2012 Class 10 Set-1 Section-c

8/12/2019 2012 Class 10 Set-1 Section-c

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CBSE X Mathematics 2012 Solution (SET 1)

Section C

Q19.Solve forx: 4x24ax+ (a

2b

2) = 0

Solution:

The given quadratic equation is 2 2 24 4 0x ax a b .

2 2 2

2

2

2

2

4 4 0

4 4 0

4 [ 2 2 2 2 ] ( ) ( ) 0

4 2 2 2 2 0

4 2 2 0

2 [2 ( )] [2 ( )] 0[2 ( )][2 ( )] 0

2 ( ) 0 or 2 ( ) 0

2 o

x ax a b

x ax a b a b

x a a b b x a b a b

x b a x a b x a b a b

x b a x a b x a b a b

x x a b a b x a b

x a b x a b

x a b x a b

x a b

r 2

or2 2

x a b

a b a bx x

Thus, the solution of the given quadratic equation is given by or2 2

a b a bx x

.

OR

Solve for 2:3 2 6 2 0x x x

Solution:

The given quadratic equation is 23 2 6 2 0x x .

Comparing with the quadratic equation ax2+ bx+ c= 0, we have

a= 3, 2 6b and c= 2

Discriminant of the given quadratic equation,

D= b24ac

2

2 6 4 3 2 24 24 0


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2 6 02 3 2

2 66

6

3

b Dx x

a

x

x

Thus, the solution of the given quadratic equation isx=6

3.

Q20.Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:Since ABCD is a parallelogram,AB = CD (1)

BC = AD (2)

It can be observed thatDR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)Adding all these equations, we obtain

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)CD + AB = AD + BC

On putting the values of equations (1) and (2) in this equation, we obtain

2AB = 2BCAB = BC (3)

Comparing equations (1), (2), and (3), we obtain

AB = BC = CD = DAHence, ABCD is a rhombus.


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OR

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles

at the centre of the circle.

Solution:

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle

at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the center of the circle.

Consider OAP and OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)OA = OA (Common side)

OAP OAS (SSS congruence criterion)

Therefore, A A, P S, O O

And thus, POA = AOS

1 = 8

Similarly,2 = 3

4 = 5

6 = 7

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360

(1 + 8) + (2 + 3) + (4 + 5) + (6 + 7) = 360

21 + 22 + 25 + 26 = 360

2(1 + 2) + 2(5 + 6) = 360

(1 + 2) + (5 + 6) = 180

AOB +

COD = 180

Similarly, we can prove that BOC + DOA = 180

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles atthe centre of the circle.


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Q21.Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm

and 8 cm. Then construct another triangle, whose sides are3

5times the corresponding sides

of the given triangle.

Solution:

Given:BC = 6 cm, C = 8 cmThe triangle to be formed is to be right angled triangle.

Steps of construction:1. Draw a line segment BC = 6 cm.

2. Draw a ray CN making an angle of 90 at C.3. With C as centre, taking 8 cm as the radius make an arc at CN intersecting it at A. Join

AB.

4. Now, ABC is the triangle whose similar triangle is to be drawn.5. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

6. Locate 5 (Greater of 3 and 5 in

3

5 ) points B1, B2, B3, B4and B5on BX so that BB1= B1B2=B2B3 = B3B4 = B4B5

7. Join B5C and draw a line through B3(Smaller of 3 and 5 in3

5) parallel to B5C to intersect BC

at C.

8. Draw a line through Cparallel to the line CA to intersect BA at A.


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9. ABC is the required similar triangle whose sides are3

5times the corresponding sides of

ABC.

Q22.In Fig. 7, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and

3.5 cm and centre O. If POQ = 30, then find the area of the shaded region.

22Use

7

Solution:

PQ and AB are the arcs of two concentric circles of radii 7 cm and 3.5 cm respectively.Let r1and r2be the radii of the outer and the inner circle respectively.

Suppose be the angle subtended by the arcs at the centre O.

Then 1 27 cm, 3.5 cm and 30r r

Area of the shaded region= Area of sector OPQArea of sector OAB

2 2

1 2

2 2

1 2

2 2

2

2

2

360 360

360

30 227 cm 3.5 cm

360 7

1 2249 12.25 cm

12 7

1 22 36.75 cm12 7

9.625 cm

r r

r r

Thus, the area of the shaded region is 9.625 cm2.


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Q23.From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same

height and same base diameter is hollowed out. Find the total surface area of the remaining

solid.22

Use

7

Solution:

It is given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm

Diameter of the cylindrical part = 12 cm

Radius (r) of the cylindrical part12

cm=6cm2

Radius of conical part = 6 cm

Slant height (l) of conical part 2 2 cmr h

2 2= 6 7 cm

= 36 49 cm

= 85 cm

=9.22cm (approx.)

