It is that branch of Mathematics which deals withthe measurement of geometric magnitudes such aslengths of lines, areas of surface and volumes of solids.It is of two kinds : Plane mensuration and Solidmensuration.

Area : It is the space enclosed under givenboundaries.

Perimeter : It is the total length of boundaries.Volume : The place covered by a solid thing is

the volume of that. It is always shown in cube unit.Formulae Area of Plane Figure

1. Triangle : A plane figure which is covered bythree sides (lines).

a) Area of right angled triangle = Base x Height

2

b) Height = 2 x Area

Base

c) Hypotenuse of a right angled triangle

= 22Base + Height

d) Area of a having unequal sides

= S S - a S - b S - c

here a, b, c are the sides of and S = a+ b + c

2;

e) Perimeter of triangle = a + b + c

f) Area of equilateral triangle = 2a 3

4

(where 'a' is the length of the side)

g) Side of equilateral triangle = Perimeter

3

OrArea x 4

3

h) Perimeter of equilateral triangle = a + a + a=3ai) Area of an Isosceles triangle whose base is ‘b’

and each of equal side is ‘a’ = b

4a² - b²4

j) Perimeter of an Isosceles triangle = 2a+b2. Quadrilateral : A four sided plane figure is calledquadrilateral.

Area= ½ x diagonal x sum of internal perpendicular(i) Rectangle : A four sided plane figure whose

every angle is of 90° andits opposite sides are equaland parallel. These sidesare called length andbreadth.

a) Area = Length x Breadth,

b) Length =Area

Breadth, c) Breadth =

Area

Length

d) Perimeter = 2(l + b)

e) Diagonal = (Length)² + (breadth)²

(ii) Area of four walls of a room= Perimeter x Height

= 2(l + b)h and h = Area of four walls

Perimeter

(iii) Square : A four sided plane figurewhose all sides are equal, andevery angle is of 90°

a) Area = Side x side = (Side)2

or Area = 1

2(Diagonal)²

b) Diagonal or hypotenuse of square = 2 x Area

or 2 (side)

c) Side= Aread) Perimeter = Side x 4(iv) Parallelogram : A four

sided plane figure whoseangles are not necessary of90° but opposite sides areequal and parallel.

MENSURATIONS-I DAY - 20

a) Area = Base x Height.b) Perimeter = Sum of all sides.(v) Trapezium : A four sided plane

figure whose any one pair ofopposite sides is parallel.

a) Area = 1

2 (Sum of sides) x Height

b) Perimeter = Sum of all sidesDiagonals of a trapezium maybe equal

(vi) Rhombus : A four sided planefigure whose all sides are equaland opposite sides are parallelbut angle is not necessary of 90°.

a) Area of rhombus = 1

2Diagonal1 Diagonal2

b) Perimeter = Side x 4

c) Side =

22

1 2DiagonalDiagonal

2 2

d) (Diagonal)2 = 2

22

1Diagonal(Side)

2

3. Polygon : Any plane figure which is covered bymore than two sides.Area of regular polygon (all sides are equal)

= 2nacot n4

(here ‘n’ stands for number ofsides and ‘a’ is for length of sideand stands for 180°)

(Note : cot 30°= 3 , cot 60° = 13 , cot 45° =1,

cot 15° = 2 + 3 )Sum of the inner angles of a polygon= (number of sides x 180°) - 360° or (2n - 4) 90°,

here 'n stands for number of sides.4. Circle : A plane figure bounded by a

curved line which covers a point fromall sides at an equal distance is calledcircle. A line drawn from the centralpoint to the circumference is called

radius.

a) Area = x r2

b) Circumference of circle = 2 r

(Use as 227 unless said and ‘r’ stands for radius.

Radius is half of diameter.)Segment of circle : Segment is a cut off part ofa circle. The boundary of a segment consists ofan arc of the circle and the chord determiningthe segment.

