Mathematics | Class 10th

CBSE Board Paper 2017-18

2

1. If x = 3 is one root of the quadratic equation x2 – 2kx – 6 = 0, then find the value of k.

CBSE Board Paper 2017-18 Set - 1

General Instructions:

1. All Questions are compulsory.

2. This question paper consists of 30 questions divided into four

sections – A, B, C and D.

3. Section A contains 6 questions of 1 mark each. Section B

contains 6 questions of 2 marks each. Section C contains

10 questions of 3 marks each. Section D contains 8 questions of

4 marks each.

4. There is no overall choice. However, an internal choice has been

provided in two questions of 1 mark, two questions of 2 marks,

four questions of 3 marks each and three questions of 4 marks

each. You have to attempt only one of the alternatives in all such

questions.

5. Use of calculator is not permitted.

Time allowed: 3 Hours Max Marks: 80

Section A

1

3

2. What is the HCF of the smallest prime number and the smallest composite number?

1

3. Find the distance of a point P (x, y) from the origin.

1

4. In an AP, if the common difference (d) = – 4, and the seventh term (a7) is 4, then find the first term.

1

5. What is the value of (cos267° – sin223°)?

1

6. Given ∆ABC ~ ∆PQR, if

, then find

1

2

8. ABCD is a rectangle. Find the values of x and y.

2

Section B

7. Given that √2 is irrational, prove that (5 + 3 √2) is an irrational number.

4

9. Find the sum of the first 8 multiples of 3.

2

10. Find the ratio in which P (4, m) divides the line segment joining the points A (2, 3) and B (6, –3). Hence find m.

11. Two different dice are tossed together. Find the probability:

(i) of getting a doublet

(ii) of getting a sum 10, of the numbers on the two dice.

2

12. An integer is chosen at random between 1 and 100. Find the probability that it is:

(i) divisible by 8. (ii) not divisible by 8.

2

2

5

Section C

13. Find HCF and LCM of 404 and 96 and verify that HCF x LCM = Product of the two given numbers.

3

14. Find all zeroes of the polynomial (2x4 – 9x3 + 5x2 + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3)

3

15. If A (–2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

3

OR

If A (–5, 7), B (–4, –5), C (–1, –6) and D (4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

16. A plane left 30 minutes late than its scheduled time and to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

3

17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonals.

3

OR

If the area of two similar triangles is equal, prove that they are congruent.

6

18. Prove that the lengths of tangents drawn from an external point to a circle are equal.

3

19. If 4 tanθ = 3, evaluate

3

OR

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

20. Find the area of the shaded region in Fig. 2, where arcs are drawn with centres A, B, C and D intersect in pairs at mid - points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. [Use π = 3.14]

3

21. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown below. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article

3

7

Section D

22. The table below shows the salaries of 280 persons:

Calculate the median salary of the data.

3

23. A motor boat whose speed is 18 km/hr in still water takes 1hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

4

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/ hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

OR

OR

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much is canvas cloth required just to cover the heap?

8

24. The sum of four consecutive numbers in an AP is 32, and the ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Find the numbers.

4

25. In an equilateral ∆ABC, D is a point on side BC such that

. Prove that 9(AD)2 = 7(AB)2

4

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

4

OR

26. Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the ∆ ABC.

27. Prove that:

4

28. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

(i) The area of the metal sheet used to make the bucket. (ii) Why should we avoid the bucket made by ordinary

plastic? [Use π = 3.14]

4

29. As observed from the top of a 100m high light house from the sea - level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use 3 = 1.732]

4

9

30. The mean of the following distribution is 18. Find the frequency f of class 19 – 21. Class Frequency 11-13 3 13-15 6

15-17 9 17-19 13 19-21 f

21-23 5 23-25 4

4

OR

The following distribution gives the daily income of 50 workers of a factory

Daily Income (Rs.) No. of workers 100-120 12

120-140 14 140-160 8 160-180 6

180-200 10

10

1. Given: x = 3 is a root of the equation f(x) = x2 – 2kx – 6 = 0.

To find: The value of k.

