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CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

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CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Time allowed: 3 hours
Maximum marks: 80

General Instructions:
Read the following instructions carefully and follow them:

  1. All questions are compulsory.
  2. This question paper contains of 30 questions.
  3. Question No. 1 – 6 in Section A are very short answer type questions carrying 1 mark each.
  4. Question No. 7 – 12 in Section B are short answer type questions carrying 2 marks each.
  5. Question No. 13 – 22 in Section C are long answer – I type questions carrying 3 marks each.
  6. Question No. 2 3- 30 in Section D are long answer – II type questions carrying 4 marks each.

Section A

Questions number 1 to 6 carry 1 mark each.

Question 1.
If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial.
Answer:
Given: Sum of the roots, α + β = -6
Product of the roots, αβ = 5
Quadratic polynomial is x2 – Sx + P = 0
= x2 – (-6)x + 5 = 0 ⇒ x2 + 6x + 5 = 0

Question 2.
A man goes 15 m due west and then 8 m due north. Find the distance of the man from the starting point.
Answer:
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 1
The man begins from O and goes to A and then to B making rt. angle ΔOAB.
OB2 = OA2 + AB2
= 152 + 82
= 225 + 64
= 289
∴ OB = +\(\sqrt{289}\)
= 17 m

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 3.
From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.
Answer:
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 2
PA = PB …[Tangents drawn from external point are equal
∴ ∠PBA = ∠PAB = 50° …[Angles equal to opposite sides
∠APB = 180° – 50° – 50° = 80° …[Angle-sum property of a A
In cyclic quad. OAPB
∠AOB + ∠APB = 180° …[Sum of opposite angles of a cyclic quadrilateral is 180°
∠AOB + 80° = 180°
∴ ∠AOB = 180° – 80° = 100°

Question 4.
In Fig., AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m. Find the length of the ladder. (Use √3 = 1.73)
v
Answer:
BD = AB – AD = 6 – 2.54 = 3.46 m
In rt., ΔDBC, sin 60° = \(\frac{\mathrm{BD}}{\mathrm{DC}}\)
\(\frac{\sqrt{3}}{2}=\frac{3.46}{D C}\)
√3 DC = 3.46 × 2 ⇒ DC = \(\frac{3.46 \times 2}{1.73}\) = 4 m
Length of the ladder, DC = 4.

Question 5.
Find the 9th term from the end (towards the first term) of the A.P. 5, 9,13, …, 185.
Answer:
Here First term, a = 5
Common difference, d = 9- 5 = 4
last term, l = 185
nth term from the end = l – (n – 1 )d
∴ 9th term from the end = 185 – (9 – 1)4
= 185 – 8 × 4
= 185 – 32 = 153

Question 6.
Cards marked with number 3, 4, 5, …., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Answer:
Total no. of cards = 50 – 3 + 1 = 48
Perfect square number cards are 4, 9, 16, 25, 36, 49, i.e., 6 cards.
∴ P(perfect square number) = \(\frac{6}{48}\) = \(\frac{1}{8}\)

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.
If ‘m’ and ‘n’ are the zeroes of the polynomial ax2 – 5x + c find the value of ‘a’ and ‘c’ when m + n = mn = 10.
Answer:
Quadratic polynomial, ax2 – 5x + c = 0
Here ‘a’ = 4, ‘b’ = -5, ‘c’ = c
m + n = 10 (Given)
\(\frac{-b}{a}\) = 10
\(\frac{-(-5)}{a}\) = 10
10a = 5
a = \(\frac{5}{10}\) = \(\frac{1}{2}\) ……. (i)

mn = 10 (Given)
\(\frac{c}{a}\) = 10
c = 10a
c = 10(\(\frac{1}{2}\)) ….From (i)
∴ c = 5

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 8.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Answer:
Let BC be the pole and EF be the tower.
Shadow AB = 4 m and DE = 28 m
In ΔABC and ΔDEF
∠1 = ∠3 …[Sun’s angle of elevation at the same time
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 4
∠2 = ∠4 ..[Each 90°
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BC}}{\mathrm{EF}}\) …..[In AA similarity, corresponding L sides are proportional
⇒ \(\frac{4}{28}\) = \(\frac{6}{\mathrm{EF}}\)
⇒ EF = \(\frac{6 \times 28}{4}\) = 42
∴ Height of the tower, EF = 42 m