Area of segment =

360 x r2 - 21

r2

sin

= r2

- sin

360 2

Sector of circle : It consists of segment andthe cone having two bases upto the centre of thecircle.

a) Area = 12 x arc x radius =

2r3600

,

b) Arc length =

0

r180

5.a)Area of path of width (Wp) outside the park oflength ‘L’ + Breadth ‘B’ is Wp[L+B+2Wp]

b) Area of path of width (Wp) inside the park oflength ‘L’ + Breadth ‘B’ is 2Wp[L+B - 2Wp]

c) Cost of Gravelling/Tilling = Total Area x unit cost.

d) No. of tiled stone = Area to be tiled/stoned

Area of a tile/Stone

e) Distance covered by a wheel =Circumference No. of Revolutions

6. If ABC is an eq. of side ‘S’ then

a) Radius of Incircle = 1

2 3

b) Area of Incircle =

(S)²12

c) Radius of circumference = 1

(S)3

d) Area of circumcircle =

(S)²3

A

B C

e)Incircle Circumcirle

2

4

1

1

Ratio of Radii

Ratio of Areas

UNIT CONVERSIONS10 mm - 1 cm10 cm - 1 decimeter10 dm - 1 meter10 m - 1 Decameter1000 m 100 dm 1 kmAlso, 1 HECTARE 10,000m2

Pythagorus TripletsP B H3 4 512 5 1324 7 2540 9 4115 8 1720 21 29

MULTIPLE CHOICE QUESTIONSIn the following questions, select the

correct choice among the alternative givenbelow the question.1. Find the area of a rectangle whose one side is 3

metres and the diagonal is 5 metres.(a) 12 sq m (b) 8 sq m(c) 16 sq m (d) 14 sq m

2. A rectangular carpet has an area of 120 m² and aperimeter of 46 m. The length of its diagonal is:(a) 15 m (b) 16 m(c) 17 m (d) 20 m

3. The length of a rectangular plot is 144 m and itsareas is same as that of a square plot with oneof its sides being 8 1 m. The width of the plot is:(a) 7 m (b) 49 m(c) 14 m (d) Data inadequate

4. Find the Sides of two squares, which containtogether 12.25 hectares, the sides of the squaresbeing in the ratio of 3:4.(a) 210 m, 280 m (b) 90 m, 120 m(c) 150 m, 200 m (d) 180 m, 240 m

5. In order to fence a square Manish fixed 48 poles,if the distance between two poles is 5 metresthen what will be the area of the square soformed?(a) 2600 cm2 (b) 2500 cm2

(c) 3025 cm2 (d) None of these6. The diagonal of a square is 6cm. Find its

perimeter.

(a) 12 2 cm (b) 6 2 cm(c) 24 cm (d) Data inadequate

7. A 5 m wide lawn is cultivated all along the outsideof a rectangular plot measuring 90 m x 40 m.The total area of the lawn is:(a) 1200 m2 (b) 1300 m2

(c) 1350 m2 (d) 1400 m2

8. A rectangular lawn 60 metres has two roads each5 metres wide, running in the middle of it, oneparallel to length and the other parallel to thebreadth. Find the cost of gravelling them at 60paisc per square metre.(a) ` 285 (b) ` 385(c) ` 2.75 (d) ` 475

9. How many paving stones, each measuring 10dm by 9 dm are required to pave a verandah 60m long and 6m broad?(a) 600 (b) 800(c) 400 (d) 450

10.How far has a bicycle travelled when its drivingwheel 35 cm in diameter, has made 6300revolutions?(a) 6930 m (b) 6390 m(c) 6920 m (d) 6830 m

11. The length of a rectangle is increased by 20%.By what per cent should the width be decreasedto maintain the same area?

