# CBSE Class 10 Mathematics Previous Year Question Paper 2020 Delhi Set - 3

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Question numbers 1 to 10 are multiple choice questions of 1 mark each.

Select the correct option.

Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is

A. (2, 0)

B. (0, 2)

C. (3, 0)

D. (2, 2)

Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are

A. (3, 5)

B. (3, -5)

C. (-3, -5)

D. (-3, 5)

Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is

A. 4

B. 3

C. 2

D. 1

Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is

A. 3

B. 4

C. 5

D. 2

Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and

A. 0 < r < b

B. 0 < r < b

C. 0 ≤ r < b

D. 0 ≤ r ≤ b

Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are

A. m, m+3

B. –m, m+3

C. m, –(m+3)

D. –m, –(m+3)

Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is

A.

B.

C. 5

D. 10

Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are

A. ±0.2

B. ±0.02

C. 0.4

D. 2

Q 9. The common difference of the A.P.

A. 1

B.

C. -1

D. 2

Q 10. The nth term of the A.P. a, 3a, 5a, …. Is

A. na

B. (2n – 1)a

C. (2n + 1)a

D. 2na

In Q. Nos. 11 to 15, fill in the blanks. Each question carries 1 mark:

Q 11. In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are ______, ______.

Q 12. In ΔABC, AB=6  cm, AC = 12 cm and BC = 6 cm, then ∠B=______.

OR

Two triangles are similar if their corresponding sides are ______.

Q 13.  In given Fig. 2, the length PB=______cm.

Q 14. In fig. 3, MN || BC and AM : MB =1 : 2, then

Q 15. The value of sin 32° cos 58° +cos 32° sin 58° is ______.

OR

The value of   is ______.

Q Nos. 16 to 20 are short answer type questions of 1 mark each.

Q 16. A die is thrown once. What is the probability of getting a prime number?

Q 17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3, then find the probability of x< 4.

OR

What is the probability that a randomly taken leap year has 52 Sundays?

Q 18. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).

Q 19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π =3.14).

Q 20. Find the class marks of the classes 20 – 50 and 35 – 60.

Q.Nos.21 to 26 carry 2 marks each.

Q 21. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper.

The following were the answers given by the students:

2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3  + 7, 7x +  , 5x3 – 7x + 2, 2x2 + 3 –  , 5x –  , ax+ bx2 + cx + d, x +  .

1. How many of the above ten, are not polynomials?
2. How many of the above ten, are quadratic polynomials?

Q 22 .A child has a die whose six faces show the letters as shown below:

The die is thrown once. What is the probability of getting (i) A, (ii) D?

Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that .

OR

In fig.5, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.

Q 24. Prove that

OR

Show that tan4θ + tan2θ = sec4θ - sec2θ

Q 25. Find the mode of the following frequency distribution:

Q 26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.

Q. Nos. 27 to 34 carry 3 marks each.

Q 27. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that .

Q 28. The area of a circular play ground is 22176 cm2. Find the cost of fencing this ground at the rate of Rs. 50 per metre.

Q 29. If the mid – point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.

OR

Find the area of triangle ABC with A (1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).

Q 30. If Fig.6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.

Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and .

OR

Solve for

Q 32. Which term of A.P. 20, 19, 18, 17,....  is the first negative term.

OR

Find the middle term of the A.P. 7, 13, 19, …., 247.

Q 33. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8cm standing water is required?

Q 34. Show that:

Q. Nos. 35 to 40 carry 4 marks each.

Q 35. The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.

OR

The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a ‘more than’ type distribution and draw its ogive.

Q 36. From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.

Q 38. Prove that  is an irrational number.

Q 39. Draw a circle of radius 3.5cm. From a point P, 6cm from its centre, draw two tangents to the circle.

OR

Construct a ΔABC with AB=6cm, BC=5cm and ∠B=60°. Now construct another triangle whose sides are   times the corresponding sides of ΔABC.

Q 40. A solid is in the shape of hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7cm and height of cone is 3.5cm, find the volume of the solid. (Take π = ).

Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is

A. (2, 0)

B. (0, 2)

C. (3, 0)

D. (2, 2)

Solution: Correct option : A

Let P(x, 0) … (Since, the point P is on x – axis)

The points A(–1, 0) and B(5, 0) are also lying on x – axis and P is equidistant from

the points A and B.