Question 9.
One zero of the polynomial x2 + 11x + k is -3, find the value of k and the other zero.
Answer:
Let P(x) = x2 + 11x + k
P(-3) = (-3)2 + 11(-3) + k
0 = 9 – 33 + k
0 = – 24 + k
⇒ k = 24
Now, polynomial is x2 + 11x + k = 0
⇒ x2 + 11x + 24 = 0
x2 + 8x + 3x + 24 = 0
x(x + 8) + 3(x + 8) = 0
(x + 3) (x + 8) = 0
∴ Other zero is x + 8 = 0
⇒ x = -8

Question 10.
If x = \(\frac{2}{3}\) and x = -3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.
Answer:
ax2 + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
α = \(\frac{2}{3}\) and β = -3 …[ Given
Sum of roots = \(\frac{-b}{a}\)
(α + β) = \(\frac{-b}{a}\)
\(\frac{2}{3}\) + (-3) = \(\frac{-7}{a}\)
\(\frac{2-9}{3}\) = \(\frac{-7}{a}\)
\(\frac{-7}{3}\) = \(\frac{-7}{a}\)
⇒ a = 3 … (i)

Product of roots = \(\frac{c}{a}\)
(α × β) = \(\frac{c}{a}\)
\(\frac{2}{3}\) × (-3) = \(\frac{b}{a}\)
-2 = \(\frac{b}{3}\) …[From (i)
⇒ b = -6
∴ a = 3, b = -6

Question 11.
In Fig., a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 5
Or
Question 11.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q (2, -5) and R(-3, 6), find the coordinates of P.
Answer:
AB = 12 cm, BC = 8 cm, CA = 10 cm …[Given
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 6
As we know,
AF = AD, CF = CE, BD = BE
Let AD = AF = x cm,
then, DB = AB – AD = (12 – x) cm
∴ BE = (12 – x) cm
Similarly, CF = CE = AC – AF = (10 – x) cm
BC = 8 cm …[Given
⇒ BE + CE = 8
⇒ 12 – x + 10 – x = 8 ⇒ 22 – 8 = 2x
⇒ 2x = 14 ∴ x = 7
∴ AD = x = 7 cm
BE = 12 – x = 12 – 7 = 5 cm
CF = 10 – x = 10 – 7 = 3 cm
Or
Let the point P be (2k, k), Q(2, -5), R(-3, 6).
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 7
PQ = PR …[Given
PQ2 = PR2 …[Squaring both sides
(2k – 2)2 + (k + 5)2 = (2k + 3)2 + (k – 6)2 …[Given
4k2 + 4 – 8k + k2 + 10 k + 25 = 4k2 + 9 + 12k + k2 – 12k + 36
⇒ 2k + 29 = 45 ⇒ 2k = 45 – 29
⇒ 2k = 16 ⇒ k = 8
Hence coordinates of point P are (16, 8).

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 12.
In Fig., AP and BP are tangents to a circle with centre O, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB.
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 8
Answer:
PA = PB …..[Tangents drawn from external Point are equal
Given: ∠APB = 60°
∠PAB = ∠PBA ……. (i)…[Angles opp. to equal sides
In ΔPAB, ∠PAB + ∠PBA + ∠APB = 180° …[Angle-sum-property of a Δ
⇒ ∠PAB + ∠PAB + 60° = 180° …[∵ From (i)
⇒ 2∠PAB = 180° – 60° = 120°
⇒ ∠PAB = \(\frac{120^{\circ}}{2}\) = 60°
⇒ ∠PAB = ∠PBA = ∠APB = 60°
∴ ΔPAB is an equilateral triangle.
Hence, PB = AB = AP = 5 cm [∵ All sides of an equilateral A are equal

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.
Not in Syllabus.
Answer:
Not in Current Syllabus.

