4 Marks Questions1. An oil drum is in the shape of a cylinder

having the following dimensions: diameteris 2 meters and height is 7 meters. Thepainter charges 3 per m2 to paint the drum.Find the total charges to be paid to thepainter for 10 drums?

Sol: Step 1: Identify what the question isabout. Here, the cost of painting 10 drums isto be calculated. For that we can do thefollowing:

Total Cost = Cost for one drum 10Step 2: Now, we have to calculate the cost forone drum. Since, the painter paints thecylinder on all sides, we have to find the totalsurface area of cylinder. This gives us thearea that the painter paints.

It is given that diameter of the (oil drum)cylinder = 2 m.

Radius of cylinder = 1 mTotal surface area of a cylindrical drum

= 2 r(r + h)22= 2 1 (1 + 7) = 50.28 m2

7

Step 3: Cost for one drum = area painted forone drum 3 = 50.28 3Step 4: Total cost for 10 drums = Cost for onedrum 10

= 50.28 3 10 = 1508.40The best approach to solving mensuration

problems is to attempt the problem step bystep, beginning with understanding what thequestion is asking us to find out. Sometimes,if we aren’t give with a figure, it helps to drawit on our own. 2. A sphere, a cylinder and a cone are of the

same radius and same height. Find theratio of their curved surface areas (CSA)?

Sol: Before solving, draw the figures. Then, solve this prob-

lem by substituting thevariable values in therespective CSAformulas and then findingthe ratio.

Let r be the commonradius of a sphere, a cone and cylinder.

Height of sphere = its diameter = 2r.Then, the height of the cone = height of

cylinder = height of sphere. (As it is given thatall of them are equal) = 2r. ...... (1)Sphere: CSA = 4r2 ...... (2)Cone: CSA = rl

Here l is the slant height. From (1) andusing the formula for slant height, we have

= r2 + h2

r2 + (2r)2 =

5r

r 5r = 5r2 ...... (3)

Cylinder: CSA = 2rh = 2r 2r = 4r2 ...... (4)Now the ratio of (2), (3) and (4) is

= 4r2 : 4r2 : 5r2 = 4 : 4 :

5

Sometimes, we will be given with twodimensional values using which we can findother measurements. The only trick to solvesuch questions is to recollect the rightformulas and substitute values cautiously. 3. A right circular cylinder has base radius

14 cm and height 21 cm. Find: i) Area of base or area of each endii) Curved surface areaiii) Total surface area iv) Volume of the right circular cylinder

Sol: r = 14 and h = 21i) Area of base or area of each end r2 =

22 (14)2 = 616 cm2

722ii) Curved surface area = 2rh = 2 7

14 21 = 1848 cm2

iii) Total surface area = 2 616 + 1848

= 3080 cm2

iv) Volume of Cylinder = r2h = 616 21 = 12936 cm3

As we can see, these problems requiremere application of the formulas. 4. The diameter of a metallic sphere is 6 cm.

It is melted and drawn into a wire havingdiameter of the cross section as 0.2 cm.Find the length of the wire.

Observation: It is important to note that evenwhen the sphere is melted to form a wire, thevolume remains same.Volume of the metal used in wire = Volume ofthe sphere .... (1)Step 1: Draw the Diagram

Step 2: Diameter of metallic sphere = 6 cm

Radius of metallic sphere = 3 cmDiameter of cross - section of cylindrical

wire = 0.2 cm.Radius of cross section of cylinder wire

= 0.1 cmLet the length of wire be 1 cm.Step 3: From (1), we can equate the twovolumes.

4 (0.1)2 h = 33

31 4 ( )2 h = 27

10 31 h = 36

10036100

h = cm

= 3600 cm. = 36 m.Therefore, the length of the wire is 36 m.5. A wooden article was made by scooping out

a hemisphere from each end of a solid cylin-der, as shown in the figure. If the height ofthe cylinder is 10 cm. and its base is of 3.5 cm,find the total surface area of the article.

Sol: Let the original heightof cylinder = 10 cm

base radius of cylinder

= 3.5 cmso base radius of hemi-sphere

= 3.5 cm(same as that of cylin-

der)The total surface area would be the sum of

curved surface area of cylinder and the sur-face areas of 2 hemispheres.surface area of cylinder = 2rhsurface area of one hemisphere = 2r2

TSA = 2rh + 2(2r2) TSA = 2rh + 4r2

TSA = 2r (h + 2r)22

TSA = 2 3.5(10 + 2 3.5) 7

= 22 (10 + 7)TSA = 374 cm2

email: help@eenadupratibha.netøŒEî¦ô¢Ù 30 ì÷Ùñô¢ª 2019

Indu KilaruSubject Expert

Writer

Mensuration

Length of Side of a Cube is..

Target-2020

TenthMathematics Paper-2

100100

1. The length of cuboid is twice the lengthof breadth. Its height is 5 cm. The volumeof cuboid is 1440 cm3. What is thebreadth?