(a)2

163

% (b)2

263

%

(c)1

333

% (d) Data inadequate

12. Find the area of a rhombus one sides of whichmeasures 20 cm and one of whose diagonals is24 cm.(a) 384 sq cm(b) 348 sq cm(c) 484 sq cm(d) Can't be determined

13. Find the no. of diagonals of a pentagon.(a) 5 (b) 10(c) 15 (d) 8

14. Find the area of an equilateral triangle each ofwhose sides measures 12 cm

(a) 36 3 sq. cm (b) 18 3 sq. cm

(c) 24 3 sq. cm (d) 30 3 sq. cm

15. A ladder is placed so as to reach a window 63 mhigh. The ladder is then turned over to theopposite of the street and is found to reach apoint 56 m high. If the ladder is 65 m long, findthe width of the street.(a) 49 m (b) 45 m(c) 40 m (d) 59 m

ANSWERS

1. (a) 2. (c) 3. (b) 4. (a) 5. (c)6. (a) 7. (d) 8. (a) 9. (c) 10. (a)

11. (a) 12. (a) 13. (a) 14. (a) 15. (a)

Hints and Explanations :1.a) Using pythogorus triplet

BC = 4Area = 3x4 = 12 sq. m

2.c) Area = 120 m² = LXBPerimeter = 2 (L + B)2 = 46 m

L2 + B2 = (L + B)2 - 2LB

[ a2 + b2 = (a2 + b)2 - 2ab]= (23) - 2(120)

L2 + B2 = 529 - 240

L2 + B2 = 289

Diagonal = 2 2L +B =17

( Diagonal = 2 2L +B )

3.b) 144 x B = 84 x 84

B = 84 x 84

144=

7056

144= 49

4.a) Area = 12.25 hectresor 12.25 x 10,000

or 122500 m2

Sides Ratio 3 : 4 or 3x : 4x

(3x)2 + (4x)2 = 122500

9x2 + 16x2 = 122500

25x2 = 122500

x2 = 122500

25= 4900

x = (4900)2 = 70Sides = 3 x 70 : 4 x 70= 210 : 280.

5.c) No. of poles fixed = 48

No. of poles fixed per side= 48

4= 12

Each side = 5 x 11 = 55

Area = (55)2 = 3025 m2.

6.a) 2 side = 6

side = 6

2

Perimeter = 4 x side

= 4 x 6

2

= 264 × x

2 2 = 12 2

7.d)Area of path outside rectangle = 2Wp[L+B+2Wp]= 2 x 5 (90+40+10)= 10(140) = 1400 sq. m

8.a) Area of Path = (L W) + (B W) - (W W)Area of Park = (60 x 5) + (40 x 5) - (5 x 5)= 300 + 200 - 25= 500 - 25 = 475 sq. m

Total cost = 60

475 ×100

= Rs. 285.

9.c)No. of Stone =Area to be (tiled) Stones

Area of (Stone)

=

60 610 9

10 10

= 60 6 10

9

= 400

= 1410.a) Distance travelled by a wheel is always equal

to Circumference x No. of revolutions

Radius = 35

2

Circumference = 2r = 22 35

2 =117 2

Dist. Traveled = 6300 x 110

= 693000 cmor 6930 m or 6.93 km

11.a) Percentage decrease = 20

100100 + 20

R = 20%

% decreased = 20

100100 + 20

= 20

100120

= 1623

12.a) (Side)² = 2 2

224 D2 2

400 = 144 + 2

2D

2

400 - 144 = AO² AD = 36 Area = ½ x 32 x 24 = 384

13.a) n = 5

No. of Diagonals = n(n - 3)

2

no. of diagnols =

5(5 - 3) 5 2

2 2= 5

14.a) 3

(12)(12) = 36 34

15.a) BC² = 65² - 63²= (65+63) (65-63) =256

BC = 16CD² = 65² - 56²

= 121 x 9

= 1089CD = 33

BD = BC + CD= 16 + 33= 49

Volume and Surface Area of Solids1. Tetrahedron : A triangular solid bounded by four

plane faces.Total surface area of regular tetrahedron

= (Side)2 3Volume of regular tetrahedron

= 1

3x area of base x height

2. Cube : A triangular solid figure bounded by sixequal square faces.Volume of cube = (Side or Edge)3,Surface area of cube = 6 (Edge)2,

