HBSE Class 10th Maths Sample Paper 2024 Answer

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HBSE Class 10th Maths Sample / Model Paper 2024 Answer

Class – 10th               Subject – Mathematics
Time : 3 Hours            Maximum Marks : 80

General Instructions :
(i) There are 5 sections A, B, C, D and E in this question paper.
(ii) Section-A consists of one mark questions from 1 to 20. 1 to 18 are Multiple Choice Questions (MCQs), One Word Answer, Fill in the blank, True/False and question numbers 19 and 20 are Assertion-Reasoning based questions.
(iii) Section-B consists of Very Short Answer Type (VSA) questions of two marks each from 21 to 25.
(iv) Section-C consists of short-answer (SA) type questions of three marks each from 26 to 31.
(v) Section-D consists of Long-Answer (LA) type questions of five marks each from 32 to 35.
(vi) Question numbers 36 to 38 in Section-E are case study based questions of four marks each. Internal choice is given in each case study question of two marks each.
(vii) All questions are compulsory. However, provision of internal choice has been made in 2 questions of Section-B, 2 questions of Section-C, 2 questions of Section-D and 3 questions of Section-E.

SECTION – A
(20 questions of 1 mark each)

1. If two positive integers p and q can be expressed as p = ab² and q = a³b ; a, b being prime numbers, then LCM(p,q) is :
(a) ab
(b) a²b²
(c) a³b²
(d) a³b³
Answer : (c) a³b²

2. Which of the following is not an irrational number ?
(a) 5 + √2
(b) 3 – √3
(c) 2 + √9
(d) 4 – √8
Answer : (c) 2 + √9

3. If 2 + √3 is one zero of the polynomial p(x) = x² – 4x + 1, then other zero of the polynomial is ……………
(a) 3 + √2
(b) 2 – √3
(c) 3 + √3
(d) 2 + √3
Answer : (b) 2 – √3
Standard quadratic equation is ax² + bx + c.
Here, a = 1, b = -4, c = 1
x = [-b ± √(b²-4ac)] ÷ 2a
= [-(-4) ± √(-4)²-4(1)(1)] ÷ 2(1)
= [4 ± √16-4] ÷ 2
= [4 ± 2√3] ÷ 2
= 2 ± √3
so, roots are 2 – √3 and 2 + √3.

4. Which term of AP : 10, 7, 4, ………… is -32 ?
(a) 12th
(b) 13th
(c) 14th
(d) 15th
Answer : (d) 15th
Here, a = 10, d = 7-10 = -3, an = -32
nth term of AP, an = a + (n-1)d
10 + (n-1)(-3) = -32
10 – 3n + 3 = -32
-3n = -32-3-10
-3n = -45
n = 15

5. Values of k for which the quadratic equation 2x² – kx + k = 0 has equal roots is
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8
Answer : (d) 0, 8
Here, a = 2, b = -k, c = k
Discriminant (D) = b² – 4ac = 0
(-k)² – 4(2)(k) = 0
k² – 8k = 0
k(k-8) = 0
k = 0, 8

6. The distance between the points (-4√3, 8) and (-√3, 5) is :
(a) 4
(b) 5
(c) 6
(d) √6
Answer : (c) 6
Here, A(-4√3, 8) and B(-√3, 5)
x1 = -4√3, x2 = -√3, y1 = 8, y2 = 5
Using Distance Formula,
AB = √(x2-x1)²+(y2-y1
= √(-√3+4√3)²+(5-8)²
= √(3√3)²+(-3)²
= √27+9
= √36 = 6

7. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer : (b) similar but not congruent

8. In Fig., if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°

Answer : (b) 70°
∠PTQ + ∠POQ + ∠P + ∠Q = 360°
Here, ∠P = ∠Q = 90° (Radius ⊥ Tangent)
∠PTQ + ∠POQ = 180°
∠PTQ + 110° = 180°
∠PTQ = 180° – 110° = 70°

9. In Fig., AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to
(a) 65°
(b) 60°
(c) 50°
(d) 40°

Answer : (c) 50°
∠ABC + ∠ACB + ∠BAC = 180°
∠ABC = 90° (angle in semicircle is 90°)
90° + 50° + ∠BAC = 180°
∠BAC = 180° – 90° – 50° = 40°
∠BAT + ∠BAC = 90° (Radius ⊥ Tangent)
∠BAT + 40° = 90°
∠BAT = 90° – 40° = 50°

10. What is the the value of (sin30° + cos30°) – (sin60° + cos60°) ?
(a) -1
(b) 0
(c) 1
(d) 2
Answer : (b) 0
(sin30° + cos30°) – (sin60° + cos60°)
= (1/2 + √3/2) – (√3/2 + 1/2) = 0

