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Time: 3 hours Maximum Marks: 80
General Instructions:
Read the following instructions carefully and follow them:
1. This question paper contains 38 questions. All questions are compulsory.
2. Question paper is divided into FIVE sections – Section A, B, C, D and E.
3. In Section A, question number 1 to 18 are multiple choice questions (MCQs) and question number 19 and 20 are Assertion -Reason based questions of 1 mark each.
4. In Section B, question number 21 to 25 are very short answer (VSA) type questions of 2 marks each.
5. In Section C, question number 26 to 31 are Short Answer (SA) type questions carrying 3 marks each.
6. In Section D, question number 32 to 35 are long answer (LA) type questions carrying 5 marks each.
7. In Section E, question number 36 to 38 are case based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study.
8. There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E.
9. Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
10. Use of calculator is not allowed.
Section – A
1. If p2 = 32/50, then p is a/an
(A) Whole number
(B) integer
(C) rational number
(D) irrational number
View AnswerAns. (C) rational number
p2 = 32/50 = 16/25 => p = 4/5
Since, P is in form of p/q where q ≠ 0
∴ p is a rational number
2. The roots of the equation x2 + 3x – 10 = 0 are:
(A) 2, -5
(B) -2, 5
(C) 2, 5
(D) -2, -5
View AnswerAns. (A) 2, -5
x2 + 3x – 10 = 0
x2 + 5x – 2x – 10 = 0
x (x + 5) – 2(x + 5) = 0
x = 2 and x = -5
3. The point on the x-axis which is equidistant from (-4, 0) and (10, 0) is
(A) (7, 0)
(B) (5, 0)
(C) (0, 0)
(D) (3, 0)
View AnswerAns. (D) (3, 0)
(-4, 0) and (10, 0) both les on x-axis as y-coordinates are 0.
So, point equidistant from (-4, 0) and (10, 0) will be
= (x1 + x2)/2 (Mid-point formula)
= (-4 + 10)/2 = 6/2 = 3
∴ Point will be (3, 0)
4. Which of the following is not an A.P?
(A) -1, 2, 0, 8, 2, 8, ……
(B) 3, 3 + √2, 3 + √2, 3 + √2, ……
(C) 4/3, 7/3, 9/3, 12/3, ……
(D) 7/3 – 4/3 = 3/3 = 1
View AnswerAns. (C) 4/3, 7/3, 9/3, 12/3, ……
Common difference (d) = 7/3 – 4/3 = 3/3 = 1
9/3 – 7/3 = 2/3
As we can see that 1 ≠ 2/3
Thus, difference between consecutive terms is not same.
Hence, this is not an A.P
5. The radius of a sphere (in cm) whose volume is 12 π cm3, is
(A) 3
(B) 3√3
(D) 32/3
(D) 31/3
View AnswerAns. (C) (D) 32/3
Volume of sphere = 4/3 πr3
12 π = 4/3 πr3
r3 = 32
r = 32/3
6. If α and β are the zeroes of the polynomial x2 – 1, then the value of (α + β) is
(A) 2
(B) 1
(C) -1
(D) 0
View AnswerAns. (D) 0
(x2 – 1)
= (x – 1)(x + 1)
x = -1 or 1
Thus α = -1 and β = 1
∴ α + β = -1 + 1 = 0
7. Shown below is a circle with centre O. The shaded sector has an angle of 280 and area A cm2
Which of these is the area of the unshaded sector?
