

 OQ bisects AOC
1
AOC
2
 90 0
 COQ 
= a right angle
( linear pair axiom)
InABC,DE II BC (Given)

AD AE

(By Thales Theorem)
BD CE
A
3X-2
D
5X-4
E
7X-5
5X-3
B

2014 - 2015
C
3x  2 5x  4

7x  5 5x  3
  3x  2 5x  3  7x  5  5x  4
 15x 2  19x  6  35x 2  53x  20
 35x 2  15x 2  53x  19x  20  6  0
 20x 2  34x  14  0
 10x 2  17x  7  0
 10x 2  10x  7  x  1  0
10x  x  1  7  x  1  0
 x  110x  7   0
www.eabhyasacademy.com
x  1or7 / 10
7.
  4and    1800
or  4  1800
1800
 360
5
  4  36   1440

 or   
PATHFINDER
B
A

O
C
D
2
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
8.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
In figure AD II EF II BC
AE DF

EB FC
AE DF


2AE FC
DF 1


FC 2
1.5 1

  FC  3cm
FC 2

9.
D
A
E
F
B
C
Let lengths of sides 3k,4k
1
.3k.4k  24
2
 6k 2  24
24
 k2 
4
6
k 2
 Area 
 Hypotenuse  9k 2  16k 2  9.22  16.22
 36  64  100  10
10.
let the angle be of measure x. Then complement of x  90 0  x
Given, x  3  90 0  x 
 x  2700  3x
 Its complement  900  67.50  22.50
11.
B  1800  A  1800  1180  620
12.
B  90 0  620  280
let the angles be x and 9x . Then
2014 - 2015
x  9x  900  10x  900
 x  90
 The angles are 90 and 810
13.
 AOB is a straight line, AOC  180 0
 3x  2x  1800
 5x  1800
 x  360
14.
Given a  b 
2
 900  b  600
3
Also a  b  180 0 (Linear pair)
www.eabhyasacademy.com
 b  600  b  1800
 2b  1200
 b  600  a  1200
15.
Let one angle be =x. Then
The other angle = 80% of x 
PATHFINDER
4x
5
3
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
Given, x 

4x
 900
5
9x
900  5
 900  x 
 50 0
5
9
 The other angle 
16.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
POR  3ROQ
4  500
 40 0
5
(Given)
(Linear pair)
POR  ROQ  1800
 3ROQ  ROQ  1800
 4ROQ  1800
 ROQ 
1800
 450
4
Now SOQ  POR
 ROQ (Vert. opp. S )
17.
 3  450  1350
Let the smaller angle be x, Then
Larger angle  5x  200
Given x  5x  20 0  100 0
 6x  1200
 x  200
 Larger angle  5  200  200  800
18.
5  1  x
(Corresponding angles)
5  7  1800
(Linear pair)
2014 - 2015
2
5x
x  1800 
 1800
3
3
1800  3
x
 1080
5
 5  1080
x
19.
a  c  d  360 0  b
( S round a pt)
 360 0  850  2750
20.
AFE  BFG  600 and BFC  BFG  FGD  1800
E
60 
B
F
I
www.eabhyasacademy.com
D
G
A
C
 FGD  1800  600  1200
1
 DGI  FGD  600
2
PATHFINDER
4
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
21.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
A  B  C  180 and A  B  C
0
A
 A  B  C  600
 1  300 , 2  300
D
FromBDC, 1  2  D  1800
 D  1800  600  1200
22.
1
B
2
2
1
C
0
0
0
Let the angle be x 0 , then x  2 180  x  or
x  3600  2x
x  2x  3600
3x  3600
x
23.
3600
 1200
3
2x  5x  8x  360 0
 or 15x  3600
3600
 240
15
 x  240
x
24.
Perimeter of right-angled isosceles triangle  2  1
or 2.x