11. If cosA = 4/5 then what is the value of tanA ?
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4
Answer : (b) 3/4
cosA = Base/Hypotenuse = 4/5
Using Pythagoras Theorem,
(Perpendicular)² + (Base)² = (Hypotenuse)²
P² + B² = H²
P² = H² – B² = 5² – 4² = 25 – 16 = 9 = 3²
P = 3
tanA = Perpendicular/Base = 3/4

12. If sinθ – cosθ = 0, then the value of (sin⁴θ + cos⁴θ) is
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4
Answer : (c) 1/2
sinθ – cosθ = 0
sinθ = cosθ
θ = 45°
sin⁴θ + cos⁴θ = sin⁴45° + cos⁴45° = (1/√2)⁴ + (1/√2)⁴ = 1/4 + 1/4 = 2/4 = 1/2

13. The minute hand of a clock is 84 cm long. What is the distance covered by the tip of the minute hand from 10:10 am to 10:25 am ?
(a) 44 cm
(b) 88 cm
(c) 132 cm
(d) 176 cm
Answer : (c) 132 cm
The time between 10:10 am to 10:25 am = 15 minutes
60 min = 360°
1 min = 360/60 = 6°
15 min = 15 × 6° = 90°
Distance = θ/360° × 2πr = 90/360 × 2 ×22/7 × 84 = 132 cm

14. Area of a sector of angle p (in degrees) of a circle with radius R is :
(a) p/180 × 2πR
(b) p/180 × πR²
(c) p/360 × 2πR
(d) p/720 × 2πR²
Answer : (d) p/720 × 2πR²
Area of sector = θ/360° × πr²

15. Volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
Answer : (d) 16 : 9
Volume of sphere = 4/3 πr³
Surface area of sphere = 4πr²
V1 : V2 = R³ : r³ = 64 : 27 = 4³ : 3³
R : r = 4 : 3
S1 : S2 = 4² : 3² = 16 : 9

16. If the Median and the Mode of a data are 11 and 17 respectively, then its mean is
(a) 7
(b) 8
(c) 9
(d) 10
Answer : (b) 8
Mode = 3(Median) – 2(Mean)
17 = 3(11) – 2(Mean)
2(Mean) = 33 – 17 = 16
Mean = 16/2 = 8

17. For the following distribution :

the sum of lower limits of the median class and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Answer : (b) 25

18. Which of the following cannot be the probability of an event ?
(a) 0.2/3
(b) 0
(c) 1/0.1
(d) 35%
Answer : (c) 1/0.1
1/0.1 = 10/1 = 10
Probability of an event can range from 0 to 1
0 ≤ P(E) ≤ 1

Direction for Questions 19 & 20: In question numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct options from (a), (b), (c) and (d) as given below :
(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 but Reason (R) is the not 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.

19. Assertion (A) : 2 is an example of a rational number.
Reason (R) : The square roots of all positive integers are irrational numbers.
Answer : (c) Assertion (A) is true but Reason (R) is false.

20. Assertion (A) : The point (0, 4) lies on y-axis.
Reason (R) : The x-coordinate on the point on y-axis is zero.
Answer : (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

SECTION – B
(5 questions of 2 marks each)

21. Solve the following pair of linear equations :
x – y = 3
x/3 + y/2 = 6
Solution :
x – y = 3 ……(i)
x/3 + y/2 = 6 ⇒ 2x + 3y = 36 ……(ii)
Multiply eqn.(i) by 3 and add in eqn.(ii), we get
5x = 45
x = 9
Putting x = 9 in eqn.(i), we get y = 6
so, x = 9 and y = 6

22. 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.
Solution :

Let x be the height of the tower.
Two triangles are similar as at the same time ∠C = ∠N
6/4 = x/28
x = 42 m
so, height of the tower = 42 m

OR

In the fig., PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that PQR is an isoscele triangle.

Solution :
PS/SQ = PT/TR (Given)
so, ST || QR (Converse of BPT)
∠PST = ∠PQR ……..(i) (Corresponding Angles)
Also ∠PST = ∠PRQ …….(ii) (Given)
∠PRQ = ∠PQR [From (i) and (ii)]
so, PQ = PR (sides opposite to equal angles)
Hence PQR is an isoscele triangle.

23. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution :

OA = 5 cm, OP = 3 cm
OP ⊥ AB
Therefore AP = √(5²-3²) = √16 = 4 cm
AB = 2AP = 2 × 4 = 8 cm

24. Evaluate the following :
[5cos²60°+4sec²30°-tan²45°] ÷ [sin²30°+cos²30°]
Solution :
= [5(1/2)²+4(2/√3)²-(1)²] ÷ [(1/2)²+(√3/2)²]
= [5/4 + 16/3 – 1] ÷ [1/4 + 3/4]
= [67/12] ÷ [1]
= 67/12

25. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Solution :
Area of minor segment = (θ/360 × πr²) – (1/2 × r² × sinθ)
= (60/360 × 3.14 × 225) – (1/2 × 225 × √3/2)
= 117.75 – 97.312
= 20.4375 cm²
Area of major segment = Area of circle – Area of minor segment
= 3.14 × (15)² – 20.4375
= 706.5 – 20.4375
= 686.0625 cm²

OR

Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution :
Circumference = 2πr = 22
2 × 22/7 × r = 22
r = 7/2 cm
Area of quadrant = ¼ × πr² = 1/4 × 22/7 × (7/2)² = 77/8 cm²

SECTION – C
(6 questions of 3 marks each)

26. Prove that √3 is irrational.
Solution :
Let, if possible, √3 be a rational no.
√3 = p/q, where p and q are co-prime integers and q ≠ 0.
⇒ 3 = p²/q²
⇒ p² = 3q² …………(i)
⇒ 3 divides p²⇒ 3 divides p also.
Let p = 3m ………..(ii) where m is any integer.
⇒ p² = 9m² …………….(iii)
From (i) and (iii)
3q² = 9m²
⇒ q² = 3m²
⇒ 3 divides q² ⇒ 3 divides q also.
Hence, p and q have 3 as common factor.
∴ p and q are not co-prime.
Hence our supposition is wrong.
∴ √3 is an irrational number.

27. Find the zeroes of the quadratic polynomial 6x² – 3 – 7x and verify the relationship between the zeroes and the coefficients.
Solution :
Standard form of quadratic polynomial is ax² + bx + c.
6x² – 3 – 7x = 6x² – 7x – 3
6x² – 9x + 2x – 3 = 0
3x(2x-3) + 1(2x-3) = 0
(2x – 3)(2x + 1) = 0
so, x = 3/2, -1/3
Zeroes of polynomial are α = 3/2, β = -1/3
Sum of zeroes = α + β = 3/2 – 1/3 = 7/6 = -(-7/6) = -b/a = -coefficient of x / coefficient of x²
Product of zeroes = α × β = 3/2 × (-1/3) = -1/2 = c/a = constant term / coefficient of x²

28. Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
Solution :
Let the number of Rs. 50 and Rs. 100 notes be x and y respectively.
50x + 100y = 2000
x + 2y = 40 ……..(i)
Also, Meena got 25 notes in all.
x + y = 25 ……..(ii)
Subtract eqn.(ii) from eqn.(i), we get
y = 15
Put y = 15 in eqn.(i), we get x = 10
so, the number of Rs. 50 and Rs. 100 notes be 10 and 15 respectively.

OR

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages ?
Solution :
Let Jacob’s age be x years and his son’s age be y years.
Five years hence(later),
x + 5 = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 10 ……..(i)
Also, five years ago(before),
x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = -30 ……..(ii)
Subtract eqn.(ii) from eqn.(i), we get
4y = 40
y = 10
Put y = 10 in eqn.(i),
x – 3(10) = 10
x = 40
Thus, present age of Jacob (x) = 40 years and
present age of Jacob’s son (y) = 10 years.

29. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution :

Given : ABCD be a parallelogram. circumscribing a circle with centre O.
To Prove : ABCD is a rhombus.
Proof : We know that the tangents drawn to a circle from an exterior point are equal is length.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS
Adding,
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
∴ AB + CD = AD + BC
or 2AB = 2AD (since AB = DC and AD = BC of parallelogram ABCD)
∴ AB = BC = DC = AD
Therefore, ABCD is a rhombus

30. If sinθ + cosθ = √3, then prove that tanθ + cotθ = 1.
Solution :
sinθ + cos θ = √3
squaring on both sides
(sinθ + cos θ)² = 3
sin²θ + cos²θ + 2sinθcosθ = 3
1 + 2sinθcosθ = 3 [∵ sin²θ + cos²θ = 1]
2sinθcosθ = 3 – 1
2sinθcosθ = 2
sinθcosθ = 2/2
sinθcosθ = 1 = sin²θ + cos²θ
1 = (sin²θ+cos²θ) ÷ (sinθcosθ)
1 = tanθ + cotθ
tanθ + cotθ = 1
Hence Proved.