Diagonal of cube = Edge x 3 ,

Edge or Side of cube = 3 volume

or d

3 or

Surface

6

3. Cuboid (Rectangular solid) : A solid figurebounded by six rectangular faces.

Volume = length x Breadth x Height = lbh

Length = Volume

Breadth x Height=

Volume

b x h

Breadth = Volume

Length x Height =

Volume

l x h

Height = Volume

Length x Height=

Volume

l x b

Diagonal of cuboid

= 2 2 2Length + Breadth + Height

= 2 2 2l +b +hWhole surface area = 2(lb + bh + lh)

4. Cylinder : A solid cylindrical figure which isequally curved from all sides, where height = hand radius = r.

hh

Volume of the cylinder = r2h= Area of base x Height

Curved surface of cylinder = 2rh= Circumference of base x height

Area of each end = r2

Total surface area of a solid cylinder = 2r (r + h)Total surface area of hollow pipe or cylindricalring = 2(R + r)(h + R - r)Where 'R' stands for outer radius and ‘r’ for innerradius and ‘h’ stands for the height of pipe.

5. Cone : A solid figure with circular base andtapering to a point.Height = h, slant height = l, Radius of base = r .

h

r

l

Slant height (l) = 2 2 h + r

Perpendicular height (h) = 2 2 l - r

Radius of base (r) = 2 2 l - r

Volume of cone = 13 r2h

Lateral surface or curved surface

MENSURATIONS-II

area of cone = rhWhole surface or surface area of cone

= r (r +l) (in whole surface area, the area ofbase is also included.)

6. Sphere : A solid circular f igure which isequidistant from its central point.

r

Volume of sphere = 43 r3

Surface area of sphere = 4r2

Volume of hemisphere =

31 4r

2 3= 32

r3

Surface area of hemisphere =

2

24 rr

2= r2

Segment : The figure bounded by a plane and thepart of sphere it cuts off.

hr

Volume = h

6(3r2 + h2)

or volume = 2h

6(3d-2h)

Surface Area = dhwhere r is the radius of segment, h is height ofsegment and d is diameter of the sphere.Sector of sphere : It is the solid composed of asegment of the sphere and the cone having thebase of the segment for its base and the centreof the sphere for its vertex.

h

r

Volume = 23 r2h, S = r

22rh - h2h +

where ‘r’ is radius of sphere, h is height of curvedsurface of segment of the sphere that forms thebase of the sector. S is surface area.

7. Prism : A polyhedron (i.e., solid bounded by planesurfaces only) whose are parallelograms andwhose ends are identical polygons in parallelplane.

Al

Volume of prism= Area of triangular plane x HeightTotal surface area= Lateral surface area + Area of endsArea of base = Calculate according to the figure

Height of prism = Volume

Area of Base

Lateral surface Area= Perimeter of base x HeightPerimeter of base= Calculate according to the figure.

8. Pyramid : A pyramid is a solid bounded by planefaces on any polygonal base, and the faces arein a triangular shape meet on a common pointon certain height. If the sides of the base areequal then it is called a right pyramid.Height : The straight line drawn perpendicularfrom the centre of the base to the vertex point iscalled height.Slant height : The distance of line which bisectsthe side of the base at a right angle from thevertex is called slant height.

P

h l

O Aa

Volume of right pyramid

= 1

3 x Area of base x Height

Slant surface area of pyramid

= 1

2 x (Perimeter of base) x Slant height

or Slant surface area of right pyramid

=

22 2na

+ h2 n

acot

4

The whole surface = Slant surface + Area of baseSlant height of right pyramid

= 2 2cot

n

2a4

h

Height of Pyramid

= Volume of Pyramid x 3

Area of Base

(For area of base please see the formula of areaof polygon, stands for 180°)

9. Polyhedron : A polyhedron (poly means many,hedron means bases) is a solid bounded by planefaces.Volume of polyhedron = Area of base x Height(Note : For area of base, see the formula ofarea of polygon)Total surface area of polyhedron= Perimeter of base x Height + 2 (Area of base)

Important Points to Remember1. If all the sides of any polygon are increased by n

times then its area will be increased by n2 timesand its volume will be increased by n3 times.