⇒ P is a midpoint of AB.

By using mid – point formula, we get

Therefore, the coordinates of the point P is (2, 0).

Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are

A. (3, 5)

B. (3, -5)

C. (-3, -5)

D. (-3, 5)

Solution: Correct option : C

The coordinates of the point which is the reflection of point (–3, 5) is (–3, –5).

Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is

A. 4

B. 3

C. 2

D. 1

Solution: Correct option : D

The point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1.

By using section formula, we get

Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is

A. 3

B. 4

C. 5

D. 2

Solution: Correct option : B

Prime factorisation of 196 = 2 × 2 × 7 × 7 = 22 × 72

The sum of the exponents of the prime factors in the prime factorisation of 196 is 2 + 2 = 4

Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and

A. 0 < r < b

B. 0 < r < b

C. 0 ≤ r < b

D. 0 ≤ r ≤ b

Solution: Correct option : C

Euclid’s Division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and 0 ≤ r < b.

Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are

A. m, m+3

B. –m, m+3

C. m, –(m+3)

D. –m, –(m+3)

Solution: Correct option: B

p(x) = x2 – 3x – m(m + 3)

Substitute x = –m

p(–m) = (–m)2 – 3(–m) – m(m + 3)= m2 + 3m – m2 – 3m = 0  ...(i)

Substitute x = m + 3

p(m + 3) = (m + 3)2 – 3(m + 3) – m(m + 3)

= (m + 3)(m + 3 – 3 – m) ...(ii)

From (i) and (ii),

–m and m + 3 are the zeroes of the given polynomial.

Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is

A.

B.

C. 5

D. 10

Solution: Correct option: D

Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are

A. ±0.2

B. ±0.02

C. 0.4

D. 2

Solution: Correct option: A

Taking square root on both the side, we get

Q 9. The common difference of the A.P.

A. 1

B.

C. -1

D. 2

Solution: Correct option: C

Q 10. The nth term of the A.P. a, 3a, 5a, …. Is

A. na

B. (2n – 1)a

C. (2n + 1)a

D. 2na

Solution: Correct option: B

Here a1 = a, d = 3a – a = 2a

We know that,

an = a1 + (n – 1)d

= a + (n – 1)2a

= a + 2an – 2a

= 2an – a = (2n – 1)a

In Q. Nos. 11 to 15, fill in the blanks. Each question carries 1 mark:

Q 11. In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are ______, ______.

Solution: In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are 30° and 45°.

In ΔO1CA,

⇒ ∠O1CA + ∠CAO1 + ∠CO1A = 180°

⇒ 90° + ∠CAO1 + 60° = 180°

⇒ ∠CAO1 = 30°

So, the angle of depression from the observing position O1 of the object A is 30°

And the angle of depression from the observing position O2 of the object A is 45°.

Q 12. In ΔABC, AB=6  cm, AC = 12 cm and BC = 6 cm, then ∠B=______.

OR

Two triangles are similar if their corresponding sides are ______.

Solution: In ∆ABC, AB = , AC = 12 cm and BC = 6 cm, then ∠B = 90°.

AB2 = = 108, BC2 = (6)2 = 36 and AC2 = (12)2 = 144

Here, AC2 = AB2 + BC2

By the converse of Pythagoras theorem, we get

m∠B = 90°.

OR

Two triangles are similar if their corresponding sides are in the same ratio.

Q 13.  In given Fig. 2, the length PB=______cm.

Solution: In given Fig 2, the length PB = 4 cm.

AB is a tangent to the internal circle at P and OP is its radius.

⇒ OP ⊥ AB

In ∆APO,

OA2 = OP2 + AP2 ... (OP ⊥ AB)

⇒ AP2 = 52 – 32 = 16

⇒ AP = 4 cm

We know that,

In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

⇒ AP = PB = 4 cm

Q 14.  In fig. 3, MN || BC and AM : MB =1 : 2, then

Solution: In fig. 3 MN || BC and AM: MB = 1: 2, then .

AM: MB = 1: 2

Let AM = x and MB = 2x

⇒ AB = AM + MB = 3x

Here, ∆AMN ~ ∆ABC ... (AA test)

Q 15. The value of sin 32° cos 58° +cos 32° sin 58° is ______.

