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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
I)
TOTAL:
SUBJECT: MATHEMATICS
50
1x7=7
Choose the best alternative from following options
1) If Tn = 2n – 1, then S2 =
(A) 1
(B)
3
(C)
(D) 5
2) The number of diagonals to be drawn in a hexagon
(A) 5
(B) 9
(C) 7
(D) 6
3) The subset of sample space is called
(A)
Trial
(B) Event
(C) Sample space
(D) Sample point
4) The formula to find Standard Deviation by step deviation method
(A)
∑
−
∑
(B)
∑
−
∑
xC (C)
∑
−
∑
(D)
∑
−
∑
5) The formula to find the distance between two points in a plane d =
(A) ( − ) + ( − )
(B) ( − ) − ( − )
( ) ( − ) +( − )
(D) ( − ) − ( − ) ,
6) If √3 tan = 1 ,then =
(A)
(B)
45
(C)
90
(D) 0
7) The degree of the polynomial x2 - 9x + 20 is
(A) 1
(B) 3
(C) -1
(D) 2
II. Answer the following
1x5=5
8) What is composite number?
A number greater than 1 and not a prime number is a composite number.
9) Draw Venn diagram to illustrate A U (B ∩ C).
10) State Thales Theorem.
If a straight line is drawn parallel to a side of a triangle,then it divides the
other two sides proportionally.
11) What are the angles in the Major segment?
Acute angle
12) What is the formula to find the Volume of a solid hemisphere?
V=
r3
III. Answer the following
2x 10 = 20
13) Verify Euler’s formula for the given graph
14)
15)
16)
17)
N = 7, R = 5, A = 10
N+R = 7+ 5 = 12
A+2 = 10 + 2 = 12
∴ N+R = A+2
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14) Sketch out the field to the following notes from the field book
(1 cm = 50m.)
To E
350
100To D
300
75To C
250
150
50 To B
150 to F
100 To G
50
From A
1cm = 50m
50m =1cm, 150m = 3cm, 250m = 5cm, 300m = 6cm, 350m = 7cm, 100m = 2cm,75m=1.5cm
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15) In a circle of radius 4.5cm draw two radii such that the angle between them is700
Construct tangents at the non-centre ends of the radii.
18) Rationalize the denominator and simplify
=
=
=
√
√
√
√
√
=
√
√
√
√
√
√
√
√
√
√
√
√
=
=
√
x
√
− √
16) Find the product of √3 x √2
√3 x √2
√3
= 3 = 3
= √3
√2
= 2 = 2
= √2
√3 x2
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√81x8
√648
17) If U = { 1, 2, 3, 4, 5, }, A = { 2, 4}, B = {3, 4, 5 } Prove that (AUB)1 = A1 ∩ B1
U = { 1, 2, 3, 4, 5, }
A = { 2, 4}
B = {3, 4, 5 }
LHS =(AUB)1
=[{ 2, 4}U{3, 4, 5 }]1
18) How many 4 digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 without
repeating any digit.
19) The number o
Thousands Hundreds
Tens
ones
6
5
4
3
P1
P1
P1
P1
6
5
4
3
=6x5x4x3 = 30
20) Solve by using the formula x2 - 7x + 12 = 0
a = 1, b = -7 , c = 12
(
=
)± (
)
±√
=
=
±√
=
± √
= ± √−
21) Write all Trigonometric ratios apply to a
sin
=
cos
=
=
cosec
=
sec
=
cot
=
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22) Prove that the line joining the points (4, 5) and (0,-2) is perpendicular to the line
passing through (2,-3) and (-5, 1).
−
Slope
= 2− 1
2
1
−2− 5
Slope of first line - m1 =
0−4
−7
m1 =
−4
m1 =
Slope of second line
m2 =
1−(−3)
−5−2
1+3
m2 =
−7
m2 =
m1xm2 = x
m1xm2 =−1
Lines are perpendicular
∴
IV. Answer the following
23) Calculate the standard deviation of the following data
X
f
0-10
7
10-20
10
20-30
15
3 x 2= 6
30-40
8
40-50
10
CI
X
f
d=x-25
fd
d2
fd2
0-10
10-20
20-30
30-40
40-50
5
15
25
35
45
7
10
15
8
10
-20
-10
0
10
20
-140
-100
0
80
200
400
100
0
100
400
2800
1000
0
800
4000
50
=
∑
−
=
−
=
– .
= √
.
=
40
8600
∑
.
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24) Prove that the tangents drawn from an external point to a circle
(a) Are equal
(b) Subtend equal angles at the centre
(c) Are equal inclined to the line joining the centre and the external point.
Given:A is the centre.B is an external point.. BP and BQ are the tangents.AP,
AQ and AB are joined
To prove : (a).
BP = BQ
(b). ∠PAB = ∠QAB
(c). ∠PBA = ∠QBA
Proof: In ∆APB and ∆AQB,
[ ∵ Radius of the same circle
AP = AQ
[ ∵ Radius drawn at the point of contact is
∠APB = ∠AQB =900
perpendicular to the tangent
AB =
∴
∴
∆APB ≡ ∆AQB
(a).
AB
[ ∵ RHS postulates
BP = BQ
(b). ∠PAB = ∠QAB
[ ∵ CPCT
(c). ∠PBA = ∠QBA
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V. Answer the following
4 x 3= 12
25) Draw direct common tangents to two circles of radii 5cm and 2.5cm having their
Centers 10cm apart and measure their lengths.
26) Draw the graph of y = x2 + x – 6
x
y
-4
6
-3
0
-2
-4
-1
-6
0
-6
1
-4
2
0
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3
6
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27) Prove that “If the square on the longest side of a triangle is equal to the sum of the
Squares on the other two sides then those two sides contain a right angle”
In triangle,If a square of a side is equal to the sum of the squares of the other two
sides,then it will be a reight angled triangle.
Given: In the ∆ABC , AB2+ BC2 = AC2
To prove : ∠ABC = 900
Construction : At B draw AB⟘BC , extend BC to D such that DB = BC
Join ‘A’ and ‘D’.
Proof: ∆ABD ಯ ∠ABC = 900 [ ∵ Construction
∴ AD2 = AB2 + BC2 [∵Phythagoras theorem
But In ∆ABC ,
AC2 = AB2 + BC2 [ ∵ Given
⇒
AD2 = AC2
∴ AD = AC
In ∆ABD and ∆ABC,
AD = AC
[ ∵ Proved
BD = BC
[ ∵ Construction
AB = AB
[ ∵ Common
∆ABD ≡ ∆ABC [ ∵ SSS Axiom
⇒ ∠ABD = ∠ABC
But, ∠ABD +∠ABC =1800 [ ∵ BDC is straight line
⇒ ∠ABD = ∠ABC = 900
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