2 1  2 1
or 2x  1
 hypotenuse  1
25.
2x
x
x
ABC is an equilateral le and AD  BC
3
BC
 AD is the medium  AD 
2
2014 - 2015
3 2 3
2
BC   2C.D.
4
4
3
2
2
 AD   4CD
4
2
 AD  3CD2
 AD2 
26.
If the angles are in the ratio 1:2:3
then the angles of the triangle are 300 ,600 and 90 0 .
The triangle becomes a right triangle.
The sides ratio opposite to the angles 300 ,600 ,900 is BC : AB : AC
A
www.eabhyasacademy.com
 AC sin300 : AC sin600 , AC
300
1 3
:
:1
2 2
 1: 3 : 2

600
B
PATHFINDER
C
5
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
27.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Area of equilateral trianlge 
1
 base  altitude
2
1
3a
a
2
2

28.
3a
4
Let AB and CD represent the poles
Draw BE  r to CD. Then
AB = 9m and CD = 14m
Now, BE = AC = 12m and
D
DE   CD  CE 
 14  AB 
 14  9 m
 5m
B
E
A
C
 BD2  BE2  DE2
2
2
 12   5   144  25  169
BD  13m
29.
Given ABCD is a rhombus.
The diagonals AC = 60 cm, BD = 45 cm
Area of rhombus
30.
=
1
1
d1 d2   60  45  1350 cm2
2
2
C  180 0  A  B
A
 1800  800  600  400
800
 x  200
D
InBDC,
DBC  y  x  180
2014 - 2015
y0
0
x0
60 0
x0
C
B
 DBC  30 
0
or y  1800  200  300  1300
1.
C  180 0  A  B
A
www.eabhyasacademy.com
 1800  800  600  400
800
 x  200
InBDC,
DBC  y  x  180
D
y0
0
x0
B
or y  1800  200  300  1300
PATHFINDER
x0
60 0
C
 DBC  30 
0
6
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
2.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
In figure l II ED II EC and transversal AB and AE make intercepts
AD AE

DB EC
ButAB  12cm, AC  16cmandEC  4cm
 AE  16  4  12cm
 Ratio 
AE 12 3


EC
4 1
AD 3
AD  DB 3  1

 

DB 1
DB
1
AD 4 12

 
DB 1 DB
12
DB 
 3cm
4
AD  AB  DB  12  3  9cm
Let AB and CD represent the poles
 Ratio 
3.
Draw BE  r to CD. Then
AB = 9m and CD = 14m
Now, BE = AC = 12m and
D
DE   CD  DE 
 14  AB 
 14  9 m
 5m
B
E
A
C
 BD2  BE2  DE2
2
2
 12   5   144  25  169
BD  13m
4.
If AD is angle bisector then
AB 2


AC 3
5.
AB BD

AC DC
2014 - 2015
 AB : AC  2 : 3  AC : AB  3 : 2
Given,
In ABC, AD is the bisector of A and AB = 4cm, AC = 5.2cm and BD = 3cm.
In ABC , AD is the bisector of A
AB BD

AC DC
4
3

 4DC  3  5.2
5.2 DC
3  5.2
DC 
 3.9cm
4
But,BC  BD  DC  3cm  3.9cm  6.9cm

www.eabhyasacademy.com
6.
In BAC and CDA
ADC  BAC = C  C
 CAD  ABD
 BAC II ADC

BC CA

or AC2  BC.CD  12  3  36  AC  36  6cm
AC CD
PATHFINDER
7
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
7.
0
0
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
(Linear pair)
0
BAC  180  112  68
In ABC


x  1800  680  90 0  1800  1580  220
8.
Angle sum property of a triangle)
By the angle sum property of a triangle we have
R  S  T  1800


 x  5x  50  250  1800
1500
 250
6
 S  5  250  50  1300
 6x  1500  x 
9.
ABC  ACB  ACD  1800
(  AB II CD , co - interior angles are supplementary)
 750  600  ACD  1800
 ACD  1800  1350  450
10.
In ABC
ext. CBD  ACB  CAB
(exterior angle property of a  )
 1350  ACB  550
 ACB  1350  550  800
 n  ACB  800
(Vertically opposite angles)
11.
ext. CBY  CAB  ACB
 1140  45.50  ACB
2014 - 2015
 ACB  1140  45.50  68.50
Now a  ACB  180 0 (Linear pair)
 a  1800  ACB
 1800  68.50  111.50
12.
BCA II DCB  A.A.axiom 
BC AC AB