OR

The value of   is ______.

Solution: The value of sin 32° cos 58° + cos 32° sin 58° is 1.

sin 32° cos 58° + cos 32° sin 58°

= sin 32° cos(90° – 32°) + cos 32° sin(90° – 32°)

= sin 32° sin 32°  + cos 32° cos 32°

= sin232° + cos232°

= 1 …. (sin2θ + cos2θ = 1)

OR

The value of  is 2.

Explanation:

Q Nos. 16 to 20 are short answer type questions of 1 mark each.

Q 16. A die is thrown once. What is the probability of getting a prime number?

Solution: A die is thrown once.

S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6

Let A be the event of getting a prime number.

A = {2, 3, 5} ⇒ n(A) = 3

⇒ P(A) =  .

Q 17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3, then find the probability of x< 4.

OR

What is the probability that a randomly taken leap year has 52 Sundays?

Solution: Here, S = {–3, –2, –1, 0, 1, 2, 3} ⇒ n(S) = 7

Let A be the event that the square of the chosen number is less than 4.

A = {–1, 0, 1}

⇒ P(A) =  .

OR

In the leap year, total number of days = 366 = 52 × 7 + 2

Total 52 weeks 52 Sundays, now the 2 remaining days will be as follows.

S = {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)}

n(S) = 7

So, the probability of not getting Sunday in the remaining 2 days =

The probability that a randomly taken leap year has 52 Sundays is .

Q 18. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).

Solution: sinA + sin2A = 1

⇒ sinA = 1 – sin2A = cos2A … (i)

⇒ sin2A = cos4A … (ii)

cos2A + cos4

= sinA + sin2A   … from (i) and (ii)

= 1      (sinA + sin2A = 1)

Q 19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π =3.14).

Solution: Here, r = 6cm, θ = 30°

Area of the sector of a circle =  × 3.14 × 6 × 6 = 9.42 cm2.

Q 20. Find the class marks of the classes 20 – 50 and 35 – 60.

Solution: Class mark of the class 20 – 50 =

Class mark of the class 35 – 60 =

Q 21. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper.

The following were the answers given by the students:

2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 +   + 7, 7x +  , 5x3 – 7x + 2, 2x2 + 3 –  , 5x –  , ax+ bx2 + cx + d, x +  .

1. How many of the above ten, are not polynomials?
2. How many of the above ten, are quadratic polynomials?

Solution:

1. From the above ten, only three are not polynomials.

They are   because in all three polynomials, the degree of x is not a positive integer.
2. From the above ten, only one is a quadratic polynomial which is 3x2 + 7x + 2 because highest degree of the variable x is 2.

Q 22 .A child has a die whose six faces show the letters as shown below:

The die is thrown once. What is the probability of getting (i) A, (ii) D?

Solution: From the given question, n(S) = 6

1. Let E be the event of getting A on the die.

As there are 2 A’s, n(E) = 2

Hence, the required probability is
2. Let F be the event of getting D on the die.

As there is only one D, n(F) = 1

Hence, the required probability is

Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that .

OR

In fig.5, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.

Solution: Given: Triangle ABC and triangle DBC are on the same base BC.

To prove:
Construction:
Draw AM perpendicular to BC from point A and draw DN perpendicular to BC from point D.

Prove:
In Δ AMO and Δ DNO
∠AOM = ∠DON                   vertically opposite angles
∠AMO = ∠DNO                   90˚ angles
⇒ ΔAMO ≈ ΔDNO                AA test of similarity
Ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides

OR

AB2 = BD2 + AD2        Pythagoras theorem
⇒ AD2 = AB2 – BD2   … (i)
AC2 = CD2 + AD2         Pythagoras theorem
⇒ AD2 = AC2 – CD2   … (ii)
AB2 – BD2 = AC2 – CD2         from (i) and (ii)
⇒ AB2 + CD2 = BD2 + AC2

Q 24. Prove that

OR

Show that tan4θ + tan2θ = sec4θ - sec2θ

Solution:

OR

LHS =  tan4 θ +  tan2 θ

= (sec2 θ - 1) sec2 θ - 1

= sec4 θ - 2sec2 θ + 1 + sec2 θ - 1

= sec4 θ -  sec2 θ

Hence,  tan4 θ +  tan2 θ =  sec4 θ -  sec2 θ.