Sol: Length = 2x, breadth = x, height = 5 V = lbh1440 = x(2x)5 10x2 = 1440x2 = 144 x = 12 breadth = 12 cm

2. Length of side of a cube is 5 cm. Itsvolume is ....

Sol: Volume of cube = a3

53 = 125 cm3

3. The lateral surface area of a cube is16 cm2. Its side is.....

Sol: LSA of cube is = 4a2

16 = 4a2 a2 = 4 a = 2

4. The radius of a sphere is 3 cm. Itsvolume is ........

4Sol: Volume = r334= 3 3 3 = 363

5. The total surface area of a cube is 24 cm2. The side length is .....

Sol: 6a2 = 24 a2 = 4 a = 2 length = 2 cm

One Mark Questions

1. Find the lateral surface area and totalsurface area of a cuboid with the followingdimensions length is 2 cm, breadth is 5 cmand height is 6 cm.

Sol: Lateral surface area = 2h(l + b)= 2(6)(2 + 5) 12(7) = 84 cm2

Total surface area = 2(lb + bh + lh)= 2(2 5 + 5 6 + 6 2)= 2(10 + 30 + 12) = 2(52) = 104 cm2

2. A company wanted to manufacture the1000 hemispherical basins from a thinsteel sheet. If the radius of hemisphericalbasin is 21 cm, find the required area ofsteel sheet to manufacture the abovehemispherical basins.

Sol: Radius of basin = 21 cmSurface area = 2r2

22= 2 21 21 = 2772 cm2

7Sheet reqired for one basin = 2772 cm2

Total area of sheet required for 1000basins = 2772 1000

= 2772000 cm2 (or) 277.2 m2

3. Find the volume and total surface area ofhemisphere with radius 4 cm.

Sol: Radius = 4 2Volume of hemisphere = r33

2 22 2816= ()(4)3 = = 134.09 cm3

3 7 21Total surface area of hemisphere = 3r2

22 1056= 3 4 4 = = 150.85 cm2

7 74. A cone of height 24 cm and radius of base

6 cm is made up of modelling clay. A childreshapes it in the form of sphere. Find theradius of sphere.

Sol: Step 1: Draw the diagrams

Step 2: We can see that since that sameclay is used, the volume remains equal.

1 4i.e., r2h = r33 3

1 4 6 6 24 = r33 3

r3 = 6 6 6 r = 6 cm 5. A toy is in the form of a cone mounted on a

hemisphere, the diameter of the base andheight of the cone are 6 cm and 4 cmrespectively. Determine surface area of toy.

Sol: Step 1: Draw the diagram Step 2: Make deductionsheight of the cone (h) = 4 cm

diameterradius of the cone (r) =

26= = 3 cm 2

radius of hemisphere = 3 cmStep 3: Formulae to be used - surfacearea of cone = rland surface area of hemisphere = 2r2

Step 4: Solve for surface area of cone = rlnow, slant height l =

r2 + h2 =

32 + 42

= 25 = 5 cm

So, rl = 3.14 3 5 = 47.1 cm2

Step 5: Solve for surface area of hemi-sphere = 2r2 = 2 3.14 32 = 56.52Step 6: total surface area of solid = 47.1 +56.52 = 103.62 cm2

6. Find the volume of the largest rightcircular cone that can be cut out of a cubewhose edge is 7 cm.

Sol: Step 1: Draw the diagramStep 2: Formulas

1volume of cone = r2h3

Step 3: substitute values Note: Diameter of cone is same as edge of

7cube i.e., d = 7; r = 2

height of cone is same as edge of cube i.e., h = 7

1 7now, V = ()2 7 = 89.83 cm23 2

Two Marks Questions

h = 24 cm

r = 6 cmsphere

63

4

r

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One & Two Marks Questions

1. Show that 4 tan2 45 sec2 60 + cos2 601

= .4

1Sol: tan 45 = 1, sec 60 = 2, cos 60 = 2

1 1 1 4(1)2 (2)2 + ( )2 = 4 4 + = 2 4 4

2. Show that tan2 sin2 = sin2 tan2 sin2 Sol: LHS: tan2 sin2 = sin2 cos2

1= sin2 [ 1]cos2

1 cos2 = sin2 [ ]cos2

sin2 = sin2 = sin2 tan2 (RHS)cos2

LHS = RHS.

3. (1 + tan2 ) (1 sin2 ) = 1 Prove it.

1Sol: sec2 cos2 = cos2 = 1cos2

... sec2 tan2 = 1sin2 + cos2 = 1

cos 67tan 404. 2. = 1 Prove it.

sin 23 cot 50Sol: cos 67 = cos (90 23) = sin 23

tan 40 = tan (90 50) = cot 50sin 23cot 50 LHS 2. sin 23 cot 50

= 2 (1) 1 = 1 (RHS)LHS = RHS.

5. Show that cot + tan = cosec .sec cos sin

Sol: LHS = cot + tan = + sin cos

cos2 + sin2 = sin .cos

1 1 1= = . sin .cos sin cos

= cosec .sec = RHS6. From the following figure show that

sec2 tan2 = 1ACSol: sec = BC

ABtan =

BC

AC ABL.H.S = sec2 tan2 = ( )2 ( )2BC BC

AC2 AB2 AC2 AB2= =

BC2 BC2 BC2

BC2= = 1 = RHS. (

...