(A) 2/7 A cm2
(B) 4/3 A cm2
(C) 4/3 A cm2
(D) 7/9 A cm2
View AnswerAns. (A) 2/7 A cm2
Angle of shaded sector = 280
Area of shaded sector = A cm2
A = θ/360 πr2
A = 280/360 πr2
A = 7/9 πr2
9A/7 = πr2
Angle of unshaded sector = 360 – 280 = 80
Angle of unshaded sector = θ/360 πr2
= 80/360 πr2
= 2/9 x 9A/7
= 2/7A cm2
8. If the zeroes of the quadratic polynomial x2 + (a + 1)x + b are 2 and -3, then
(A) a = -7, b = -1
(B) a = 5, b = -1
(C) a = 2, b = -6
(D) a = 0, b = -6
View AnswerAns. (D) a = 0, b = -6
x2 + (a + 1) x + b ….(i)
Zeroes of given quadratic polynomial are 2 and -3
∴ α = 2 and β = -3
Then, sum of roots (α + β) = 2 + (-3) = -1
Product of roots (αβ) = 2 x -3 = -6
∴ Quadratic polynomial
=> x2 = (α + β) x + αβ= 0
=> x2 + 1x – 6 = 0 ……. (ii)
From equation (i) and (ii)
a + 1 = 1
a = 0
And b = -6
9. In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of ∠AQB is
(A) 62.5°
(B) 125°
(C) 55°
(D) 90°
View AnswerAns. (A) 62.5°
∠PAC = 90 (Tangent is perpendicular to the radius through point of contact)
∠PBA = 90
∠APB = 55 (Given)
So ∠APB + ∠PAC +∠PBA + ACB = 360 ( sum of all angles of quadrilateral)
∠ACB = 360 – 235
= 125°
∠ACB = 2∠AQB (Angle subtended by an arc at centre is double the angle subtended by it at any other point of contact)
∠AQB = 125/2 = 62.5°
10. Two concentric circles are centered at O (-4, 3). The ratio of the area of inner circle to that of the outer circle is 1:9. Points A and B lie on the boundaries of the inner and outer circle, respectively, as shown below.
The coordinates of point B are (3, 5). Which of the following are the coordinates of A?
(A) (2/3, 13/3)
(B) (-5/3, 11/3)
(C) (-9/4, 14/4)
(D) (-33/10, 16/5)
View AnswerAns. (B) (-5/3, 11/3)
Here OA/OB = m1/m2
Let A = (x, y), B = (x2, y2) and O (x1, y1)
=> x1 = -4, y1 = 3 and x2 = 3 and y2 = 5
11. cos2 θ/sin2 θ – 1/ sin2 θ, in simplified form, is
(A) tan2 θ
(B) sec2 θ
(C) 1
(D) -1
View AnswerAns. cos2 θ/sin2 θ – 1/ sin2 θ
(cos2 θ – 1)/ sin2 θ
We know that, sin2 θ + cos2 θ= 1
cos2 θ – 1 = – sin2 θ
Substitute value of cos2 θ – 1
– sin2 θ/sin2 θ = -1
12. The LCM of smallest 2-digit number and smallest composite number is
(A) 12
(B) 4
(C) 20
(D) 40
View AnswerAns. (C) 20
Smallest two-digit number = 10
Smallest composite number = 4
Factor of 4 = 22
Factor of 10 = 2 x 5
∴ LCM of 4 and 10 => 22 x 5 = 20
13. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23 is
(A) 7/90
(B) 1/9
(C) 4/45
(D) 9/89
View AnswerAns. (C) 4/45
Prime number less than 23 = 2, 3, 5, 7, 11, 13, 17, 19
∴ Discs having prime number less than 23 = 8
Total number of discs = 90
Required probability = 8/90 = 4/45
14. For any natural number ‘n’, (125)n is always divisible by
(A) 10
(B) 8
(C) 6
(D) 5
View AnswerAns. (D) 5
Any number whose unit digit is 5, its expansion will always end with 5. Thus, number ending with 5 is divisible by 5.
15. Which of the following is an empirical formula?
(A) Mode = 3 Median – 2 Mean
(B) Mean = ½[3 Median – Mode]
(C) Median = 1/3[Mode + 2 Mean]
(D) All of the above
View AnswerAns. (D) All of the above
16. If an event cannot occur, then its probability is
(A) 1
(B) ¾
(C) ½
(D) 0
View AnswerAns. (D) 0
Probability of an impossible event is 0.
17. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is
(A) ½
(B) 3/13
(C) 4/13
(D) 1/13
View AnswerAns. (B) 3/13
Total number of cards = 52
Total number of face cards = 12
P (Probability of getting a face card) = 12/52 = 3/13
18. In the given figure, DE || BC. If AD = 3 cm, AB = 7 cm and EC = 3 cm, then the length of AE is
(A) 2 cm
(B) 2.25 cm
(C) 3.5 cm
(D) 4 cm
View AnswerAns. (B) 2.25 cm
In ∠ABC, DE || BC
Then, AD/DB = AE/EC
(By Basic Proportionality Theorem)
¾ = AE/3
AE = 9/4 = 2.25 cm
Directions: Choose the correct option. In the question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R)
(A) Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A)
(B) Both Assertion (A) and Reason (R) are true; and Reason (R) is not the correct explanation of Assertion (A)
(C) Assertion (A) is true but Reason (R) is false
(D) Assertion (A) is false but Reason (R) is true
View AnswerAns. (A) Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A)
In case of Assertion:
The origin is the only point which is equidistant from (2, 3) and (-2, -3) as it is the only point which lies on perpendicular bisector of line segment which joins these two points.
Therefore, Assertion is true.
In case of Reason:
On the other side, midpoint of line joining (2, 3) and (-2, -3) is also origin.
Therefore, Reason is true
Thus both Assertion and Reason are true and Reason is correct explanation of Assertion.
19. Assertion (A): The origin is the only point equidistant from (2,3) and (-2, -3)
Reason (R): The origin is the midpoint of the line joining (2,3) and (-2, -3)
20. Assertion (A): n2 – n is divisible by 2 for every positive integer.
Reason (R): √2 is not a rational number.
View AnswerAns. (B) Both Assertion (A) and Reason (R) are true; and Reason (R) is not the correct explanation of Assertion (A)
In case of Assertion:
If n = 2a
Then n2 – n = 4a2 – 2a = 2a(2a – 1) which is divisible by 2
Therefore, Assertion is true
In case of Reason:
√2 is an irrational number, thus reason is also true. Hence both assertion and reason are true but reason is not correct explanation of assertion.
Section – B
21. In the given figure DE ||AC and DF || AE
Prove that BF/FE = BE/EC
View AnswerAns. In ΔABC, DE || AC
So, using basic proportionality theorem, we get
BD/DA = BE/EC …. (i)
In ΔBAE, DF || AE
So, using basic proportionality theorem, we get
BD/DA = BF/FE ….. (ii)
From (i) and (ii), we get
BE/EC = BF/FE
22. Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the sun is such that tan θ = 6/7
View AnswerAns.
In Right ΔABC
tan θ = AB/BC = P/B …..(i)
But tan θ = 6/7 (Given)
Substitute value of tan θ and height of pole AB = 18 m in equation (i)
6/7 = 18/BC
=> BC = 18 x 7/6 = 21 cm
Here, the length of the shadow = 21 cm
23. If α and β are zeroes of the quadratic polynomial p(x) = x2 – (k + 6)x + 2(2k – 1), then find the value of k, if α + β = αβ/2
View AnswerAns. Given α and β are the zeroes of polynomial x2 – (k + 6) x + 2(2k – 1)
So, Sum of zeroes, α + β = k + 6 (α + β = -b/a) …. (i)
Product of zeroes αβ = 2(2k – 1) (αβ = c/a) ….. (ii)
Also, α + β = αβ/2 (Given)
K + 6 = 2(2k – 1)/2
2k + 12 = 4k – 2
k = 7
24. (A) Show that 6n can not end with digit 0 for any natural number ‘n’
View AnswerAns. If 6n ends with 0 then it must have 5 as a factor.
But, 6n = (2 x 3)n
= 2n x 3n
This shows that 2 and 3 are the only prime factors of 6n
According to the fundamental theorem of arithmetic prime factorization of each number is unique.
So, 5 is not a factor of 6n
Hence 6n can never end with the digit 0.
OR
(B) Find the HCF and LCM of 72 and 120
View AnswerAns.
HCF (72, 120) = Product of common terms with lowest power
= 23 x 31 = 8 x 3 = 24
LCM (72, 120) = Product of prime factors with highest power
= 23 x 32 x 5 = 8 x 9 x 5 = 360
Thus HCF and LCM of 72 and 120 are 24 and 360 respectively.
25. If a number is added to twice its square, then the resultant is 21. Write the quadratic representation of this situation.