Explanation:

We know that if x = a is a root of a polynomial f(x), then f(a) = 0.

So, we can say that f(3) = 0

f(3) = 32 – 2k(3) – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ 3 – 6k = 0

⇒ - 6k = -3

⇒ 6k = 3

⇒ k = 1/2

Hence, k = 1/2

2. Smallest Prime Number = 2

Prime Factorization of 2 = 1 × 2

Smallest Composite Number = 4

Prime Factorization 4 = 1 × 2 × 2

we know that H.C.F is the highest common factor or the common factors of two numbers.

Therefore,

H.C.F of 2 and 4 is 1 × 2 = 2.

Solutions (Set-1)

Section A

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3. Given: Point P (x, y)

To Find: Length OP, where point O (0, 0)

Explanation:

From the figure,

Using Pythagoras Theorem in ΔPOQ,

(∵ PQR is a right triangle with OP as the hypotenuse)

Hence, the distance between point P and origin is .

4. Given:

• d = – 4

• a7 = 4

To find: a (first term)

Explanation:

Now, in an AP we know,

an = a + (n – 1)d

Where an is the nth term in the AP, d is the common difference, and a is the first term in the AP.

Putting values in the above equation for n = 7

12

a7 = a + 6d

⇒ 4 = a + 6( – 4)

⇒ 4 = a - 24

a = 28

∴ The first term is 28.

5. Consider (cos267° – sin223°),

Formula Used:

..... (1)

Putting θ = 67° in (1) we get,

sin223° = cos267°

Putting this in the given equation (cos267° – sin223°),

⇒ sin223° – sin223°

= 0

Hence, (cos267° – sin223°) = 0

6. Since ∆ABC ~ ∆PQR,

As we know,

Two triangles are similar if they have all their angles equal and corresponding sides are in the same ratio.

So,

Theorem Used: The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides.

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7. We will prove this statement by contradiction.

Let us assume that (5 + 3√2) is a rational number. This means that:

Where “a” and “b” have an HCF as 1 which means a and b are co-primes.

Now,

Since a and b are integers.

a - 5b will also be an integer.

This implies that is a rational number.

This is a contradiction as it is given in the question that √2 is an irrational number, which means that it cannot be expressed as a fraction of two numbers “a” and “b” with HCF as 1.

∴ Our initial hypothesis is wrong.

Hence, (5 + 3√2) is an irrational number.

Section B

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8. ABCD is a Rectangle.

This means:

AB = DC and AD = BC

(∵ opposite sides of Rectangle are equal)

x + y = 30 .... (1)

x – y = 14 .... (2)

Add (1) and (2) to get,

(x + y) + (x – y) = 30 + 14

⇒ 2x = 44

⇒ x = 22

Putting the value of x in equation (1) to get,

22 + y = 30

⇒ y = 30 – x

⇒ y = 30 – 22

= 8

⇒ x = 22 and y = 8

9. To Find: Sum of first 8 multiples of 3

Explanation:

Multiples of 3 are 3,6,9, ......

This means we have to find the sum 3 + 6 + 9 + ......+ (up to 8 multiples)

This sum can be found out using AP whose:

• First Term (a) = 3

• Common Difference (d) = 3

• Number of Terms (n) = 8

As we know,

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where,

n = number of terms,

a = first term

d = common difference.

Putting the necessary values,

Required Sum

= 4[6 + 7(3)]

= 4 (6 + 21)

= 4(27)

Required Sum = 108

10. The required line is shown below:

Using the section formula,

Coordinates of P which divides AB having points A(x1, y1) and B(x2, y2) in, say k:1, are given by:

According to our question,

For, A(2, 3) and B(6, –3), let the ratio be k: 1,

We get:

Equating the coefficients, we get,

⇒ 6k + 2 = 4k + 4 ⇒ 2k = 2 ⇒ k = 1

and also,

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Putting the value of k, we get

⇒ m = 0

The ratio is 1:1 and the value of m is 0.