2. If the radius of the circle is increased by n timesthen the area will be increased by n2 times.

3. If the radius of the sphere is increased by n timesthen the curved surface area of the sphere willbe increased by n2 times and the volume will beincreased by n3 times.

4. If the diagonal of a square is increased by n timesthen its area will be increased by n2 times.

5. If all sides of any polygon are increased by 25%then its area will be increased by 56.25% andthe volume will be increased by 95.31%.

6. If all the sides of any polygon are increased by50% then its area will be increased by 125% andits volume will be increased by 237.5%.

7. The ratio of the area of a square to that of thesquare drawn on its diagonal is 1 : 2.

5. If the radius and height of a cylinder and coneare same then the volume of the cylinder will bethree times than that of cone.

9. If the volume and radius are same than the heightof the cone will be 3 times to that of the cylinder.

10. The ratio of volume of sphere and curved surface

area is

3

2

43

r

4 r=

r

3= r : 3

11. The ratio of area of circle and its circumference

is =

2r

2 r=

r

2 = r : 2

12. If the sides of a triangle are in the ratio of 3 : 4 :5 and 5 : 12 : 13 then the triangle will be a rightangled triangle and the biggest side will be thehypotenuse.

13. The number of small cubes cut out from a big cube

=

3Edge of big cube

Edge of small cube

14. If the volumes of cube, sphere, polyhedron,cylinder and tetrahedron are equal, then the wholesurface area in the ascending order are : sphere,cylinder, polyhedron, cube and tetrahedron.

15. If a circle is inscribed in an equilateral triangle,then the radius of the circle will be 1/3rd of theheight of the triangle.

16. If an equilateral triangle is inscribed in a circle,then the radius of the circle will be 2/3rd of theheight of the triangle.

17. The greatest cube is cut out from a right circularcone, the side of the cube

x = 2r x h

2r + h where r and h be the radius and

height of the cone.Example : Two of the side faces of a pyramid areequilateral triangles and the other two side faces areright angled triangles. If the length of each side ofthe equilateral triangles is 6 metres, then find thevolume of the pyramid.Sol : The base in such type of pyramid is rectangularbecause the opposite faces of the pyramid areequilateral triangles and the remaining faces are rightangled (on the top) triangles.

APB = 90°

AB = 36 + 36 = 6 2 m and AD = 6 m. (Given)

O

P

90º

D

A B6 2

Area of base of pyramid = 6 2 x 6 = 36 2 m2

Diagonal of base = 72 + 36 = 6 3

AO = 6 3

2= 3 3

Height of the pyramid = OP = 2 2AP - AO

= 36 - 27 = 3

So, volume of pyramid= 16 x 6 2 x 3

3= 36 2 m3.

Area of four walls = 2 (L + B) x H

MULTIPLE CHOICE QUESTIONS1. A river 2 metres deep and 45 metres wide is

flowing at the rate of 3 km per hour. Find howmuch water runs into the sea per minute.(a) 5000 cu in (b) 5400 cu m(c) 4500 cu m (d) None of these

2. A wooden box of dimensions 8 m x 7 m x 6 m isto carry rectangular boxes of dimensions 8cm x7cm x 6cm. The maximum number of boxes thatcan be carried in the wooden box, is :(a) 9800000 (b) 7500000(c) 1000000 (d) 1200000

3. A tank contains 60000 cubic metres of water. Ifthe length and breadth are 50 metres and 40metres respectively, find the depth.(a) 50 metres (b) 25 metres(c) 30 metres (d) 20 metres

4. A swimming bath is 21 m long and 15 m broad.When a number of men dive into the bath, theheight of the water rises by one cm. If the averageamount of water displaced by one of the men be0.1 cubic m, how many men are there in thebath ?(a) 32 (b) 46(c) 42 (d) 36

5. A school room is to be built to accommodate 70children, so as to allow 2.2 sq metres of floorand 11 cub metres of space for each child. If theroom be 14 metres long, what must be its breadthand height?(a) 12 metres, 5.5 metres(b) 11 metres, 5 metres(c) 13 metres, 6 metres(d) 11 metres, 4 metres