Q 25. Find the mode of the following frequency distribution:

Solution: Here, the maximum frequency is 10, and the class corresponding to this is 30 – 35.

So the modal class is 30 – 35.

Therefore, l = 30, h = 5, f0 = 9, f1 = 10 and f2 = 3

Mode =

Hence, the mode of the above data is 30.625.

Q 26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.

Solution: h = 14 cm and r = 6 cm

As the circular cone of same height and same radius is removed from the cylinder, therefore we have

Volume of the remaining solid

= Volume of cylinder – volume of cone

=  πr2h - πr2h

= πr2h

=

= 1056 cm3

Q. Nos. 27 to 34 carry 3 marks each.

Q 27. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that .

Solution:

Lengths of the tangents from an external point are equal.

Therefore, AQ = AR, BQ = BP and CP = CR

Perimeter of ΔABC = AB + BC + CA

= AB + BP + PC + AR – CR

= AB + BQ + PC + AQ – PC

= AQ + AQ

⇒ AB + BC + CA = 2AQ

(AB + BC + CA) = AQ

Q 28. The area of a circular play ground is 22176 cm2. Find the cost of fencing this ground at the rate of Rs. 50 per metre.

Solution: Area of a circular playground = 22176 cm2

⇒ πr2 = 22176

⇒ r2 = 22176

⇒  r2 = 22176 ×

⇒ r = 84 cm

Circumference of a playground = 2πr2 = 2 ×  × 84 = 528 cm = 5.28 m

The cost of fencing the playground at Rs. 50 per metre

= 5.28 × 50 = Rs. 264

Q 29. If the mid – point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.

OR

Find the area of triangle ABC with A (1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).

Solution: The coordinates of the midpoint of the line segment joining A(3, 4) and B(k, 6) are

It is given that x + y – 10 = 0

3 + k = 10

k = 7

Hence, the value of k is 7.

OR

In ∆ABC, coordinates of A are (1, –4)

Let (a, b) and (x, y) be the coordinates of points B and C respectively.

Midpoint of AB is (2, –1) and of AC is (0, –1)

Midpoint of AB =

⇒ a = 3, b = 2

Midpoint of AC =

⇒ x = -1, y = 2

Therefore, B(3, 2) and C(-1, 2)

Area of Δ ABC =  x1(y2 - y3)+ x2(y3 - y1) + x3(y1 - y2)

1(2 - 2 ) + 3(2 + 4) - 1(-4 - 2)

0 + 18 + 6 = 12

Hence the area of ΔABC is 12 square units.

Q 30. If Fig.6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.

Solution: Given that ΔABC ~ ΔDEF

⇒ 4x – 2 = 18

⇒ 4x = 20

⇒ x = 5

Then the sides of triangles are

AB = 2x – 1 = 9, BC = 2x + 2 = 12, AC = 3x = 15

DE = 18, EF = 3x + 9 = 24, DF = 6x = 30

Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and .

OR

Solve for

Solution: 2x + y = 23 …(i)

4x – y = 19 …(ii)

Adding (i) and (ii) we get

⇒ 6x = 42

⇒ x = 7

Put it in (i) we get

⇒ 14 + y = 23

⇒ y = 9

5y – 2x = 45 – 14 = 31

OR

According to the paper, condition given is x ≠ -4, 7 which implies that the equation should be  .

⇒ -30 = x2 - 3x -28

⇒ x2 - 3x + 2 = 0

⇒ (x - 2)(x - 1) = 0

⇒ x = 2 or x = 1x

Q 32. Which term of A.P. 20, 19, 18, 17,....  is the first negative term.

OR

Find the middle term of the A.P. 7, 13, 19, …., 247.

Solution: 20, 19, 18, 17,...

The above sequence is an A.P.

⇒ a1 = 20 and d =

Consider, an < 0

⇒ a1 + (n – 1)d < 0

⇒  20 + (n - 1) ×  < 0

⇒ 80 - 3(n - 1) < 0

⇒ 80 – 3n + 3 < 0

⇒ 83 < 3n

⇒ n >

Hence, 28th term of the given sequence is first negative term.