DC BC BD
B
 BC2  AC  DC  6 
27
 81
2
 BC  81  9
A
 AB  27  3 3
www.eabhyasacademy.com
13.
x  y  1800
6
C
27
2
D
[linear pair]
 450  y  1800
y  1800  450  1350
z  x  450 [vertically opposite angles]
u  y  1350 [vertically opposite angles]
 y  1350 , z  450 ,u  1350
PATHFINDER
8
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
14.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
G is the centroid of ABC
A
E
F
G
B
C
D
 AG 
2
2
2
AD,BG  BE and CG  CF
3
3
3
 AG  BG  CG 
15.
2
 AD  BE  CF 
3
Let ABC be a trianlge right angled at B. Let the shortest side AB be x meters. Then, hypotenuse
 AC    2x  6  m
InABC, AC2  AB2  BC2
2
  2x  6  x 2   2x  4 
2
 4x 2  24x  36  x 2  4x 2  16x  16
 x 2  8x  20  0
A
  x  10  x  2  0
 x  10, x  2
AB  x  10m
C
+6
2x
x
2x+4
B
BC  2x  4  2 10   4  24m
AC  2x  6  2 10   6  26m
16.
Let a man start from O. First, the man goes 18m due east and reaches A. Then, he goes 24m due north
reaching B
2014N- 2015
Now,inABO
OA  18m, AB  24m,
B
2
24m
2
OB2  18    24 
 324  576
OB2  900  OB  900m
 OB  30m
17.
W o 18m A
S
E
Hence, the man is at a distance of 30m from the starting point
Let AB and CD be two poles
AB  12m,CD  7m
Draw a line CE toBD
InAEC, AC2
A
2
www.eabhyasacademy.com
 12  12  7 
5cm
 CE2  AE2
2
 144  25
 169
C
AC  13
D
PATHFINDER
12cm
7cm
7cm
E
B
9
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
18.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
1
Area of triangle   base  height
2
1
  2R   R
2
 R2

19.
  4and    1800

1800
 360
 or   
5
  4  36   1440
20.
B
A
or  4  1800

O
C
If AD is angle bisector then
D
AB BD

AC DC
AB 2

AC 3
 AB : AC  2 : 3

 AC : AB  3 : 2
21.
In figure AD EF BC
AE DF

EB FC
AE DF


2AE FC
DF 1


FC 2
1.5 1

  FC  3cm
FC 2

22.
F
B
2014 - 2015

1
1
BOC  AOC  OP bisects BOC
2
2

1
1
 BOC  AOC  POC  BOC
2
2

www.eabhyasacademy.com
E
D
C
POQ  POC  COQ
1
 1800
2
23.
A

 OQ bisects AOC
1
AOC
2
 90 0
 COQ 
= a right angle
( linear pair axiom)
 PQ R II  LM N and 3 PQ = LM, MN = 9cm
PQ QR
1 QR

 
 QR  3cm
LM MN
3
9
24.
AB  3, AC  5, B  900  BC  4
A
BD2  AB,BC  3.4  12
BD  12
PATHFINDER
D
B
C
10
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
25.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Let one angle of a parallelogram be x 0
Let the other angle of parallelogram be
2x 0
3
Since sum of adjacent angles of a parallelogram is 180 0
So, x 
2x 0
5x
 180 
 1800
3
3
 x  360  30  1080
The smallest angle of the parallelogram =
26.
2
2
x   108  720
3
3
Let AB = BC = x
Given AC  8 2
AB 2  BC2  AC2