OR

Prove that :
1 + [cot²θ/(1+cosecθ)] = cosecθ
Solution :
LHS = 1 + [cot²θ/(1+cosecθ)]
= 1 + [(cosec²θ-1)/(1+cosecθ)] [∵ cot²θ = cosec²θ – 1]
= 1 + [(cosecθ-1)(cosecθ+1)/(1+cosecθ)]
= 1 + (cosecθ-1)
= cosecθ = RHS
Hence Proved.

31. All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that card has a value
[Hint : In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed,
then total no. of remaining cards = 52 – 3×4 = 40]
(i) 7
Solution :
No. of favourable outcomes to card value 7= 4 because card value 7 may be of a spade, a diamond, a club or a heart
P(card value 7) = (no.of favourable outcomes to the event) ÷ (Total no. of possible outcomes) = 4/40 = 1/10

(ii) greater than 7
Solution :
Cards having value greater than 7 are from 8, 9 or 10
No. of favourable outcomes = 3 × 4 = 12
P(card having value greater than 7) = 12/40 = 3/10

(iii) less than 7
Solution :
Cards having value less than 7 are from 1, 2, 3, 4, 5 or 6
No. of favourable outcomes = 6 × 4 = 24
P(card having value less than 7) = 24/40 = 3/5

SECTION – D
(4 questions of 5 marks each)

32. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed ?
Solution :
Let original speed of the train be x km/h.
Then, time taken to travel 63 km = 63/x hours
New speed = (x+6) km/hr
Time taken to travel 72 km = 72/(x + 6) hours
ATQ,
63/x + 72/(x+6) = 3
(63x+378+72x)/(x²+6x) = 3
135x + 378 = 3x² + 18x
3x² – 117x – 378 = 0
x² – 39x – 126 = 0 [∴ divide both side by 3]
x² – 42x + 3x – 126 = 0
x(x-42) + 3(x-42) = 0
(x – 42)(x + 3) = 0
x = -3 or x = 42
As the speed cannot be negative, x = 42
Thus, the average speed of the train is 42 km/h.

OR

A motor boat 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.
Solution :
Let the speed of the stream be x km/h.
The speed of the boat upstream = (18-x) km/h
And the speed of the boat downstream = (18+x) km/h
We know that, time = distance/speed
Time taken to go upstream = 24/(18-x) hours
Also, time taken to go downstream = 24/(18+x) hours
ATQ,
24/(18-x) – 24/(18+x) = 1
24(18+x) – 24(18-x) = (18-x)(18+x)
x² + 48x – 324 = 0
x² + 54x – 6x – 324 = 0
x(x+54) – 6(x+54) = 0
(x + 54)(x – 6) = 0
x = -54 or 6
Since x is the speed of the stream, it cannot be negative.
so, the speed of the stream as 6 km/h.

33. Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Solution :

To Prove : AD/DB = AE/EC
Construction : Draw EM ⊥ AB and DN ⊥ AC. Join B to E and C to D.
Proof : In ∆ADE and ∆BDE,
ar(∆ADE) ÷ ar(∆BDE) = (½×AD×EM) ÷ (½×DB×EM) = AD/DB ……….(i)
In ∆ADE and ∆CDE,
ar(∆ADE) ÷ ar(∆CDE) = (½×AE×DN) ÷ (½×EC×DN) = AE/EC ……….(ii)
Since, DE || BC [Given]
∴ ar(∆BDE) = ar(CDE) ……….(iii)
[∆’s having same base and between the same parallel lines then they are equal in area]
From eqn.(i), eqn.(ii) and eqn.(iii),
AD/DB = AE/EC
Hence Proved.

34. A juice seller was serving his customer using glasses as shown in the figure. The inner diameter of the cylindrical glass was 5 cm but bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent and actual capacity of the glass. [Use π = 3.14]

Solution :
The inner radius of the glass = 5/2 cm = 2.5 cm
Height of the glass = 10 cm
The apparent capacity of the glass = πr²h = 3.14 × 2.5 × 2.5 × 10 = 196.25 cm³
Volume of hemisphere = ⅔πr³ = ⅔ × 3.14 × 2.5 × 2.5 × 2.5 = 32.71 cm³
The actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= 196.25 – 32.71 cm³
= 163.54 cm³

OR

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution :

Radius of base of cylindrical portion = 2 m
Height of the cylindrical portion = 2.1 m
Slant height of conical top = 2.8 m
Curved surface area of cylindrical portion = 2πrh = 2 × 22/7 × 2 × 2.1 = 26.4 m²
Curved surface area of conical portion = πrl = 22/7 × 2 × 2.8 = 17.6 m²
Total curved surface area Area of canvas used = 2πrh + πrl = 26.4 + 17.6 = 44 m²
Cost of canvas = Rate × Surface area = 500 × 44 = Rs.22000