OR

7, 13, 19, …, 247

Given sequence is in A.P.

a1 = 7, d = 6 and an = 247

⇒ a1 + (n – 1)d = 247

⇒ 7 + (n – 1)6 = 247

⇒ 6(n – 1) = 240

⇒ n – 1 = 40

⇒ n = 41

The middle term should be a21.

a21 = a1 + 20d = 7 + 20 × 6 = 127

The middle term of the given sequence is 127.

Q 33. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8cm standing water is required?

Solution: Area of the face of canal = 6 × 1.5 m2

Speed of water flow = 10 km/h = 10000 m/h

So the water flow through the canal in 1 hour

= 6 × 1.5 × 10000

= 90000 m3

Water flow through the canal in 30 min = 90000/2 = 45000 m3

The height of standing water = 8 cm = 0.08 m

Area can irrigate = 45000 × 0.08 = 562500 m2

Q 34. Show that:

Solution:

Q 35. The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.

OR

The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a ‘more than’ type distribution and draw its ogive.

Solution:

⇒ 704 + 20f = 720 + 18f

⇒ 2f = 16

⇒ f = 8

OR

Q 36. From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

Let BC be the building, AB be the tower and D be the point on ground from where elevation angles are to be measured.

In ΔBCD,

CD = 20 m …(i)

In ΔACD,

Hence, the height of the tower is

Q 37. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Solution: Let the time taken to fill the pool by pipe A be x hours and by pipe B be y hours.

Thus, the quantity of water filled by the pipe A in one hour is   and by the pipe is  .

In 1 hour the quantity of water filled by both the pipes is

It takes 12 hours to fill the pool for both the pipes.

…(i)

Let A be the pipe with larger diameter and B be the pipe with smaller diameter.

In 4 hours the quantity of water filled by pipe A is  .

In 9 hours the quantity of water filled by pipe B is  .

According to the question,

Multiplying (i) by 4 we get

4m + 4n =   …(v)

Subtracting (v) from (iv) we get

n =

Put it in (iii) we get

Hence, x = 20 and y = 30

Hence, the pipe A and pipe B fills the pool in 20 hours and 30 hours respectively.

Q 38. Prove that  is an irrational number.

Solution: Let us assume that  is a rational number

We can write it in the form of  where q ≠ 0.

Also p and q are co-prime numbers.

Here p is divisible by 5

p = 5k where k is a positive integer

Hence, p2 = 25k2

Substituting 5q2 = p2

⇒ 5q2 = 25k2

⇒ q2 = 5k2

q is divisible by 5

From this we can say that p and q have a common factor 5

It contradicts to our assumption p and q are co-prime.

Hence,   is an irrational number.

Q 39. Draw a circle of radius 3.5cm. From a point P, 6cm from its centre, draw two tangents to the circle.

OR

Construct a ΔABC with AB=6cm, BC=5cm and ∠B=60°. Now construct another triangle whose sides are   times the corresponding sides of ΔABC.

Solution: Steps of construction:

1. Draw a line segment OP = 6 cm
2. With centre O and radius 3.5 cm, draw a circle.
3. Draw the perpendicular bisector of OP intersecting it at M.
4. With centre M and radius OM, draw a circle which intersects the circle at T and S.
5. Join PT and PS which are the required tangents.

OR

Steps of construction:
Step 1: Draw a line segment AB = 6 cm
Step 2: Draw an angle of 60° at B so that ∠XBA = 60°
Step 3: With centre B and radius 5 cm, draw an arc intersecting XB at C
Step 4: Join AC, Δ ABC is the required triangle.
Step 5: Draw a line BY below BC which makes an acute angle ABY.
Step 6: Mark B1, B2 and B3 at equal distances from B such that BB= B1B2  = B2B3
Step 7: Join AB3
Step 8: Draw B2A parallel to AB3
Step 9: Draw C' A'  parallel to CA through A’ intersecting BC at C’

ΔA’BC’ is the required similar triangle.

Q 40. A solid is in the shape of hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7cm and height of cone is 3.5cm, find the volume of the solid. (Take π = ).

Solution: Hemisphere: r = 7 cm

h = 3.5 cm, r = 7 cm

Volume of cone =
Volume of hemisphere =
Volume of solid = 718.6 - 179.6 = 898.2 cm3

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