2
2x 2  8 2  x  8cm
27.
The adjacent angles of a rhombus are supplementary
 2x  350  x  5  180 0
 3x  1800  300
 x
28.
2100
 700
3
A  B  1800
 3x  10    5x  30   180
8x  20 180
8x 160
x  20
2014 - 2015
0
 A  C   3x  10   500
B  D   5x  30   1300
29.
x  150  3x  250  180 0
 4x  400  1800
 4x  1400  x 
1400
4
x  350
 XOP  x  150  350  150  50 0
www.eabhyasacademy.com
30.
z  1800  1080  350  150  50 0
But, 4x  5x  z  (  corresponding angles are equal)
 9x  z 0  x 
720
 80
9
 q0  5x 0  400 , p0  4x 0  320 (  vertically opposite angles are equal)
 p  q  z  320  400  720  1440
PATHFINDER
11
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
BRAIN TWISTERS
:
1.
x:y=3:7
Total ratio x + y = 10
 x  y  1800
x
y
2.
3.
4.
3
 180  54  z  540
10
7
180  126  a  1260
10
1800 (or) supplementary
From figure
2x  7  3x  13  1800
MND 1800   3x  13
5x  20  180 0
= 180 - 109
x  320
= 710
From figure,

AB AG
2 x

  x4
BC GF
3 6

BC GF 3 6

   y 8
CD FE
4 y
:
5.
In a parallelogram ABCD, we know that
A  C and B  D
 A  C  1150 and
B  D  650 or
2014 - 2015
In ABCD A  B  1800
1150  B  1800
B  1800  1150  650
 C  A  1150 , D  B  650
6.
Exterior angle of a regular polygon =

360
n
360
360

 1200
x
10  x
3
3

1
x 10  x
www.eabhyasacademy.com
3 10  x   3x  x 10  x 
30  3x  3x  10x  x 2
 x 2  10x  30  0
x  5  or  6
PATHFINDER
12
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Interior angle of a regualr polygon = 180 
= 180 
360
n
360
6
= 120
7.
By intercept theorem
ST PQ
x 4

   x 6
TU QR
9 6
Both A and R are true R is the correct explanation of A.
8.
In a parallelogram A, B, C, D. we know that the sum of any two adjacent angles are equall t 180 0
A  B  1800 ; A  D  1800 ; C  D  1800 ; B  C  1800
A
F
9.
G
E
B
C
Let G be the point of their intersection.
Since BE = CF, we have BG 
2
2
BE  CF  CG  BG  CG
3
3
Also, GE = BE - BG = CF - CG = GF
Now in  s BGF and CGE , we have
BG = CG
GF = GE
and BGF  CGE
  BGF   CGE
[Vertically opposite angles]
[By SAS Congruence Asiom]
2014 - 2015
 BF  CE
1
1


 BF  2 AB, CE 2 AC


 AB  AC
10.
 P is the orthocentre of  ABC
 AD  BC , BE  CA and CF  AB
A
F
B
P
E
D
C
Now, in  PBC ,
 AD  BC ,  PD  BC
www.eabhyasacademy.com
 CF  AB ,  BF  CP
Now altitudes BF and PD of  PBC intersect in A.
 A is the orthocenter of  PBC
PATHFINDER
13
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
:
0
11.
3x
60  5
3 
 60  x 
 1000 = obtuse angle
1)  x   30  90 
5
5
3


0
5y
200  5
 5y 
 200  y 
 2400 = reflex angle
2) 
  20  180 
6
5
 6 
0
5z  3z
z z
0
 400  8z  40 15  z  750 = acute angle
3)     40 
15
 3 5
4) 4a  3600  a  900 = right angle
12.
a) Th sum of interior angle of polygon =  2n  4   900
b) The sum of exterior angle of polygon = 3600
c) Each interior angle of regular polygon =
d) Each exterior angle of regular polygon 
13.
 2n  4  900
n
3600
n
Here G is the centroid of the  ABC and it divides each median in the ratio 2 : 1
A
F
B
G
E
C
D
a, b)
AG 2 AG  GD 2  1 AD
1
1
 