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Base area of the circular part

= 2rh+ rl+ r2

2 2 2

2 2 2

2

22 22 222 6 7cm 6 9.22cm 6 6cm

7 7 7

264cm 173.86cm 113.14cm

551cm

OR

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket

is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is

24 cm, then find the radius and slant height of the heap.

Solution:


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Height (h1) of cylindrical bucket = 32 cm

Radius (r1) of circular end of bucket = 18 cm

Height (h2) of conical heap = 24 cmLet the radius of the circular end of conical heap be r2.

The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conicalheap.Volume of sand in the cylindrical bucket = Volume of sand in conical heap

2 2

1 1 2 2

1

3r h r h

2 2

2

1 18 32 24

3r

2 2

2

22 2

2

1 18 32 24

3

3 18 32

18 424

r

r

r2= 18 2 = 36 cm

Slant height = 2 2 2 2 236 24 12 (3 2 ) 12 13 cm

Therefore, the radius and slant height of the conical heap are 36 cm and 12 13 cm respectively.

Q24.The angles of depression of two ships from the top of a light house and on the same side ofit are found to be 45 and 30. If the ships are 200 m apart, find the height of the light

house.

Solution:

The given situation can be represented diagrammatically as,


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Here, AB is the light house and the ships are at the points C and D; his the height of the light

house and BC =x.

In right angled ABC:AB

tan 45AC

tan 45

1

h

x

h

x

x h

In right angled ABD,

ABtan30BD

tan30200

tan 30200

1

2003

h

x

hx h

h

h

h


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200 3

3 200

3 1 200200

m3 1

3 1200

3 1 3 1

h h

h h

h

h

h

2003 1

2

100 3 1

h

h

Hence, the height of the light house is 100 3 1 m.

Q25.A point P divides the line segment joining the points A (3, 5) and B (4, 8) such that

AP K

PB 1 . If P lies on the linex+y= 0, then find the value of K.

Solution:

The given points are A (3,5) and B (4, 8).

1 1 2 2Here, 3, 5, 4 and 8x y x y .

SinceAP K

PB 1 , the point P divides the line segment joining the points A and B in the ratio K:1.

The co-ordinates of P can be found using the section formula 2 1 2 1,mx nx my ny

m n m n

.

Here, m = K and n= 1

K 4 1 3 K 8 + 1 5 4K+3 8K 5Co-ordinates of P , ,

K + 1 K + 1 K+1 K+1

It is given that, P lies on the linex+y= 0.


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4K+3 8K 50

K+1 K+1

4K 3 8K 50

K +14K 2 0

4K 2

1K

2

Thus, the required value of K is1

2.

Q26.If the vertices of a triangle are (1,3), (4,p) and (9, 7) and its area is 15 sq. units, find the

value(s) ofp.

Solution:

Given, vertices of a triangle are (1,3), (4,p) and (9, 7).

1 1

2 2

3 3

1 , 3

4 ,

9 , 7

x y

x y p

x y

Area of given triangle

= 1 2 3 2 3 1 3 1 21

2 x y y x y y x y y

11 7 4 7 3 9 3

2

17 40 27 9

2

110 60

2

5 6

p p

p p

p

p

Here, the obtained expression may be positive or negative.

5 6 15 or 5 6 156 3 or 6 3

3 or 9

p pp p

p p


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Q27.A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at

random from the box, then find the probability that it will be (i) a blue card (ii) not ayellow card (iii) neither yellow nor a blue card.

Solution:

Number of red cards = 100

Number of yellow cards = 200

Number of blue cards = 50Total number of cards = 100 + 200 + 50 = 350

Number of blue cardsi P a blue card

Total number of cards

50

350

1

7

Number of cards other than yellowii P not a yellow card


Number of red cards + Number of blue cards


100 50

350

150

350

37

Number of cards which are neither yellow nor blue

iii P Neither yellow nor a blue cardTotal number of cards

Number of red cards

Total number o

f cards

100

350

2

7

Q28.The 17th

term of an AP is 5 more than twice its 8th

term. If the 11th

term of the AP is 43,

then find its nth

term.


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Solution:

Let abe the first term and dbe the common difference of the given A.P.According to the given question,

17th

term = 2 8th

term + 5

17 8i.e., 2 5

17 1 2 8 1 5 (as 1 )

16 2 7 5

16 2 14 5

2 5 ... 1

n

a a

a d a d a a n d

a d a d

a d a d

a d

Also, 11th

term, a11 = 43

11 1 43

10 43

2 5 10 43 Using 1

12 48

4

2 5 2(4) 5 8 5 3

a d

a d

d d

d

d

a d

Thus, nth

term of the AP, 1na a n d On putting the respective values of aand d, we get

3 1 4 3 4 4 4 1na n n n

Hence, nth

term of the given AP is 4 1n .

2012 Class 10 Set-1 Section-c

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