11. Let S be a set of all possible outcomes when 2 die are tossed

together.

Number of elements in S i.e. n(S) = 36

As we know,

(i) A: getting a doublet

Favorable outcomes = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

n(A) = 6

(ii) B: getting a sum 10

Favorable outcomes = {(4,6), (5,5), (6,4)}

n(B) = 3

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12. S = (2,3, ......99)

Total possible outcomes between 1 and 100 = 98

As we know,

(i) Now, numbers divisible by 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.

So, total favorable observations = 12

P(divisible by 8)

(ii) P(not divisible by 8) = 1 – P(divisible by 8)

{As Probability of compliment of an event = 1 - Probability of an event}

13. Prime Factorising the given numbers:

404 = 2 × 2 × 101

96 = 2 × 2 × 2 × 2 × 2 × 3

Section C (Solutions)

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Now,

H.C.F = Multiplication of common factors of numbers. Therefore,

HCF = 2 x 2

= 4

L.C.M = Multiplication of all the factors of numbers. Therefore,

LCM = 2 x 2 × 101 × 2 × 2 × 2 × 3

= 9696

HCF x LCM = 9696×4

= 38784

Product of Two Numbers = 404×96

= 38784

So,

HCF x LCM = Product of Two Numbers

Hence, Verified.

14. Given:

• f(x) = (2x4 – 9x3 + 5x2 + 3x – 1)

• Zeroes = (2 + √3) and (2 – √3)

Given the zeroes, we can write the factors = [x – (2 + √3)] and [x – (2 – √3)]

{Since, If x = a is zero of a polynomial f(x), we can say that x - a is a factor of f(x)}

Multiplying these two factors, we can get another factor which is:

((x – 2) - √3) ((x – 2) + √3) = (x – 2)2 – (√3)2

(Using (a2 – b2) = (a + b) (a – b))

⇒x2 + 4 – 4x – 3 = x2 – 4x + 1

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So, dividing f(x) with (x2 – 4x + 1)

f(x) = (x2 – 4x + 1) (2x2 – x – 1)

Solving (2x2 – x – 1), we get the two remaining roots by the formula:

where f(x) = ax2 + bx + c = 0

Here a = 2, b = -1 and c = -1

(using Quadratic Formula)

15. Given:

Parallelogram ABCD

A (– 2, 1), B (a, 0), C (4, b), D (1, 2)

Now, Midpoint of AC = Midpoint of BD

(∵ diagonals of a parallelogram bisect each other)

Using Midpoint Formula i.e.

The midpoint of the segment joining and is given

by

Putting values,

The midpoint of AC

The midpoint of BD

Now,

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⇒ a + 1 = 2 ⇒ a = 1

And,

⇒ 1 + b = 2 ⇒ b = 1

⇒ a = 1 and b = 1

So, coordinates are A( – 2, 1), B(1, 0), C(4, 1), D(1, 2).

Using Distance Formula i.e.

Distance between (x1, y1) and (x2, y2 ) is given by:

D =

Now,

And,

OR

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Ar (ABCD) = Ar (ABC) + Ar (ADC)

The area of the triangle with points (x1 , y1), (x2 , y2) and (x3 , y3) is:

• Ar(Δ ABC)

• Ar (Δ ADC)

So, Ar(ABCD) = Ar(Δ ABC) + Ar (Δ ADC)

= 72 sq. units

16. Let the usual Speed = x km/h

To Find: Usual Speed (x)

Given: Distance = 1500km

Formula Used:

Case 1: With Usual Speed,

Time = 1500/x

Case 2: Increased speed = (x + 100) km/h

Time Taken (with fast speed) =

According to the question, the plane leaves 30 minutes late and reaches on time, which means,

Time taken with usual speed - Time taken by fast speed = 30 minutes

= 1/2 hour

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After cross multiplying, we get,

⇒ 1500(2) (x + 100) = x (x + 100) + 1500(2)(x)

⇒ x2 + 100x – 300000 = 0

Using Quadratic Formula,

For an equation ax2 + bx + c = 0,

Comparing the equation x2 + 100x – 300000 = 0 with ax2 + bx + c = 0, we get,

a = 1, b = 100 and c = -300000

As speed cannot be negative, x = 500 km/h

So, the usual speed is 500 km/h.