 3  GD  AD   7.2 2.4 cm
GD 1
GD
1
GD
3
3
 AG AD  GD  7.2  2.4  4.8 cm
c, d)
2014 - 2015
CG 2 3.4
 
 2  GF  1.7 CM  CF  CG  GF  3.4  1.7  5.1cm
GF 1
GF
BG 2
BG  GE 2  1
 

GE 1
GE
1
:
14.
A : B  4:5
Sum of the ratio = 4 + 5 = 9
But A  B  1800
www.eabhyasacademy.com
 A
4
 1800  800
9
5
B   1800  100
9
C  A  800 , D  B  1000
A  80
15.
B  100
PATHFINDER
14
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
16.
17.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
D  C  100  80  200
In  ABC , we have CAB  ABC  180 0 and ABC  900
 320  900  ACB  180 0  ACB  580
18.
We know that the diagonals of a rectangle are equal and bisect each other
 OA  OB  OBA  OBA  320
ABC  900  OBA  OBC  90 0
320  OBC  90 0
 OBC  580
19.
In a right angle triangle the square of hypotenuse is equal to sum of the squares of opposite side and
adjacent side i.e., AC2  AB 2  BC2
20.
The medians AD, BE and CF pass through G
G is the cventroid of  ABC . Hence it will drive each median AD, BE and Cf in the ratio 2
A
F
E
G
C
B
Now,
BG 2
8
2
 
  GE  4  BE  BG  GE  8  4  12
GE 1
GE 1
21.
GC 2
GC 2
AG 2 AG
 
  GC  10 ;
 
 1  2  1 3
FG 1
5
1
GD 1 GD
22.

AG  GD
AD
13.5
3
 3
 3  GD  4.5
GD
GD
GD
:
2014 - 2015
23.
The number of lines of symmetry for a parallelogram is zero.
24.
In a rectangle ABCD, AC2  AB 2  BC2
D
C
cm
10
b
A 8 cm B
Given AB = 8 cm, AC = 10 cm, BC = b cm  b 2  102  82  36 , breadth = 6 cm
A
F
25.
4
B
www.eabhyasacademy.com
E
G
cm
D
C
BG 2 BG  GE 2  1
 

GE 1
GE
1

BE
 3  BE  3GE  3  4  12cm
GE
 BG  BE  GE 12  4  8cm
PATHFINDER
15
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Let ABCD be the IIgm and let A  C  1500
(  opp. angles are equal
1.
as
A  C

A  C  1500
2A  1500
1500
 750
2
A 
A  C  750

Now A  C  1800
(  adjacent angles aJ’e supplementary)
D
C
A
B
75  D  1800

D  1800  75  1050
Now B  D  1050
(  Opposite angles are euqal)
2.
We have ADB  CBD  650
(  alternate interior angles are equal)
D
2014 - 2015
C
650
750
A
B
But in ADB ,
A  ABD  ADB  1800
(  sum of three angles of a trianlge is 180 0 )
750  ABD  650  1800
ABD  1400  180 0
ABD  400
www.eabhyasacademy.com
3.
CDB  ABD  40 0
(  alternarte interior angles are equal)
Let ABCD be the rhombus whose diagonals AC and BD are of lengths of 8 cm and 6 cm respectively. Let
AC and BD intersect at O. Since the diagonals of rhombus bisect at right angles.
AO 
1
1
AC   8  4cm
2
2
and
BO 
1
1
BD   6  3cm
2
2
In
AOB
PATHFINDER
16
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
By pythagoras theorem,
D
C
O
A
B
AB 2  OA 2  OB 2
 4 2  32
 16  9
 25
AB = 5
 length of each side of the rhombus = 5 cm.
4.
Sum of all exterior angles of a regular polygon = 3600
Each exterior angle = 450
Number of exterior angles =
350
8
45
 Number of sides in the given polygon = 8.
5.
Let the measure of the fourth angle be x 0
We know that the sum of the angles of a quadrilateral is 3600
 54 + 80 + 116 + x = 360
 250  x  360
 x   360  250  110
6.
2014 - 2015
0
0
0
Let the measures of angles of the given quadrilateral be  2x  ,  3x  ,  5x  and  8x 
0
We know that the sum of the angles of a quadrilateral is 3600
 2x  3x  5x  8x  360
 18x  360  x  20
So, the measures of angles of the given quadrilateral are
0
0
 2  20  ,  3  20  ,  5  20 
0
and  8  20 
0
i.e., 400 ,600 ,100 0 and 160 0
7.
Let the measures of each of the equal angles be x 0
We know that the sum of all the angles of a quadrilateral is 3600
www.eabhyasacademy.com

