Practice Paper – TERM II (2021 – 2022)
Class – IX
Mathematics (041)
Time: 2 hours Maximum Marks: 40
General Instructions:
1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
3. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
4. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study-based questions.
5. Use of calculators is not permitted.
SECTION – A
1. The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Surface Area = 4πr²
= 4 x 22/7 x 3.5 x 3.5 = 154 m2
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
OR
In fig ∠ABC = 690, ∠ACB = 310, Find ∠BDC
In ΔABC
∠ABC + ∠BAC + ∠BCA = 1800
690 + ∠BAC + 310 = 1800
∠BAC = 800
∠BAC = ∠BDC (angle on the same segment)
So ∠BDC = 800
3. Diagonal AC of a parallelogram ABCD bisects ∠A. Show that it bisects ∠C also.
For proving ΔADC & ΔABC to be congruent
∠ACD = ∠ACB (by CPCT)
4. If x2 – x – 42 = (x + k)(x + 6), then the value of k is
x2 – 7x +6x – 42 = 0
x(x – 7) + 6(x – 7) = 0
(x + 6)(x – 7) = 0
On comparison k = -7
OR
Find the value of x + y + z, if x2 + y2 + z2 = 18 and xy + yz + zx = 9
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= x2 + y2 + z2 + 2(xy + yz + zx)
= 18 + 2 x 9
= 36
(x + y + z) = √36 = 6
5. ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 550 and ∠BAC = 450, find ∠BCD
∠BCA = ∠BDC = 45 (Angles in the same segment)
In ΔBCD
∠DBC + ∠BDC + BCD = 180 (Angle sum property of a triangle)
55 + ∠BDC + 45 = 180
∠BDC =80
6. Construct an equilateral triangle, given its side and justify the construction.
Draw equilateral triangle Justify the construction
SECTION B
7. Find the possible expressions for length and breadth of the following rectangle in which area is given
Area: 25a2 – 35a + 12
25a2 – 20a – 15a + 12
5a(5a – 4) – 3(5a – 4)
(5a – 3)(5a – 4)
l = (5a – 3) b = (5a – 4)
8. Construct a triangle ABC in which BC = 7 cm, ∠B = 750 and AB + AC = 13 cm.
For constructing base BC = 7cm and ∠B = 75
AB + AC = 13 cm and join CX
And complete triangle construction
OR
Construct a triangle PQR in which QR = 6 cm, ∠Q = 600 and PR – PQ = 2 cm
9. Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following case
P(x) = x3 – 3x2 + 3x + 1, g(x) = x + 2
x + 2 = 0, x = -2
Substituting value of x = -2 in eq
P(-2) = (-2)3 + 3 x (-2)2 + 3 x (-2) + 1
= -8 + 12 – 6 + 1
= -1
No, it is not a factor.
10. A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹ 5 per 100 cm2
Radius of the Dome = (17.6 x 7)/(2 x 22) = 2.8 m
C.S.A of Dome = 2πr²
= 2 x 22/7 x 2.8 x 2.8 = 49.28 m2
Cost of painting 100 cm2 = Rs 5
1m2 = Rs 500
49.28m2 = Rs 500 x 49.28
Rs 24640
SECTION C
11. Factorise: 2x2 + y2 + 8z2 – 2√2 xy + 4√2yz – 8xz
OR
Without actually calculating the cubes, find the value of:
(28)3 + (-15)3 + (-13)3
Let a = 28, b = -15, c = -13
a + b + c = 28 – 15 – 13 = 0
a3 + b3 + c3 = 3abc
(28) 3 + (-15) 3 + (-13) 3 = 3 x 28 x (-15) x (-13)
= 16380
12. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
To prove AECF is a parallelogram
In ΔABP
AE = EB & EQ || AP
PQ = QB (By converse of mid-point theorem) ….. (1)
In ΔDQC
DF = FC & PF || QC
DP = QB (By converse of mid-point theorem) …..(2)
From (1) and (2)
PQ = QB = DP
Hence, line segments AF and EF trisect the diagonal BD
13. Four friends Rahul, Arun, Ajay and Vijay went for a picnic at a hill station. They decided to make a conical tent at a park. They were carrying 200 m2 cloth with them. They made the conical tent with height of 8 m and a diameter of 12 m. The remaining cloth was used for floor.
a) Find the volume of air in the tent?
r = 6 m, h = 8 m
Volume of air in tent = 1/3πr²h= (22 x 6 x 8)/(7 x 3) = 301.7 m3 approx
b) Find the cloth left for the floor after making the tent?
l = √ (r2 + h2) = √(62 + 82) = 10 m
CSA of tent = πrl = (22 x 6 x 10)/7 = 188.6 m2
Remaining cloth = (200 – 188.6) = 11.4 m2
14. One day Rahul visited a park along with his friend. There he saw a game of chance that consists of spinning an arrow that comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these equally likely outcomes.
a) Find the probability that the arrow will point at a number greater than or equal to 2.
7/8
b) Find the probability that the arrow will point at a number divisible by 3.
2/8