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Answer key prepared by: Prathap S M( 974 665 3636) HST Maths, GHSS Puthoor, Kollam (Dt)
Qn Sub Qn Description Step
Marks
Score
1 a 45 1
2 b 3√2 1
2 a 4 (Numbers 2,4,6,8) 1
2 b
Probability of a prime number =
1
4
1
3 (b) (0, 5) is a point on the y axis
(c) Distance from (0,5) to the x axis is 5
2
4 Sum of first 25 odd numbers = 625 2
5 a AB = 4cm. Hence BC = 3cm (Using Pythagoras) 2
3 b Cos C = 3/5 1
6 a Draw the rectangle , mark the points left and top right. 1 3
b (7,5) and (3,8) 2
7 3
8 a Base edge( a ) = 10 cm 1
3 b l=√e
2
−(a/2)
2
=√132
−(5)
2
=√144=12 cm
2
9 a Area of the plane surface = 40/4 = 10cm^2 1
3 b Total surface area of hemisphere = 3 x 10 = 30 cm^2 1
c Ratio between the surface area = 30 :40 = 3:4 1
10 a PB = PB +BC = 9 cm 1
3 b PQ x PR = PB x PC = 4 x 9 = 36 1
c PA=√36=6 cm 1
11 a
Volume=1
3
π r
2
h=324 π cm
3 1
3
b
Volume of sphere with radius 1 cm
4
3
π(1)
3
=4
3
π cm
3
Number of spheres =
324 π
4π
3
=243
1
1
12 Rough Figure 1
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SECOND TERM EXAM 2019
MATHEMATICS
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Answer key prepared by: Prathap S M( 974 665 3636) HST Maths, GHSS Puthoor, Kollam (Dt)
h
x
=tan54 h = x tan 54 .......................(1)
h
(x+20)
=tan 27 h = (x+20) tan 27......................(2)
From (1) and (2) x= 20 tan 27
(tan50 – tan 27)
=11.72 m
h = 11.72 ×1.38 = 16.17 m
OR
h = 20 x 0.8 = 16 m
1
1
1
4
13 a 28,24,20.... 1
4
b X8=28+7(−4)=0 1
c S15=15∗x8=15∗0=0 1
d S15=0 1
14 a <POS = 90 1
b
OP =
6
tan 35=60
7
Area of Rhombus =
1
2
d1 x d2 = = 1
2
x12 x
120
7
= 720
7 sq cm =102.85 sq cm
1
1
1
1
4
15 a 100 m 1
4 b 252
−x
2
=525
x= 10
length = 35 m, breadth = 15 m
1
1
1
16 4
17 a Base perimeter = 2π r=60 π 1
4
b l=√h2
+(r)
2
=√402
+302
=√2500=50 cm
1
2
Answer Key_Maths_10th_Second Term Examination_December_2019 2 of 4
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Answer key prepared by: Prathap S M( 974 665 3636) HST Maths, GHSS Puthoor, Kollam (Dt)
Curved surface area = 1/2x 60π x50=1500π cm
2
18 a AC = 3 cm ( Perpendicular bisects the chord) 1
4 b In a triangle with angles 30, 60 90, the sides opposite to them are in the ratio
1: √3 :2 , Radius = 2√3 cm
2
c Radius = 6 cm 1
19 a AB =BC = CD = AD = 5 units 3
4 b Rhombus 1
20 a
[
a
2 ]
2 1
4
b
e=√l
2
+
a
2
2 2
c l=√17cm 1
21 a <BPC = 40 1
4 b <APC =90+40 = 130 1
c < C =10 1
d <A = x, <APC = 90+ x, <C = 180 – (x+ 90+x) = 90-2x 1
22 a Slant height = 36 cm 1
5 b Ratio of central angles = 120 : 240 = 1: 2 1
c
r1=120
360
x 36=12 , r2=240
360
x 36=24
2
d Ratio of radii = 12: 24 = 1: 2 1
23 a <Q = 180 – 110 = 70 1 5
b Central angles 110, 120,130
Construction of circle, and tangents, triangle
4
24 a 1+ 2+ 3+.... +10 = 55 1 5
b 1+ 2+ 3+ ..... + n = 231
n(n+1) = 462
product of two consecutive natural number is 462
n= 21
4
25 a PS = 1/2 x 8 = 4 cm 1
5 b SR = 4 √3 , In triangle PSQ , PS = SQ = 4 cm
QR = 4+ 4 √3 ,PQ = 4 √2
1
1
c Draw the triangle 2
Answer Key_Maths_10th_Second Term Examination_December_2019 3 of 4
Page 4 of 4
Answer key prepared by: Prathap S M( 974 665 3636) HST Maths, GHSS Puthoor, Kollam (Dt)
26 a 2
5
b AC = 4 √2 1
c (6,6) (2,2) 2
27 a AC = 10 cm 1
5 b Area = 1/2 x 6 x 8 = 24 sq cm 1
c Semi perimeter = 24/2 = 12
radius = 24/12 = 2 cm
1
2
28 a Radius = 2 cm 1
5 b PD=√3 x 1=√3 1
c Draw the given figure,
Draw an equilateral triangle with one side CD
3
29 a 0 1
6
b 90 1
c 1 1
d 1 1
e 45 1
f 1 1
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