115 + 45 + x + x = 360
 160  2x  360
 2x   360  160   200  x  100
Hence, the measure of each of the equal angles is 100 0
PATHFINDER
17
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
8.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Let the lengths of two sides of the parallelogram be 4x cm and 3x cm respectively.
Then, its perimeter = 2(4x + 3x) cm = 14 x cm
9.
56
4
14

14x  56  x 

one side = (4 x 4) cm = 16 cm and other side = (3 x 4) cm = 12 cm.
Let ABCD be the given rectangle in which length aB = 8 cm and diagonal AC = 10 cm
D
C
10
A
cm
B
8 cm
Since each angle of a rectangle is a right angle, we have ABC  900
From the right  ABC , we have
AB 2  BC2  AC2
[Pythagoras’ theorem]



2
 BC2  AC2  AB2  10    8
10.
2
 cm  100  64 cm  36cm
2
2
2
 BC  36 cm  6cm
Hence, breadth = 6 cm
In s AOQ and BOP, we have
P
O
A
B
Q
2014 - 2015
OAQ  OBP
(Each equal to 90 0 )
AOQ  POB
 By AA - similarity,,
(Vertically opp. s )
 AOQ~  BOP

AO OQ AQ


BO OP
BP

AO AQ 10 AQ
10  9



 AQ 
 15cm
BO BP
6
9
6
A
B
O
D
11.
M
C
www.eabhyasacademy.com
N
From Similar s ABN and DMN
AB AN BN
2 AN


 
DM DN MN 1 DN
....
(1)
....
(2)
From similar s AOB and COM
AB AO OB
2 AO OB


 

MC OC OM
1 OC OM
PATHFINDER
18
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Again from similar s AON and BOC
AO ON ON


 2  ON:OB  2:1
OC OB
OB
12.
By the converse of basic proportionality theorem
if
CD CE

, then DE II AB
DA EB

x3
x

8x  9 3x  4
  3x  4  x  2  x  8x  9
 3x 2  13x  12  8x 2  9x
 5x 2  4x  12  0
 5x 2  10x  6x  12  0
 5x  x  2  6  x  2  0
  5x  6  x  2  0
x
13.
6
or 2
5
Equilateral triangles are similar trianlges.
In similar triangles, the ratio of their corresponding sides is the same as the ratio of their medians.
 The ratio of the sides is 3 : 2
14.
Let the shortest side of the triangle be x m.
Then hypotenuse = (2x + 6)m
2014 - 2015
Third side = (2x + 6) - 2 = (2x + 4) m
By Pythagora’s Theorem.
2
 2x  6    2x  4 
2
 x2
 4x 2  24x  36  4x 2  16x  16  x 2
 x 2  8x  20  0
 x 2  10x  2x  20  0
  x  10  x  2  0
 x 10 or  2
 x cannot be negative, x = 10
www.eabhyasacademy.com
 Hypotenuse = (2 x 10 + 6)m = 26 m
15.
3600
Exterior angle =
number of sides
=
PATHFINDER
3600
30
 51
7
7
19
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
16.
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Interior angle of a regular polygon of ‘n’ sides =
Exterior angle of a regular polygon of ‘n’ sides =
Given,
n  2  1800
n
 n  2  8 
17.
8
n  2  1800
n
360 0
n
3600
n
3600
 16  n  18
1800
Let the angle of the quadrilateral be 3x, 7x, 6x and 4x
Then, 3x+7x+6x+4x = 3600
 20x  3600  x  180
 The angles A, B, C and D are respectively 3  180 , 7  180 , 6  180 and 4  180 , i.e., 540 , 1260 , 1080
and 720
B
C
0
126
108
720
540
D
A
18.
0
Which shows that A  B  1800 and C  D  1800  Ad II BC
Hence ABCD si a trapezium.
Diagonals of a rectangles are equal and bisect each other.
0
 In AOB , AOB  162  3x
(vert, opp s )
OBA  OAB  2x (OA  OB )