View AnswerAns. Let the number be x, then
According to given information 2x2 + x = 21
Therefore, Quadratic equation = 2x2 + x – 21 = 0
OR
A quadratic equation with integral coefficients has integral roots. Justify your answer.
View AnswerAns. No, a quadratic equation with integral coefficients may not have an integral roots.
Suppose, quadratic equation 4x2 – 8x + 3 = 0 have integral coefficients.
Section – C
26. In the given figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, then prove that ΔABD ~ ΔECF
Ans. Given AB = AC
∠B = ∠C …. (i)
AD ⊥ BC
∠ADB = 90 …..(ii)
EF ⊥ AC
∠EFC = 90 …..(iii)
In ΔABD and ΔECF
∠B = ∠C From (i)
∠ADB = ∠EFC = 90 (From ii and iii)
ΔABD ~ ΔECF (by AA criterion)
27. (A) Examine that the list of numbers 13, 10, 7, 4 …… form an A.P
View AnswerAns. a2 – a1 = 10 – 13 = -3
a3 – a2 = 7 – 10 = -3
a4 – a3 = 4 – 7 = -3 and so on
As, the difference between two consecutive terms is same,
Therefore, the list forms an AP
(B) If it forms an A.P, then write the next two terms.
Next two terms are 4 + (-3) = 4 – 3 = 1
And 1 + (-3) = 1-3 = -2
28. (A) A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle 90°. Find the total area cleaned at each sweep of the blades. (Take π= 22/7)
View AnswerAns.
OR
(B) A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle.
View AnswerAns.
29. Find the ratio in which y-axis divides the line segment joining the points (6, -4) and (-2, -7). Also find the point of intersection.
View AnswerAns. Let the ratio in which the line segment joining A(6, -4) and B(-2, -7) is divided by the y-axis be k : 1. Let the coordinate of point on y-axis be (0, y)
Therefore,
= (-7 x 3 – 4)/(3 + 1) = -25/4
Therefore, the given line segment is divided by the point (0, -25/4) in the ratio 3 : 1
OR
Show that the points (7, 10), (-2, 5) and (3, -4) are vertices of an isosceles right triangle.
View AnswerAns. Let the given points are P(7, 10), Q(-2, 5) and R(3, -4)
Now using distance formula we find distance between these points i.e., PQ, QR and PR
Distance between points P(7, 10) and Q(-2, 5)
PQ = √[(- 2 – 7)2 +(5 – 10)2] =√(81 + 25) = √106
Distance between points Q(-2, 5) and R(3, -4)
QR = √[(3 +2)2 +(-4 – 10)2] =√(16 + 196) = √212
Now,
PQ2 + QR2 = 106 + 106
= 212 = PR2
PQ2 + QR2 = PR2
There points form an isosceles right triangle because sides PQ and QR are equal.
30. In a single throw of a pair of different dice, what is the probability of getting
(i) a prime number on each dice?
View AnswerAns. Total possible cases when two dice are thrown together = 6 x 6 = 36
Favourable cases when both numbers are prime are (2,2), (2,3), (2,5),(3,2), (3,3), (3,5), (5,2), (5,3), (5,5) = 9 outcomes
P(a prime number on each dice)=Favourable cases/Total cases
= 9/36 = ¼
(ii) a total of 9 or 11?
View AnswerAns. Favourable cases when sum of numbers are 9 or 11 are (3, 6) (4, 5) (5, 4) (5, 6) (6, 3) (6, 5) = 6 outcomes
P(a total of 9 or 11) = Favourable cases/Total possible outcomes
= 6/36 = 1/6
31. Prove that the lengths of two tangents drawn from an external point to a circle are equal.
View AnswerAns.
Given: PA and PB are the lengths of two tangents drawn from an external point P to circle C with radius r
To prove: PA = PB
Construction: Join OP, AO and BO
Proof:
In the triangles OAP and OBP
OA = OB (radii)
OP = OP (common side)
∠OAP = ∠OBP = 90 (Right angle)
By RHS congruency
ΔOAP ≅ ΔOBP
Therefore, PA = PB (CPCT)
Section – D
32. Sum of the areas of two squares is 544 m2. If the difference of their perimeters is 32 m, find the sides of the two squares.
