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BELTHANGADY TALUK MATHS TEACHERS WORKSHOP
MINIMUM STUDY LEVEL QUESTION PAPER
Time: 2hour
TOTAL:
SUBJECT: MATHEMATICS
50
1x7=7
I . Choose the best alternative from following options
1) If x, A, y are in H.P., then A = --(A)
(B)
(C)
(D)
2) The value of 250P0 is ---(A) 250
(B)
0
(C) 1
(D) 250!
3) The probability of sure event --(A)
(B)
0
(C)
(D)
4) The formula to find coefficient of variation.
(A)
(B)
5) If sin
=
(A)
and cos
x100
=
(B)
∑
(C)
,then tan
(C)
(D)
= ----(D)
6) Find the distance between the points (2, 8) and (6, 8).
(A)
(B)
1
(C)
2
7) if x = 1 ,then the value of g(x) = 7x2 + 2x + 14 is
(A) 14
(B) 10
(C) 23
(D) -23
(D)
−1
II. Answer the following
1x5=5
8) Write symbolically ‘Intersection of sets is distributive over union of sets”
A∩(BUC) = (A∩B)U(A∩C)
9) What is the distance between the centers of circles of radius R and r’ touch each other?
externally?
d =R+r
10) State Thales Theorem.
Thales theorem :If a straight line is drawn parallel to a side of a
triangle, then it divides the other two sides proportionally.
11) Write a formula to find the total surface area of a cylinder.
Total Surface area of a Cylinder = 2 r(r + h)
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12) The HCF of A and B is H and LCM is L, write the relation between them.
HxL=AxB
III. Answer the following
2x 10 = 20
12) Construct a pair of tangents to a circle of radius 3.5cm from a point 3.5cm away from
the circle.
13) Rationalize the denominator and simplify
√
√
=
=
=
=
=
√
√
x
√
√
(
√ ) √
√
√
√
√
√
√
√
√
√
√
14. Find the product. (6√2 -7√3 )x (6√2 -7√3 )
= 6√2 − 7√3
= 6√2 + 7√3 - 2 6√2 7√3
= 36x2 + 49x3 - 84√6
= 72 + 147 -84√6 = 219 - 84√
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12) If A = {1, 2, 3 }, B= {2, 3, 4, 5,} ,C = {2,4,5,6 } are the subsets of
U = {0,1,2,3,4,5,6,7,8,9} then Verify (AUB)1 = A1 ∩B1 .
U = {0,1,2,3,4,5,6,7,8,9}
A = {1, 2, 3 }, B= {2, 3, 4, 5,} ,C = {2,4,5,6 }
LHS (AUB)1 ={ 1, 2, 3, 4, 5}1 = {0, 6, 7, 8, 9 }
RHS = A1∩B1 ={0, 4, 5, 6, 7, 8, 9}∩{0, 1, 6, 7, 8, 9 }
= { 0, 6,7, 8, 9, }
LHS = RHS
∴(AUB)1 = A1 ∩B1 .
=
13) If
, then find the value of n’.
= 5. − 1
⇒ 3.n(n-1)(n-2)(n-3) = 5.(n-1)(n-2)(n-3)(n-4)
3.n = 5(n-4)
3.n = 5n - 20
2n = 20
3.
n = 10
17) If 6
= 360 and 6 = 15, find the value of r
r! =
6
6
360
15
r! =
r! = 24
r! = 4!
r=
18) Find the value of
+
45+
⇒
⇒
√
+
+
-
.
.
45 -
√
-
60 .
- √3
⇒ + -3
⇒ 1-3
⇒ −
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19) Find the distance between the points (-4, 5) and (-12, 3).
20)The distance between two points d = ( − ) + (
d = [−12 − (−4)] + (3 − 5)
d = [−12 + 4] + (−2)
d = [−8] + (−2)
d = √64 + 4
d = √68
d= .
−
)
units
21) Verify Euler’s formula for the given graph
N = 4, R = 5, A = 7
N+R = 4+ 5 = 9
A+2 = 7 + 2 = 9
∴ N+R = A+2
22) Sketch out the field to the following notes from the field book
(1 cm = 30m.)
To D
120To C
300
210
150
90
135To B
From A
1cm = 30m
⇒ 90m = 3cm, 120m = 4, 135m = 4.5cm, 150m = 5cm, 210m = 7cm ,300m = 10cm
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IV. Answer the following
23) Calculate the standard deviation of the following data
X
10
20
30
40
f
8
12
20
10
3 x 2= 6
50
7
X
f
d=x-30
fd
d2
fd2
10
20
30
40
50
60
8
12
20
10
7
3
-20
-10
0
10
20
30
-160
-120
0
100
140
90
400
100
0
100
400
900
3200
1200
0
1000
2800
2700
60
3
60
50
10900
12) 24) prove that “If two circles touch each other externally, the centers and the point
of contact is collinear”
If two circles touch each other externally, thecentres and the point of
contact are collinear.
Given:A and B are the centres of touching circles. P is the point of contact.
To prove : A,P,and B are collinear.
Construction: Draw the tangent XPY.
Proof:In the figure
∠APX = 900……………..(1) ∵Radius drawn at the point
of contact is
∠BPX = 900 ………… ..(2) perpendicular to the tangent
∠APX + ∠BPX = 900 +900 [ by adding (1) and (2)
∠APB = 1800
[ APB is a straight line
∴ APB is a straight line
∴ A, P andB are collinear.
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V).Answer the following
4 x 3= 12
25) Draw direct common tangents to two circles of radii 4.2cm and 2.2cm having their
centers 8cm apart.
26) solve the quadratic equation x2 + 5 x+ 6 = 0 graphically.
x
0
-1
-2
-3
-4
y
6
2
0
0
2
-5
6
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27) Prove that “In a right angled triangle, the perpendicular to the hypotenuse from the
Right angled vertex divides the original triangle into two right angled triangles, each
of which is similar to the original triangle”
.
Given: In ∆ABC ( i) ∠ABC = 900 (ii) BD⟘AC
To Prove: (i) . ∆ADB ~ ∆ABC
(ii). ∆BDC ~ ∆ABC
(iii). ∆ADB ~ ∆BDC
Proof: ∆ADB ಮತು ∆ABC ಗಳ ,
(i). ∠ADB = ∠ABC = 900
[ ∵ Given
(ii). ∠BAD = ∠CAD
[ ∵ Same angle
(iii). ∠ABD = ∠ACB
[ ∵ Third angle of The ∆
∴ ∆ADB ~ ∆ABC ……………(1)
[∵ Equiangle ∆s
In ∆BDC and ∆ABC ,
(i).∠BDC = ∠ABC = 900
[∵
(ii). ∠BCD = ∠ACB
[ ∵ Same angle
(iii). ∠DBC = ∠BAC
[ ∵ Third angle of The ∆
∴ ∆BDC ~ ∆ABC …………..(2)
Given
[∵Equiangle ∆s
From (1) and (2)
∆ADB ~∆BDC
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