AOB  OBA  OAB  180 0
 1620  3x  2x  2x  180 0
2014 - 2015
 x  180
19.
PSR  PQR  680
(opp, s of a IIgm are equal)
0
0
PTS  180  139  410
(PTQ is a straight line)
 RST  PTS  410 (SR OO PQ alt. s are equal)
20.
0
0
0
 y  PSR  RST  68  41  27
 Opp s of a rhombus are equal
2x  150  3x  300  x  450
In ABCD , CD = CB  CBD  BDC
Also, DCB  3  450  300  1050
1800  1050 750

 37.50
2
2
In a kite, the two pairs of adjacent sides are equal,
i.e., PS = PQ and sR = RQ
In PSQ
www.eabhyasacademy.com
 BDC 
21.
PSQ  PQS 
PATHFINDER
1800  700 1100

 550
2
2
20
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
 QSR  90.50  35.50
 RQS  QSR  35.50
(  RQ = RS)
In SRQ ,
R  RQS  QSR  1800
 R  35.50  35.50  1800
 R  1800  710  1090
22.
In AOB
D
C
O
480
A
B
BO = OA (Diagonals of a rectangle are equal and bisects each other)
 OAB  OBA  480
(isos,  property)
 DBA  OBA  480
 DBC  CBA  DBA
= 900  480  420
23.
DAB  1800  ADC  180 0  640  1160
 AB II DC, co  int, s are supp 
DAB, DA  AB  ABD  ADB  isos.  prop 
In
 ADB 
1800  1160 640

 320
2
2
 BDC  640  320  320
2014 - 2015
Hence, in DBC
0
DBC  180   BDC  BCD
= 180 0   320  540   180 0  860  940
24.
In BAD
BDA  BAD  57 0
In BDC
BCD  BDC  660
 D  570  660 1230 and
A  D  57 0  1230  1800
Also, D  C  1230  660  1890
Hence, by the property that co-int, angles are supplementary  lines are parallel, we have
AB II DC and AD not II BC. i.e., ABCD is a trapezium
www.eabhyasacademy.com
25.
We have, SPQ  600
S
P
PATHFINDER
A
R
Q
21
MATHEMATICS-CHAPTERSOLUTIONS
File No.25/22/15/12/2014
P  Q  1800
VIII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
(adj, s of a IIgm are supp.)
 Q  1800  600  1200
Now PQ II SR and AP is the transversal

SAP  APQ  300
(  AP bisects SPQ )
SAP SAP  SA  SP
(isos,  property)
....
(1)
Also, RAQ  AQP  600
(PQ II SR, AQ is transvrsal, alt s and AQP 
1
PQR  600 )
2
 ROQ  RAQ (AQ bisects PQR )
 RA = RQ
(isos,  property
....
(2)
 From equationo (1) and (2)
AS = AR
(  SP = RQ, sides of a llgm)
Also, in  ARQ, ARQ  600
  ARQ is equilateral
 AR  RQ  AR  SP
 The incorrect statement is AQ = PQ
NOTE :-
Level - I - Question No.4 - Entered the Wrong Answer : Key 3
2014 - 2015
But Correct Answer
: Key 2
Levle - II - Entered the Wrong Question No.25
But Correct Question is
25) If an angle of a parallelogram is two - third of its adjacent angle, the smallest angle of the
parallelogram is
2) 720
3) 720
4) 1080
www.eabhyasacademy.com
1) 540
PATHFINDER
22
MATHEMATICS-CHAPTERSOLUTIONS