17. Given:

• Square ABCD

• Equilateral Triangles BEC and AFC

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In ∆BEC and ∆AFC,

∠BEC = ∠AFC = 60°

∠BCE = ∠ACF = 60°

(∵ ∆BEC and ∆AFC both are equilateral triangles)

So, by AA rule,

∆BEC ~ ∆AFC

So,

(∵ Ratio of areas of similar triangles equals the ratio of the square of their corresponding sides)

Now,

Hence, the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonals.

Given:

• ∆ ABC ∆ PQR

• Ar(∆ ABC) = Ar(∆ PQR)

We know,

(∵ Ratio of areas of similar triangles equals to the ratio of the square of their corresponding sides)

OR

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⇒ BC = QR and AB = PQ AC = PR

∴ ∆ ABC ≅ ∆ PQR by SSS rule.

18.

Given:

• Circle with centre A

• Tangents IB and IF

To Prove: IB = IF

Proof:

In ∆ ABI and ∆ AFI

• AB = AF = Radius of the circle

• ∠ ABI = ∠ AFI = 90°

(∵ tangents are perpendicular to the radius)

• AI = AI (common side)

So, by RHS rule, ∆ ABI ≅ ∆AFI

⇒ BI = FI

(∵ corresponding sides of congruent triangles are equal)

Hence, the lengths of tangents drawn from an external point to a circle are equal.

19. Given:

We can picture this using the right triangle below:

25

Now, Using the Pythagoras Theorem i.e.

AC = 5k

Now, Using the formula,

Putting the values in the expression, we get,

Hence,

Given:

• tan 2A = cot (A – 18°)

• 2A is Acute

We know,

OR

26

Putting θ = 2A

Now, 2A is acute.

(A – 18°) and (90° – 2A) are acute.

⇒ 90° – 2A = A – 18°

20. Given:

• Square ABCD with side 12 cm

• 4 quarter circles on each of the four vertices, each with a radius(r) 6 cm

Required Area = Ar(PQRS)

(area enclosed within red arcs)

= Area of Square ABCD – Area of Four quarter circles

= Area of Square (side 12cm) – Area of a full circle (radius 6cm)

= (12×12) – π (6)2

(Area of Square = (Side)2; Area of Circle = πr2)

= 144 – 3.14(36) = 144 - 113.04

= 30.96 cm2

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21. Given:

• Cylinder with height(h) = 10cm and diameter(2r) = 7cm

• Hemispheres cut from the top and bottom

Total Surface Area = Curved Surface Area of Cylinder + Curved Surface Area of 2 hemispheres

= CSA of Cylinder + Total Surface Area of a Sphere

CSA of Cylinder = 2πrh where “r” is the radius of the cylinder

“h” is the height of the cylinder

TSA of Sphere = 4πr2 where “r” is the radius of the sphere

So,

Total Surface Area of the Article = 2πrh + 4πr2

= 2π (3.5) (10) + 4π (3.5)2

= 70π + 4π (12.25)

= 70π + 49π

= 119π

= 17(22)

= 374 sq. cm

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Given:

• Height of Cone(h) = 3.5m

• Radius of the base(r) = 12m

The volume of the cone

where “r” is the radius, “h” is the height of the cone

= 24 × 22 = 528 m3

And, Curved Surface Area of the Cone = πrl

where “l” is the slant height of the cone

Now,

(using the Pythagoras theorem)

OR

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= 12.5m

CSA of Cone = π (12) (12.5)

= 150π

= 21.428 × 22

= 471.42 m2

So, 528 m3 is the volume of the rice and, 471.42 m2 of canvas cloth is required to cover the heap.

22.