View AnswerAns. Let the side of first square and second square be x and y.
Then,
Area of first square = x2
Area of second square = y2
x2 + y2 = 544 …. (i)
And perimeter of first square = 4x
Perimeter of second square = 4y
Now, 4x – 4y = 32 ….(ii)
From equation (ii), we get
4(x – y) = 32
x – y = 32/4
x – y = 8
or x = 8 + y …. (iii)
Substituting this value of x in equation (i), we get
x2 + y2 = 544
(8 + y)2 + y2 = 544
y = -20 or y = 12
Since side of a square cannot be negative, therefore y = 12
Substituting y = 12 in equation (iii), we get x = 8 + y = 8 + 12 = 20
Therefore,
Side of first square = x = 20 cm
And,
Side of second square = y = 12 cm
OR
A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
View AnswerAns. Let the speed of the stream be x km/h
Therefore, speed of the boat upstream = (18 – x)km/h
And the speed of the boat downstream = (18 + x) km/h
The time taken to go upstream = distance /speed
= 24/(18 + x) hours
According to the question,
24/(18 – x) – 24/(18 + x) = 1
24(18 + x) – 24(18 – x)/[(18 + x)(18 – x)] = 1
x2 + 48x – 324 = 0
Using the quadratic formula, we get
x = (-48 + 60)/2 or x = (-48 – 60)/2
x = 12/2 or x = -108/2
x = 6 or x = -54
Since x is the speed of the stream, it cannot be negative
So, we ignore the root x = -54. Therefore, x = 6 gives the speed of the stream as 6 km/h
33. Eliminate θ from the following equations.
(i) x = a sec θ, y = b tan θ
View AnswerAns. x = a sec θ (given)
sec θ = x/a ….(i)
y = b tan θ (given)
tan θ = y/b ….(ii)
As we know (sec2 θ – tan2 θ) = 1
Thus substituting value from (i) and (ii) we get
x2/a2 – y2/b2 = 1
(ii) x = k + a cos θ, y = h + b sin θ
View AnswerAns. cos θ = (x – k)/a ….. (iii)
Sin θ = (y – h)/b ….. (iv)
As we know, sin2 θ + cos2 θ = 1
(x – k)2/a2 + (y – h)2/b2 = 1
34. State and prove BPT or state and prove Thale’s theorem
View AnswerAns. If a line is drawn to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: ΔABC where DE || BC
To prove: AD/DB = AE/EC
Construction: Join BE and CD
Drawn DM ⊥ AC and EN ⊥ AB
Now,
OR
In a mathematics class, a teacher drew the following figure where TQ/QR = 1/3. She then asked,” What is the sufficient condition required to prove that ΔTQP ~ ΔRQS?”
Darsh said that is sufficient if it is given that TP/SR = 1/3
Bhargav said that it is sufficient if it is given that ∠P = ∠S
Tanvi said that it is sufficient if it is given that PQ/QS = 1/3
Examine whether each of their responses is correct or incorrect. Give reasons.
View AnswerAns. Darsh’s answer is incorrect
(i) Because in order to apply any of the similarity criterion, at least one more price of information is necessary.
(ii) Bhargav’s answer is correct
In ΔPTQ and ΔQSR
∠TQP = ∠RQS (Vertically opposite angles)
∠P = ∠S (Given)
∠T = ∠R ( Angle sum property)
Hence ΔPTQ ~ ΔQSR
(iii) Tanvi’s answer is correct
In ΔTQP and ΔRQS
TQ/TR = 1/3 (give)
∠TQP = ∠RQS (Vertically opposite angles)
PQ/QS = 1/3 (Given)
Hence ΔTQP ~ ~RQS (SAS Criterion)
35. A solid is in the shape of a right circular cone surmounted on a hemisphere, the radius of each of them being 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.