35. The median of the following data is 525. Find the values of x and y, if total frequency is 100.

Solution :

n = 100 ⇒ n/2 = 50
76 + x + y = 100
x + y = 24 ……..(i)
Median = 525  (median class is 500-600)
l = 500, f = 20, cf = 36+x, h = 100
Median = l + (n/2 – cf)/f × h
525 = 500 + (50-36-x)/20 × 100
525 – 500 = (14-x) × 5
25 = 70 – 5x
5x = 70 – 25 = 45
x = 45/5 = 9
From eqn.(i), we get 9 + y = 24
y = 24 – 9 = 15

SECTION – E
(case study based questions 4 marks each)

36. Rahul wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month.

Based on the above information, answer the following questions :
(i) Find the amount paid by him in 30th instalment.
Solution :
Monthly instalment paid by Rahul are 1000, 1100, 1200, ……….. 30 terms.
a = 1000, d=1100-1000 = 100, n = 30
nth term of AP, an = a + (n-1)d
a30 = 1000 + (30-1)×100 = 1000 + (29)×100 = 1000 + 2900 = 3900
So, the amount paid by him in 30th instalment = ₹ 3900

(ii) Find the amount paid by him in 30 instalments.
Solution :
Total amount of all 30 instalments = 1000 + 1100 + 1200 + ……. + 3900
Here, a = 1000, d = 1100-1000 = 100, n = 30
Sum of n terms, Sn = n/2 [2a + (n-1)d]
S30 = 30/2 [2×1000 + (30-1)×100]
= 15 [2000 + 2900]
= 15 × 4900 = 73500
Total amount of all 30 instalments = ₹ 73500

(iii) What amount does he still have to pay after 30th instalment ?
Solution :
So, the loan amount left after 30th instalment = 118000 – 73500= ₹ 44500
Hence, he still has to pay ₹ 44500 after 30th instalment.

OR

If total instalments are 40 then amount paid in the last instalment.
Solution :
a40 = a + 39d = 1000 + 39×100 = 1000 + 3900 = 4900
Amount paid in last instalment = ₹ 4900

37. Resident welfare Association (RWA) of a society put up three electric poles A, B and C in a society’s park. Despite these three poles, some parts of the park are still in dark. So, RWA decides to have one more electric pole D in the park.

Based on the above information, answer the following questions :
(i) Find the position of the pole C.
Solution : C(7, 4)

(ii) Find the distance of the pole B from corner O of the park.
Solution : Here, B(4,9) and O(0,0)
x1 = 4, x2 = 0, y1 = 9, y2 = 0
Using Distance Formula,
AB = √(x2-x1)²+(y2-y1
= √(0-4)²+(0-9)²
= √(-4)²+(-9)²
= √16+81
= √97 units

(iii) Find the position of the fourth pole D so that four points A, B, C and D form a parallelogram.
Solution :
A(1,5), B(4,9), C(7,4) are three vertices of parallelogram ABCD and let D(x,y) be the fouth vertex.
Mid-point of diagonal AC = Mid-point of BD
[(7+1)/2, (5+4)/2] = [(x+4)/2, (9+y)/2]
x = 4, y = 0
∴ D(4,0)

OR

Find the distance between poles A and C.
Solution :
Here, A(1,5) and C(7,4)
x1 = 1, x2 = 7, y1 = 5, y2 = 4
Using Distance Formula,
AB = √(x2-x1)²+(y2-y1
= √(7-1)²+(4-5)²
= √(6)²+(-1)²
= √36+1
= √37 units

38. A group of students of class X visited India Gate on an educational 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 (formely called the Kingsway), is about 138 feet (42 metres) in height.

Based on the above information answer the following questions :
(i) What is the angle of elevation if they are standing at a distance of 42 m away from the monument ?
Solution :

tanθ = P/B = 42/42 = 1 = tan45°
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.
Solution :

tan60° = P/B = 42/x
√3 = 42/x
x = 42/√3 = 14√3 m

(iii) If the altitude of the Sun is at 60°, then find the height of the vertical tower that will cast a shadow of length 20 m.
Solution :

tan60° = P/B = h/20
√3 = h/20
h = 20√3 m

OR

The ratio of the length of a rod and its shadow is 1 : 1. Find the angle of elevation of the Sun.
Solution :

tanθ = P/B = L/X = 1/1 = 1 = tan45°
θ = 45°

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