Where N = total samples

C.F = Cumulative frequency of class preceding median class

f = Frequency of the median class

h = width of the Class interval

l = Lower limit of median class

(Note: The median class is the class under which observation falls)

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, which falls in (10 – 15) which is now

the median class

Putting necessary values in the formulae

= 10 + 3.421 = 13.421 thousand rupees

= Rs 13421

23. Let Speed of the stream be x km/h

Formula Used:

Going Upstream:

Speed = Speed of boat in still water - speed of stream

Speed = (18 - x) km/h

Distance = 24 km

Going Downstream:

Speed = Speed of boat in still water + speed of stream

Section D (Solutions)

31

Speed = (18 + x) km/h

Distance = 24 km

According to the question,

Time taken Upstream = Time taken downstream + 1hr

⇒ x2 + 48x – 324 = 0

⇒ x2 + 54x – 6x – 324 = 0

⇒ x (x + 54) – 6(x + 54) = 0

⇒ (x – 6) (x + 54) = 0

⇒ X = 6 or – 54

Since “x” is the speed, it has to be non-negative

∴ x = 6 km/h

So, the speed of the stream is 6km/h.

Let Original Average Speed be x km/h

According to the question, the total time taken is 3 hrs

⇒

⇒

OR

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⇒ 3x2 – 117x – 378 = 0 ⇒ x2 – 39x – 126 = 0

⇒ x2 – 42x + 3x – 126 = 0 ⇒ x (x – 42) + 3(x – 42) = 0

⇒ (x + 3) (x – 42) = 0 ⇒ x = – 3 or 42

As “x” is speed, it has to be non-negative.

∴ x = 42 km/h

So, the original average speed is 42 km/h.

24. Let the numbers be: (a – 3d), (a – d), (a + d), (a + 3d)

(Note the common difference is “2d”)

According to Question,

Sum of four consecutive terms in A.P = 32

(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32 ⇒ a = 32/4

⇒ a = 8

So, the numbers are: (8 – 3d), (8 – d), (8 + d), (8 + 3d)

Also,

The ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Therefore,

⇒

(Using (a + b) (a – b) = a2 – b2)

960 - 135 d2 = 448 - 7 d2

⇒ 128d2 = 512 ⇒ d2 = 4

⇒ d = ±2

So, putting d = 2 and a = 8

Numbers are (8 – 3(2)), (8 – 2), (8 + 2), (8 + 3(2))

= (8-6), (6), (10), (8+6)

= 2,6,10,14

numbers are: 2, 6, 10, 14

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So, putting d = -2 and a = 8

Numbers are (8 – 3(-2)), (8 – (-2)), (8 + (-2)), (8 + 3(-2))

= (8+6), (8+2), (8-2), (8-6) = 14,10,6,2

Numbers are: 14,10,6,2

Hence, the required numbers are 2, 6, 10, 14.

25.

Given:

• Equilateral Triangle ABC with side say “x”units

•

Now, let’s construct AE ⊥ BC

In ∆ AEC and ∆ AEB

• AC = AB = x

• ∠ AEC = ∠ AEB = 90°

• AE = AE (common side)

So, by RHS rule, ∆ AEC ≅ ∆ AEB

∴ CE = BE

(∵ Corresponding sides of congruent triangles are equal)

34

As

[given in the question]

BD

So,

Using Pythagoras Theorem,

(Hypotenuse)2 = (Base)2 + (Perpendicular)2

AE2 = AC2 – EC2

Similarly, AD2 = AE2 + ED2

And, AB2 = x2

∴

Hence, Proved.

Given:

• Square ABCD with side (a + b) units

• Points E, F, G, H on the sides such that they divide the segment in the same proportion.