Question 14.
Prove that √3 is an irrational number. Hence prove that √3 – 5 is also an irrational number.
Or
Question 14.
Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers.
Answer:
Let us assume, to the contrary, that √3 is rational.
That is, we can find integers a and b (≠ 0)
such that √3 = \(\frac{a}{b}\).
Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are coprime.
So, by√3 = a.
Squaring on both sides and rearranging, we get 3b2 = a2.
∴ a2 is divisible by 3 and a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3 b2 = 9c2, that is, b = 3c2. This means that b2 is divisible by 3 and so b is also divisible by 3.
∴ a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So we conclude that √3 is irrational.
√3 is an irrational number …[Proved
5 is a rational number
∴ √3 – 5 is an irrational number.
Or
306 =2 × 32 × 17
657 = 32 × 73
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 9
HCF = 32 = 9
LCM = 2 × 32 17 × 73
= 22338
L.H.S. = LCM × HCF
= 22338 × 9 = 201042
R.H.S. = Product of two numbers
= 306 × 657 = 201042
∴ L.H.S. = R.H.S.

Question 15.
Solve for x and y: \(\frac{2}{x}\) + \(\frac{2}{3 y}\) = \(\frac{1}{6}\); \(\frac{3}{x}\) + \(\frac{2}{y}\) = 0, (x ≠ 0, y ≠ 0) and hence find the value of ‘a’ for which y = ax – 4
Or
Question 15.
Solve for x and y: px + qy = p – q; qx – py = p + q
Answer:
\(\frac{2}{x}\) + \(\frac{2}{3 y}\) = \(\frac{1}{6}\) ……….. (i)
\(\frac{3}{x}\) + \(\frac{2}{y}\) = 0 ……….. (ii)
Let \(\frac{1}{x}\) = M and \(\frac{1}{y}\) = N
Then (i) 2M + \(\frac{2 \mathrm{~N}}{3}\) = \(\frac{1}{6}\) ⇒ 12M + 4N = 1
(ii) 3M + 2N = 0
Multiplying (ii) by 2 and subtracting from (i)
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 10
Now putting the value of M in (ii), we get
3(\(\frac{1}{6}\)) + 2N = 0 ⇒ \(\frac{1}{2}\) + 2N = 0
⇒ 2N = – \(\frac{1}{2}\) ⇒ N = –\(\frac{1}{4}\) …..II
Now revaluing M and N, we get
M ⇒ \(\frac{1}{x}\) = \(\frac{1}{6}\) ⇒ x = 6
N ⇒ \(\frac{1}{y}\) = \(\frac{-1}{4}\) ⇒ y = -4
Now putting these values in y = ax – 4, we get
-4 = a(6) – 4 ⇒ -4 = 6a – 4
⇒ 6a = -4 + 4 = 0 ∴ a = 0
Or
px + qy = p – q ……….. (i) × p
qx – py = p + q ………. (ii) × q
Multiplying (i) by p and (ii) by q and adding
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 11
Putting the value of x in (i), we get
p(1) + qy = p – q ⇒ qy = p – q – p
qy = -q ⇒ y = -1
∴ x = 1 ;y = -1

Question 16.
The distribution below gives the weights of 30 students of a class. Find the median weight of a student.
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 12
Answer:
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 13
\(\frac{n}{2}\) = \(\frac{30}{2}\) = 15
Median class is 55 – 60
As Median = l + (\(\frac{\frac{n}{2}-C f}{f}\) × h)
∴ Median = 55 + \(\) × 5 = 55 + \(\frac{5}{3}\)
= 55 + 1.\(\overline{6}\) = 56.\(\overline{6}\)

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 17.
Not in Current Syllabus
Answer:
Not in Current Syllabus.

