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

Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2016 (Delhi) to familiarize themselves with the exam format and marking scheme.

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 x ² – Sx + P = 0
= x ² – (-6)x + 5 = 0 ⇒ x ² + 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:

The man begins from O and goes to A and then to B making rt. angle ΔOAB.
OB ² = OA ² + AB ²
= 15 ² + 8 ²
= 225 + 64
= 289
∴ OB = +\(\sqrt{289}\)
= 17 m

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:

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)

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 9 ^th 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
n ^th term from the end = l – (n – 1 )d
∴ 9 ^th 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 ax ² – 5x + c find the value of ‘a’ and ‘c’ when m + n = mn = 10.
Answer:
Quadratic polynomial, ax ² – 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

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

∠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 x ² + 11x + k is -3, find the value of k and the other zero.
Answer:
Let P(x) = x ² + 11x + k
P(-3) = (-3) ² + 11(-3) + k
0 = 9 – 33 + k
0 = – 24 + k
⇒ k = 24
Now, polynomial is x ² + 11x + k = 0
⇒ x ² + 11x + 24 = 0
x ² + 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 ax ² + 7x + b = 0, find the values of a and b.
Answer:
ax ² + 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.

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

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).

PQ = PR …[Given
PQ ² = PR ² …[Squaring both sides
(2k – 2) ² + (k + 5) ² = (2k + 3) ² + (k – 6) ² …[Given
4k ² + 4 – 8k + k ² + 10 k + 25 = 4k ² + 9 + 12k + k ² – 12k + 36
⇒ 2k + 29 = 45 ⇒ 2k = 45 – 29
⇒ 2k = 16 ⇒ k = 8
Hence coordinates of point P are (16, 8).

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.

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 3b ² = a ² .
∴ a ² 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 b ² = 9c ² , that is, b = 3c ² . This means that b ² 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 × 3 ² × 17
657 = 3 ² × 73

HCF = 3 ² = 9
LCM = 2 × 3 ² 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)

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

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.

Answer:

\(\frac{n}{2}\) = \(\frac{30}{2}\) = 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.\(\overline{6}\)

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 = \(\frac{22}{7}\)]

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 + 2πr ² – πr ²
= 6(side) ² + πr ²
= 6(6) ² + \(\frac{22}{7}\) × \(\frac{7}{4}\) × \(\frac{7}{4}\) = 216 + \(\frac{77}{8}\)
= 216 + 9.625 = 225.625 cm ²

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.

Answer:

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 1 ^st term = a, Common difference = d
Given: S ₇ = 49, S ₁₇ = 289
We know, S _n = \(\frac{n}{2}\) [2a + (n – 1)d]
S ₇ = \(\frac{7}{2}\) (2a + 6d)
49 × \(\frac{2}{7}\) = 2a + 6d
2a + 6d = 14 ………… (i)
S ₁₇ = \(\frac{17}{2}\) (2a + 16d)
289 × \(\frac{2}{17}\) = [2a + (17 – 1)d]
2a + 16d = 34 ……….. (ii)
Solving (i) and (ii), we get

From (i), 2a + 6(2) = 14
2a + 12 = 14
2a = 14 – 12 = 2 ⇒ a = \(\frac{2}{2}\) = 1
S _n = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{n}{2}\)[2(1) + (n – 1)2] = \(\frac{n}{2}\)(2 + 2n – 2)
S _n = \(\frac{2 n^2}{2}\) = n ² (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 cm ²
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 = πr ² h
= \(\frac{22}{7}\) × 7 × 7 × 30 = 4,620 cm ³

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

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),

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 sin ² \(\) + sin ² \(\) = 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. = sin ² \(\frac{A}{2}\) + sin(\(\frac{B+C}{2}\))
= sin ² \(\frac{A}{2}\) + (90° – \(\frac{A}{2}\)) …[From (i)
= sin ² \(\frac{A}{2}\) + cos ² \(\frac{A}{2}\)
= 1 …..[∵ sin ² θ + cos ² θ
= 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
⇒ x ² + 250x – 7,50,000 = 0
⇒ x ² + 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

….[∵ Thief runs = n mins Policeman runs = (n -1) mins
Here, a = 100, d = 110 – 100 = 10, ‘n’ = n – 1
S _n = \(\frac{n}{2}\) [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
⇒ 10n ² + 180n – 10n – 180 – 200n = 0
⇒ 10n ² – 30n – 180 = 0
⇒ n ² – 3n – 18 = 0 …[Dividing both sides by 10
⇒ n ² – 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.

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.

Answer:
∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact
We have, OP = 5 cm, OT = 13 cm

In rt. ΔOPT,
OP ² + PT ² = OT ² …[Pythagoras’ theorem
⇒ (5) ² + PT ²
⇒ (5) ² + (PT) ² = (13) ²
⇒ PT ² = 169 – 25 = 144 cm ²
⇒ 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,
AE ² + ET ² = AT ² …[Pythagoras’ theorem
⇒ x ² + (8) ² = (12 – x) ²
⇒ x ² + 64 = 144 + x ² – 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:

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)

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)

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 cm ²
ar(ΔOAP) = ar(ΔOBP) = \(\frac{25}{2}\)√3 cm ²
Area of minor sector OACB = \(\frac{\theta}{360^{\circ}}\) πr ²
= \(\frac{120^{\circ}}{360^{\circ}}\) × 3.14 × (5) ² …[From (i)
= \(\frac{78.5}{3}\)
= 26.1\(\overline{\mathbf{6}}\) or 26.17 cm ² (approx.)