AB2 + BC2 = AC2)BC2

1 17. sin (A B) = , cos (A + B) = 2 2

0 < A + B 90, A > B then find thevalues of A and B.

1Sol: sin (A B) = = sin 30

2

A B = 30............... (1)1cos (A + B) = = cos 602

A + B = 60............. (2)

(1) + (2) A B = 30A + B = 60

2A = 9090 A = = 452

Substitute A value in eq. (2)45 + B = 60 B = 60 45 = 15

Target-2020

TenthMathematics Paper-2

100100

P. VenugopalSubject Expert

Writer

Trigonometry

A

C B

12

Mark Questions

x y.sec 1. If tan = then =

y x.cosec y x y

A) 1 B) C) D) x

x2+y2

x2 + y2

4 1 sin A

2. If cos A = then = 5 1 + sin A

1 1 5A) 1 B) C) D) 2 4 4

3. If tan = 1 sin = 1 1

A) 0 B) 1 C) D) 2 2

5. sin 36. cos 54 + cos 36.sin 54 =A) 0 B) 1 C) 2 D) None of these

6. log10 (cot 45) =

A) 0 B) 1 C) 10 D) 2

7. If cot = 3 then 1 + sin =

1 3A) B) 2 C) 3 + 1 D) 2 2

8. If 2 cos x = 3 then x =A) 0 B) 30 C) 45 D) 60

Answers: 1-A 2-B 3-D 4-A 5-B 6-A 7-D 8-B

60B

h

x D

A

C30

2000 m

From figure x =

4.

A) 1000 m B) 4000 m C) 2000 m D) 500 m

cos sin + 11. Show that

cos + sin 1= cosec + cot .

cos sin + 1Sol: LHS =

cos + sin 1Divide numerator and denominator bysin cos sin + 1

sin cot 1 + cosec = cos + sin 1 cot + 1 cosec

sin ( cosec2 cot2 = 1)

cot + cosec [cosec2 cot2 ]=

cot cosec + 1cot + cosec [1 (cosec cot )]

= cot cosec +1

= cot + cosec = RHS LHS = RHScos2

2. Solve the equation = 3cot2 cos2 cos2

Sol: cot2 cos2 = cos2 sin2

1= cos2 ( 1)sin2 1 sin2

= cos2 ( )sin2 cos2

= cos2 . = cot2 cos2 sin2

cos2 = 3cot2 cos2 cos2 1 = 3 = 3

cot2 cos2 cot2

tan2 = 3 tan = 3 tan 60 =

3 = 60.

1 cosec2 A sec2 A3. If tan A show that 7 cosec2 A + sec2 A

3=

41Sol: tan A = 7

8 cosec A = 1

8

sec A = x2 = (7 )2 + (1)2

7

x2 = 8 x =

8 cosec2 A sec2 A

LHS = cosec2 A + sec2 A

8

8 ()2 ()21 7

= 8

8 ()2 + ()21 7

8 56 88 7 7

= = 8 56 + 8 8 + 7 7

48 3= = = R.H.S

64 44. If sec tan = a, write all the trigono-

metric ratio's in terms of 'a'.Sol: sec tan = a ............ (i)

1sec + tan = ................. (ii)a

1 a2 + 1(i) + (ii) 2 sec = a + = a a

a2 + 1 sec = .........(1)2a

2acos = ............. (2)

a2 + 1

2asin = 1 cos2 = 1 ( )2a2 + 1

(a2 + 1)2 4a2

= (a2 +1)2

(a2 1)2 a2 1

= = (a2 + 1)2 a2 + 1

a2 1 sin = ............(3)a2 + 1

a2 1sin a2 + 1 a2 1

tan = = = cos 2a 2a

a2 + 1

a2 1 tan = ........ (4)2a

2a cot = ............(5)

a2 1

a2 + 1cosec = ............. (6)a2 1

5. The angles of elevation of the top of atower from two points B and C at distancesof a and b respectively from the base andin the same straight line with it are comple-mentary. Prove that the height of the toweris

ab.

Sol: In ABDhtan = ....... (1)a

In ACDhtan (90 ) = b

hcot = .......... (2)b

h htan cot =

a bh2

1 = ab

h2 = ab h = ab

Four Marks Questions

B A

Cx =

8

7

1

Da b

h

(90 )

C

A

B

51. If sec = then find the value of 4

tan .1 + tan2

2. Show that 2 tan2 60 + 1 = 7

sin 17 cos 673. Find the value of 5. + 2.

cos 73 sin 23sin 15 6 .cos 75

4. From the figure C = 90, BC = 20 cm,

AB = 29 cm Find the valuesi) sin A ii) cos A iii) tan A iv) sin B v) cos B vi) tan B

cot A + cosec A 15. Show that

cot A cosec A + 1= cot A + cosec A

6. If 5 cot = 3 then find the value of

5 sin 3 cos ( )4 sin + 3 cos

C

A

B20

29

The height of the Tower is?

Additional Problems

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