Now,

In ∆ EAF and ∆ HBE,

OR

35

∠ A = ∠ B

EA = HB = b

AF = BE = a

SAS rule: Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

So, by SAS rule, ∆ EAF ≅ ∆ HBE

Similarly, ∆ HBE ≅ ∆ GCH and ∆ GCH ≅ ∆ FDG

∴ ∆ EAF ≅ ∆ HBE ≅ ∆ GCH ≅ ∆ FDG

EF = FG = GH = HE = c

(∵ corresponding sides of congruent triangles are equal)

Also, ∠ AEF = ∠ BHE, which implies,

∠ AEF + ∠ BEH + ∠ FEH = 180°

(Straight line measures 180°)

⇒ (∠ BHE + ∠ BEH) + ∠ FEH = 180°

(∵ ∠ BHE = ∠ AEF)

⇒ ∠ FEH = 180° – (∠ BHE + ∠ BEH)

= 180° – 90° = 90°

(∵ ∆ BEH is a Right Triangle)

Similarly, ∠ EHG = ∠ HGF = ∠ GFE = 90°

EFGH is a square with side “c”

Area of square = (a + b)2

Also, Area of square = Area of Constituent figures

= 2ab + c2

⇒ (a + b)2 = 2ab + c2 ⇒ a2 + b2 + 2ab = 2ab + c2

36

⇒ a2 + b2 = c2

This proves the Pythagoras Theorem, i.e. square of the hypotenuse is equal to the square of other two sides.

26. Given in Δ ABC,

• Length of side BC = 6 cm.

• Length of side AB = 5 cm.

• ∠ ABC = 60°.

Steps of Construction:

1. Draw a line segment BC of length 6 cm.

2. With B as center, draw a line which makes an angle of 60° with BC.

Construction of 60° angle at B:

a. With B as centre and with some convenient radius draw an arc which cuts the line BC at D.

b. With D as radius and with same radius (in step a), draw another arc which cuts the previous arc at E.

c. Join BE. The line BE makes an angle 60° with BC.

37

3. Again with B as centre and with radius of 5 cm, draw an arc which intersects the line BE at point A.

4. Join AC. This is the required triangle.

5. Now, from B, draw a ray BX which makes an acute angle on the opposite side of the vertex A.

6. With B as center, mark four points B1, B2, B3 and B4 on BX such that they are equidistant. i.e. BB1 = B1B2 = B2B3 = B3B4.

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7. Join B4C and then draw a line from B3 parallel to B4C which meets the line BC at P.

8. From P, draw a line parallel to AC and meets the line AB at Q. Thus, ΔBPQ is the required triangle.

39

27.

Proof:

LHS=

(Taking sinA common from numerator and cosA common from denominator)

⇒

Replacing 1 with

⇒ tanA

= RHS

Hence, proved.

28.

(i) Given: R = 15cm; r = 5cm; h = 24cm

Curved Surface Area of Frustum

Where, R = Larger radius of the frustum

r = Smaller radius of the frustum

h = Height of the frustum

40

CSA of the frustum

Area of the base of the bucket = Area of circle

where “r” is the radius of the circle

= π (5)2 = 3.14(25)

= 78.5 sq.cm

So, the total area of the bucket = CSA of the frustum + Area of the circle

= 1632.8 + 78.5 = 1711.3 sq. cm

1711.3 sq. cm of metal sheet is required to make the bucket.

(ii) We should use biodegradable materials for the bucket which can be decomposed by the natural processes of the environment. Ordinary plastic is not biodegradable and can cause soil pollution. It can also harm animals if consumed.

29.

Given:

• AB = 100m

• ∠ EAD = 30°

• ∠ EAC = 45°

To find: CD, i.e. the distance between the boats

In ∆ ABC,

∠ ACB = ∠ EAC = 45°

(∵ alternate angles over parallel lines are equal)

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tan∠ACB = tan45° = 1

In ΔABC,

⇒ AB = BC = 100m

In ∆ ABD,

∠ ADB = ∠ EAD = 30°

(∵ alternate angles are equal)

⇒ BD = 100√3 m

So, CD (distance between the boats) = BD – BC = 100(√3 – 1) m

= 73.2 m (approx.)

30.

Mean

According to question,

Mean = 18

Cross-multiplying, we get,

⇒ 704 + 20f = 720 + 18f ⇒ 2f = 16

⇒ f = 16/2 ⇒ f = 8

42

Plotting the above points on the graph, we get,

OR

43