View AnswerAns. Radius = 7 cm
Height = 2 x Radius = 14 cm
Volume of cone = 1/3 πr2h
Volume of hemisphere = 2/3 πr3
Volume of solid = Volume of cone + Volume of hemisphere
= 1/3 πr2h + 2/3 πr3
= 1/3 πr2(h + 2r)
= 1/3 x 22/7 x 7 x 7 x (14 + 2 x 7)
= 154/3 x 28
= 4312/3 = 1437.33 cm3
Section – E
36. Read the following passage and answer the questions given at the end:
Diwali Fair
A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in figure.
Prizes are given when a black marble is picked. Shweta plays the game once.
(i) What is the probability that spinner stops on an odd number?
View AnswerAns. Numbers on spinner = 1, 2, 4, 6, 8, 10
Even numbers on spinner = 2, 4, 6, 8, 10
Odd numbers on spinner = 1
Therefore, P (spinner stops at odd number) = 1/6
(ii) What is the probability that she will be allowed to pick a marble from the bag?
View AnswerAns. Shweta will pick black marble, if spinner stops on even number.
Therefore,
n(Even number) = 5
n(Possible number) = 6
(a) P(Shweta allowed to pick a marble) = P (Even number)
= n(Even number)/n(Possible number) = 5/6
Therefore, the probability of allowing Shweta to pick a marble is 5/6.
OR
Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 balls out of which 6 are black?
View Answer(b) Since, prizes are given, when a black marble is picked
Number of black marbles = 6
Total number of marbles – 20
Therefore,
P(getting a prize) = P(a black marble)
= n(Black marbles)/n(Total marbles)
= 6/20
= 3/10
Therefore, the probability of getting prize is 3/10
(iii) Which mathematical concept is used here?
View AnswerMathematical concept used here is ‘Probability’
37. A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹9000 and from batch II is ₹26,000. Assume that each poor child pays
₹ x per month and each rich child pays ₹ y per month.
Based on the above information, answer the following questions:
(i) Represent the information given above in terms of x and y.
(ii) Find the monthly fee paid by a poor child.
View AnswerAns. Monthly fees paid by each poor children = x Rs
Monthly fees paid by each rich children = y Rs
(i) For batch I
20x + 5y = 9000 ….. (i)
For batch II
5x + 25y = 26000 …. (ii)
Multiply equation (i) by 5 we get
100x + 25y = 45000 …. (iii)
Subtract (ii) from (iii)
x = 200
Thus monthly fees paid by poor child
= Rs 200
OR
Find the difference in the monthly fee paid by a poor child and a rich child.
View AnswerAns. Substitute value of x in equation (i)
20 x 200 + 5y = 9000
5y = 9000 – 4000
Y = 5000/5 = 1000
Monthly fee paid by rich child = Rs 1000
Difference in monthly fee paid by poor child and a rich child = 1000 – 200 = Rs 800
(iii) If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II?
View AnswerAns. Poor children = 10
Rich children = 20
Total monthly collection of fees from batch II
=> 10 x 200 + 20 x 1000
= 2000 + 20000 = Rs 22000
38. A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kings way), is about 138 feet (42 metres) in height.
(i) What is the angle of elevation if they are standing at a distance of 42 m away from the monument?
View AnswerAns. Let AB be the monument of height 42 m and C is the point where they are standing such that BC = 42 m
Now, in ΔABC,
Tan θ = AB/BC => tan θ = 42/42 = 1
Θ = 45
(ii) They want to see the tower at an angle of 60. So, they want to know the distance where they should stand and hence find the distance.
View AnswerAns. In ΔABC,
tan 60 = AB/BC
√3 = 42/BC
BC = 42/√3 = 42√3/3 = 14√3
14 x 1.732 = 24.248 = 24.25 m
OR
If the altitude of the Sun is a t 60, then the height of the vertical tower that will cast a shadow of length 20 m is
View AnswerLet AB be the height of the tower.
Then, in ΔABC, tan 60 = AB/BC => √3=h/20
=>h =20√3 m
(iii) If the elevation of the angle of top of the India Gate is 30 and the height of the India Gate be 42 metre, Find the distance between point and foot of the India Gate.
View AnswerAns.
Let the distance between foot of the India Gate and point be x
Since, tan 30 = AB/BC = 42/x
k/√3 = 42
k = 42√3 m