Question 18.
In Fig., is a decorative block, made up of two solids – a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block. [Use π = \(\frac{22}{7}\)]
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 14
Answer:
Side of cube = 6 cm,
Radius (r) = \(\frac{3.5}{2}\) = \(\frac{7}{4}\) cm
Total surface area of the block
= Total surface area of cube + C.S. Area of hemisphere – Area of base
= 6(side)2 + 2πr2 – πr2
= 6(side)2 + πr2
= 6(6)2 + \(\frac{22}{7}\) × \(\frac{7}{4}\) × \(\frac{7}{4}\) = 216 + \(\frac{77}{8}\)
= 216 + 9.625 = 225.625 cm2

Question 19.
In Fig. 6, ABC is a triangle coordinates of whose vertex A ai (0, -1). D and E respectively are the mid-points of the sides AB an AC and their coordinates are (1, 0) and (0, 1) respectively. Find the coordinates of point B, C and F.
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 15
Answer:
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 16
Let B (p, q), C (r, s) and F(x, y)
Mid-point of AB = Coordinates of D
(\(\frac{0+p}{2}, \frac{-1+q}{2}\)) = (1, 0)
\(\frac{p}{2}\) = 1
p = 2
\(\frac{-1+q}{2}\)
– 1 + q = 0
q = 1
∴ B(p, q) = B(2, 1)
Mid-point of AC = Coordinates of E
(\(\frac{0+r}{2}, \frac{-1+s}{2}\)) = (0, 1)
\(\frac{r}{2}\) = 0
⇒ r = 0
\(\frac{-1+s}{2}\) = 1
– 1+ s = 2
⇒ s = 3
∴ C(r, s) = C(0, 3)
Coordinates of F = Mid-point of BC
(x, y) = (\(\frac{p+r}{2}, \frac{q+s}{2}\)) = (\(\frac{2+0}{2}, \frac{1+3}{2}\))

Question 20.
If the sum of first 7 terms of an A.P is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
Answer:
Let 1st term = a, Common difference = d
Given: S7 = 49, S17 = 289
We know, Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
S7 = \(\frac{7}{2}\) (2a + 6d)
49 × \(\frac{2}{7}\) = 2a + 6d
2a + 6d = 14 ………… (i)
S17 = \(\frac{17}{2}\) (2a + 16d)
289 × \(\frac{2}{17}\) = [2a + (17 – 1)d]
2a + 16d = 34 ……….. (ii)
Solving (i) and (ii), we get
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 17
From (i), 2a + 6(2) = 14
2a + 12 = 14
2a = 14 – 12 = 2 ⇒ a = \(\frac{2}{2}\) = 1
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{n}{2}\)[2(1) + (n – 1)2] = \(\frac{n}{2}\)(2 + 2n – 2)
Sn = \(\frac{2 n^2}{2}\) = n2 (Hence proved)

Question 21.
The sum of the radius of base and height of a solid right circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 sq. cm, find the volume of the cylinder.
[Use π = \(\frac{22}{7}\) ]
Answer:
Let the radius and height of cylinder be r and h respectively.
r + h = 37 cm …(i) [Given
Total surface area of cylinder = 1,628 cm2
2πr(r + h) = 1,628 ⇒ 2πr(37) = 1,628
2πr = \(\frac{1,628}{37}\) = 44 ⇒ 2 × \(\frac{22}{7}\) × r = 44
r = \(\frac{44 \times 7}{2 \times 22}\) = 7 cm
From (i), 7 + h = 37; h = 37 – 7 = 30 cm
Volume of cylinder = πr2h
= \(\frac{22}{7}\) × 7 × 7 × 30 = 4,620 cm3

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 22.
In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?
Answer:
Two dice can be thrown in 6 × 6 i.e., 36 ways.
(i) “a prime number on each dice” can be obtained as (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5), i.e., 9 ways.
P(a prime no. on each dice) = \(\frac{9}{36}\) = \(\frac{1}{4}\)

(ii) “a total of 9 or 11” can be obtained as
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 18
i.e., 6 ways.
∴ P(a total of 9 or 11) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 24.
2 tables and 3 chairs together cost ₹2,000 and 2 chairs and 3 tables together cost ₹2,500. Find the cost of 1 chair and 3 tables together.
Answer:
Let the cost of one table = ₹ x
Let the cost of one chair = ₹ y
According to the Question,
2x + 3y = 2,000 ……… (i)
3x + 2y = 2,500 ……….. (ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i),
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 19
Putting the value of y in (i),
2x + 3(200) = 2,000 ⇒ 2x + 600 = 2,000
⇒ 2x = 2,000 – 600 = 1,400
∴ x = 700
∴ Cost of 1 chair and 3 tables together
= 3x + y = 3(700) + 200
= 2,100 + 200 = ₹ 2,300

