Annexure - ‘ E ’ SYLLABUS/CURRICULUM MATHEMATICS (041) S.A.-II (2012-13) CLASS-X Second Term Marks : 90

UNITS MARKS

II. ALGEBRA (Contd.) 23

III. GEOMETRY (Contd.) 17

IV. TRIGONOMETRY (Contd.) 08

V. PROBABILITY 08

VI. COORDINATE GEOMETRY 11

VII. MENSU RATION 23

TOTAL 90

The Question Paper will include value based question(s) to the extent of 3-5 marks.

UNIT II : ALGEBRA (Contd.)

3. QUADRATIC EQUATIONS (15) Periods

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solution of the quadratic equations (only real roots) by factorization, by completing the square and by using quadratic formula. Relationship between discriminant and nature of roots.

Problems related to day to day activities to be incorporated.

4. ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms and their application in solving daily life problems.

UNIT III : GEOMETRY (Contd.)

2. CIRCLES (8) Periods

Tangents to a circle motivated by chords drawn from points coming closer and closer to the point.

1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.

2. (Prove) The lengths of tangents drawn from an external point to circle are equal.

3. CONSTRUCTIONS (8) Periods

1. Division of a line segment in a given ratio (internally)

2. Tangent to a circle from a point outside it.

3. Construction of a triangle similar to a given triangle.

UNIT IV : TRIGONOMETRY

3. HEIGHTS AND DISTANCES (8) Periods

Simple and believable problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30o, 45o, 60o.

UNIT V : STATISTICS AND PROBABILITY

2. PROBABILITY (10) Periods

Classical definition of probability. Connection with probability as given in Class IX. Simple problems on single events, not using set notation.

UNIT VI : COORDINATE GEOMETRY

1. LINES (In two-dimensions) (14) Periods

Review the concepts of coordinate geometry done earlier including graphs of linear equations. Awareness of geometrical representation of quadratic polynomials. Distance between two points and section formula (internal). Area of a triangle.

UNIT VII : MENSURATION

1. AREAS RELATED TO CIRCLES (12) Periods

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60o, 90o & 120o only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.)

2. SURFACE AREAS AND VOLUMES (12) Periods

(i) Problems on finding surface areas and volumes of combinations of any two of the following: cubes,

cuboids, spheres, hemispheres and right circular cylinders/cones. Frustum of a cone.

(ii) Problems involving converting one type of metallic solid into another and other mixed problems. (Problems

with combination of not more than two different solids be taken.)

RECOMMENDED BOOKS

1. Mathematics - Textbook for class IX - NCERT Publication

2. Mathematics - Textbook for class X - NCERT Publication

3. Guidelines for Mathematics laboratory in schools, class IX - CBSE Publication

4. Guidelines for Mathematics laboratory in schools, class X - CBSE Publication

5. A hand book for designing mathematics laboratory in schools - NCERT Publication

6. Laboratory manual - Mathematics, secondary stage - NCERT Publication.

Design of Sample Question Paper Mathematics (047) Summative Assessment-II Class X- (2013)

Type of Question Marks per question Total no. of Questions Total Marks

M.C.Q 1 8 8 SA-I 2 6 12 SA-II 3 10 30 LA 4 10 40 TOTAL 34 90

The Question Paper will include value based question(s) to the extent of 3-5 marks

Weightage

S.No. Unit No. Topic Weightage

1 II Algebra (contd.)

[Quadratic Equations A.P.]

2 III Geometry (contd.) [Circles, Constructions] 17

3 IV Trigonometry (contd.)

[Height and Distances]

4 V Probability 08

5 VI Coordinate Geometry 11

6 VII Mensuration 23

Total 90

MCQ- 1 Mark

Q.1 The tenth term of an A.P. -1.0, -1.5, -2.0, ............... is

(a) 3.5 (b) 5.5

Q.2 A sphere and a cone of height ‘h’ have the same radius and same volume, then r: h is

(a) 4:1 (b) 1:4

SA-II - 2 Marks

Q.3

9 K.M.

6.5 K.M.

12 K.M.

A helicopter has to make an emergency landing as shown in the figure. What is the probability of a safe landing?

Q.4.

In the figure, quadrilateral ABCD circumscribes the circle. Find the length of the side CD.

SAMPLE QUESTIONS MATHEMATICS (047) S.A.-II (2012-13) CLASS-X

JUNGLE

2 K.M.

SA-III - 3 Marks

Q.5 The sum of first, third and seventeen term of an A.P. is 216. Find the sum of the first

13 terms of the A.P.

Q.6 The point R (p-3q, q) divides the line segment joining the points A (3, 5) and B (6, 8)

in the ratio 2:1. Find the co-ordinates of R.

Q.7 Draw a triangle ABC with side BC= 7cm, AB= 6cm and ∠ABC = 60°. Construct a triangle whose sides are of the corresponding sides of ∆ ABC. Also write steps of construction.

LA - 4 Marks

Q.8 A cone of height 3.25 cm is surmounted by a hemisphere having same base. If the

diameter of the base is 3.5 cm, then find the curved surface area. (Take π = )

Q.9 If the sum of the roots of the equation

Kx2 - 2 + 1 = 0

is , then find the roots of the equation.

Q.10 A D

100 m 80 m

B E C

In ∆ DCE

tan θ – tan (90 - θ) = 0

Also AE = 100 m and DC = 80 m.

Find BC.

600

θ 100

ANSWER KEY

1. (c) 2. (b)

3. Length of a Jungle = (12 – 9) km = 3 km

Breadth of a Jungle = (6.5 –2) km = 4.5 km

Area of Jungle = 13.5 sq. km

Area of Total field = 12 x 6.5 = 78 sq. km

P (Safe landing) =

= (1)

4. AE = AH (length of tangents from external points are equal)

x = 4 – x 2x = 4 x = 2 DH = (5 – 2) = 3 cm DH = DG = 3 cm (1) CF = CG

2y – 3 = y

y = 3

DC = DG + GC = 3 + 3 = 6 cm

5. a + a + 2d + a + 16d = 216 (1)

3a + 18d = 216

a + 6d = 72

= 2a + (13 – 1) d

= 2a + 12d

101

= x 2 a + 6d

= 13 x 72 = 936 (1)

6. p – 3q =

p – 3q =

p – 3q = 5 (1) (1 mark)

Also q =

q =

= =

7 (2) (1 mark)

Substituting (2) in (1) we get

p – 3 x 7 = 5 p = 26 (1 mark)

7. correct construction (2 marks)

Steps of construction (1 mark)

Curved surface area of the hemisphere = (4πr2)

= (2 x x x ) cm

(1 mark)

102

Slant height of the cone (l) = =

= 3.7 cm (approx.) (1 mark)

CSA of cone = πrl = x x 3.7 cm

(1 mark)

Total Curved Surface Area = x (3.5 + 3.7) cm

= x 7.2 = 39.6 cm2 (approx.) (1 mark)

9. Sum of the roots =

K = 2 (1)

Now the quadratic equation is

- 2 x + 1 = 0

D = b2 - 4ac = 8 - 8 = 0 (1)

Roots are real and equal

x =

= =

Roots are , (2)

10. tan θ = tan (90 - θ)

tan θ = cot θ → θ = 45° (1)

In ∆ DCE

tan 45° =

103

EC = 80 m (1)

In ∆ ABE

= cos 60°

BE = 50 (1)

BC = BE + EC

= 50 + 80

= 130 m (1)

104