Question 25.
In ΔABC, show that sin2 \(\) + sin2 \(\) = 1.
Answer:
In ΔABC, ∠A + ∠B + ∠C = 180° …[Angle sum property of a Δ
∠B + ∠C = 180° – ∠A
Here, \(\frac{\angle B+\angle C}{2}\) = \(\frac{180^{\circ}-\angle \mathrm{A}}{2}\)
\(\frac{\angle B+\angle C}{2}\) = (90° – \(\frac{\angle \mathrm{A}}{2}\)) ………. (i)
L.H.S. = sin2\(\frac{A}{2}\) + sin(\(\frac{B+C}{2}\))
= sin2\(\frac{A}{2}\) + (90° – \(\frac{A}{2}\)) …[From (i)
= sin2\(\frac{A}{2}\) + cos2\(\frac{A}{2}\)
= 1 …..[∵ sin2θ + cos2θ
= R.H.S.

Question 26.
Not in Current Syllabus.
Answer:
Not in Syllabus.

Question 27.
A passenger, while boarding the plane, slipped from the stairs and got hurt. The pilot took the passenger in the emergency clinic at the airport for treatment. Due to this, the plane got delayed by half an hour. To reach the destination 1500 km away in time, so that the passengers could catch the connecting flight, the speed of the plane was increased by 250 km/hour than the usual speed. Find the usual speed of the plane.
Or
Question 27.
A thief runs with a uniform speed of 100 m/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/ minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief.
Answer:
Let the usual speed of the plane = x km/hr
Then the increased speed of the plane = (x + 250) km/hr
Distance = 1,500 km
According to the question,
\(\frac{1,500}{x}\) – \(\frac{1,500}{(x+250)}\) = \(\frac{30}{60}\) ….[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\) 30 mins \(\frac{30}{60}\) = \(\frac{1}{2}\) hour
⇒ \(\frac{1,500(x+250-x)}{x(x+250)}\) = \(\frac{1}{2}\)
⇒ x(x + 250) = 3,000 × 250
⇒ x2 + 250x – 7,50,000 = 0
⇒ x2 + 1,000x – 750x – 7,50,000 = 0
⇒ x(x + 1,000) – 750(x + 1,000) = 0
⇒ (x + 1,000) (x – 750) = 0
⇒ x + 1,000 = 0 or x – 750 = 0
⇒ x = -1,000 (reject) or x = 750
as the speed of plane can not be negative.
∴ Speed of plane = 750 km/hr.
Or
Let total time be n minutes.
Total distance covered by thief in n minutes
= Speed × Time
= 100 × n = 100 n metres
Total distance covered by policeman
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 20
….[∵ Thief runs = n mins Policeman runs = (n -1) mins
Here, a = 100, d = 110 – 100 = 10, ‘n’ = n – 1
Sn = \(\frac{n}{2}\) [2a + (n – 1)d]
∴ 100n = \(\frac{(n-1)}{2}\) [2(100) + (n – 1 – 1)(10)]
⇒ (n – 1) [200 + 10n – 20] = 200n
⇒ (n – 1) [10n + 180] = 200n
⇒ 10n2 + 180n – 10n – 180 – 200n = 0
⇒ 10n2 – 30n – 180 = 0
⇒ n2 – 3n – 18 = 0 …[Dividing both sides by 10
⇒ n2 – 6n + 3n – 18 = 0
⇒ n(n – 6) + 3(n – 6) = 0
⇒ (n + 3) (n – 6) = 0
⇒ n + 3 = 0 or n – 6 = 0
⇒ n = -3 (reject) or n = 6
Since n (time) can not be negative. .
∴ Time taken by policeman to catch the thief = n – 1 = 6 – 1 = 5 minutes.

CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions

Question 28.
In Fig., O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 21
Answer:
∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact
We have, OP = 5 cm, OT = 13 cm
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 22
In rt. ΔOPT,
OP2 + PT2 = OT2 …[Pythagoras’ theorem
⇒ (5)2 + PT2
⇒ (5)2 + (PT)2 = (13)2
⇒ PT2 = 169 – 25 = 144 cm2
⇒ PT = \(\sqrt{144}\) = 12 cm
OP = OQ = OE = 5 cm …[radius of the circle
ET = OT – OE = 13 – 5 = 8 cm
Let, PA = x cm, then AT = (12 – x) cm
PA = AE = x cm …[Tangents drawn from an external point
In rt. ΔAET,
AE2 + ET2 = AT2 …[Pythagoras’ theorem
⇒ x2 + (8)2 = (12 – x)2
⇒ x2 + 64 = 144 + x2 – 24x
⇒ 24x = 144 – 64 = 80
⇒ x = \(\frac{80}{24}\) = \(\frac{10}{3}\) cm
AB = AE + EB = AE + AE = 2AE = 2x
∴ AB = 2(\(\frac{10}{3}\)) = \(\frac{20}{3}\) cm = 6\(\frac{2}{3}\) cm = 6.67 cm
or 6.\(\overline{\mathbf{6}}\) cm

Question 29.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of the flying bird. (Take √3 = 1.732)
Answer:
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 23
Let BC be the tree and BD & AB are x and y respectively.
In rt. ΔABC, tan 45° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ 1 = \(\frac{80}{y}\) ⇒ y = 80 m ………… (i)
In rt. ΔADE, tan 30° = \(\frac{\mathrm{DE}}{\mathrm{AD}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{80}{x+y}\)
⇒ x + y = 80√3
⇒ x + 80 = 80√3 ….[From (i)
⇒ x = 80√3 – 80
⇒ x – 80(√3 – 1)
⇒ x = 80(1.732 – 1) ….[√3 = 1.732
⇒ x = 80(0.732)
∴ CE, x = 58.56 m
Hence, speed of bird = \(\frac{\text { Distance }}{\text { Time }}\)
\(\frac{\text { CE }}{\text { Time }}\) = \(\frac{58.56 \mathrm{~m}}{2 \mathrm{sec} .}\)
= 29.28 m/sec.

Question 30.
An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 9) From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. (Use π = 3.14 and √3 = 1.73)
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 24
Answer:
∠OAP = 90° …[Tangent is ⊥ to the radius through the point of contact
In rt. ΔOAP,
cos θ = \(\frac{\mathrm{OA}}{\mathrm{OP}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = cos 60°
∴ θ = 60°
∠AOB = 60° + 60° = 120° ……… (i)
CBSE Class 10 Maths Question Paper 2016 (Delhi) with Solutions 25
Reflex ∠AOB = 360° – ∠AOB
α = 360° – 120° = 240°
∴ The length of the belt that is still in contact with the pulley = ADB = lenght of major arc
= (\(\frac{\alpha}{360^{\circ}}\))2πr
\(\frac{240^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 5
= \(\frac{62.8}{3}\) 20.9\(\overline{\mathbf{3}}\) cm
In rt. ΔOAP, sin 60° = \(\frac{\mathrm{AP}}{\mathrm{OP}}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{\mathrm{AP}}{10}\)
⇒ 2AP = 10√3
⇒ AP = 5√3 cm
Area of ΔOAP = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × AP × OA
= \(\frac{1}{2}\) × 5√3 × 5 = \(\frac{25}{2}\)√3 cm2
ar(ΔOAP) = ar(ΔOBP) = \(\frac{25}{2}\)√3 cm2
Area of minor sector OACB = \(\frac{\theta}{360^{\circ}}\) πr2
= \(\frac{120^{\circ}}{360^{\circ}}\) × 3.14 × (5)2 …[From (i)
= \(\frac{78.5}{3}\)
= 26.1\(\overline{\mathbf{6}}\) or 26.17 cm2 (approx.)


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