6. Find the length of the longest pole that can beput in a room 10 metres by 8 metres by 5 metres.(a) 12 m (b) 16m(c) 13 m (d) 14m

7. A cylindrical iron rod is 70 cm long, and thediameter of its end is 2 cm. What is its weight,reckoning a cubic cm of iron to weigh 10 grams?(a) 4 kg (b) 4.2 kg(c) 2.2 kg (d) Data inadequate

8. The area of the curved surface of a cylinder is4400 cm2 and the circumference of its base is110 cm.Find the height and the volume of thecylinder.(a) 40 cm, 38500 cu cm(b) 45 cm, 38500 cu cm

(c) 10 cm, 58500 cu cm(d) 45 cm, 38560 cm

9. Six spherical balls of radius r are melted and castinto a cylindrical rod of metal of same radius.The height of rod will be:(a) 4r (b) 6r(c) 8r (d) 12r

10. A rectangular sheet with dimension 11mx8m isrolled into a cylinder so that the smaller sidebecomes the height of the cylinder. What is thevolume of the cylinder so formed?(a) 77 cu m (b) 87 cu m(c) 66 cu m (d) Data inadequate

11. A copper sphere of diameter 12 cm is drawninto a wire of diameter 2 cm. Find the length ofthe wire,(a) 288 cm (b) 284 cm(c) 286 cm (d) None of these

12. Surface of two cubes are in the ratio of 1 : 5.Find the ratio of their volumes.

(a) 1 : 5 5 (b) 5 5 : 1(c) 5 : 1 (d) 125 : 1

13. The radii of two cylinders are in the ratio of 2:3and their heights are in the ratio 5:3. The ratio oftheir volumes is:(a) 27 : 20 (b) 20 : 27(c) 4 : 9 (d) 9 : 4

14. If 1 cm3 cast iron weights 21 gm, then weightof a cast iron pipe of length 1 m with a bore of3 cm and in which thickness of metal is 1 cm is :(a) 21 kg (b) 24.2 kg(c) 26.4 kg (d) 18.6 kg

15. A cone, whose height is 1/16 of its radius, ismelted to form a sphere. Find the ratio of radiusof the sphere to that of the cone.(a) 1 : 3 (b) 1 : 4(c) 2 : 3 (d) 1 : 2

ANSWERS

1. (c) 2. (c) 3. (c) 4. (d) 5. (b)6. (d) 7. (c) 8. (a) 9. (c) 10. (a)

11. (a) 12. (a) 13. (b) 14. (c) 15. (b)

Hints and Explanations :

1.c) Volume of water fell into sea in 60 mins= 2 x 45 x 3000

Volume of water / min = 270000

60= 4500 cu m.

2.c) No. of boxes = volume of bigger box

volume of smaller box

= 800 x 700 x 600

8 x 7 x 6= 106.

3.c) Volume = 6000050 x 40 x h = 60000

h = 60000

50 x 40= 30.

4.d) L = 24 mB = 15 m

Volume = 24 x 15 x 1

10

Let x be the no. of men

Here volume of H2O

Increased = Volume of H2O displaced

24 x 15x 1

10= x x

1

10

x = 36.5.b) Total area = 2.2x 70= 14 x B

11 = B or B = 11 mts.also total space = 11 x 70 = 14 x 11 x h

h = 5 mts.

6.d) D = 2 2 210 x 8 x 5

= 100 x 64 x 25

= 189 = 14 cm

22 7 7× × ×80

7 2 2= 3080.

=

7.c)Volume= 22

×1×707

=220 cu. cm

weight = 220 x 10= 2200 grams= 2.2 kg.

8.a) 2RH = 4400 sq. cm

2R = 110

= 2RH = 110 H = 4400

H = 440

11= 40

R =

110

2 x

Volume =

110 110 × × 42 2

= 38500 cu cm.

9.c) 346× r

3= 2hr

h = 8r

10.a) r2 = 11, h = 8, r =

11

2

11

8

11

8

V =

11 11× × × 8

2 2

= 77 m3.11.a) Here volume of sphere = volume of wire

Diameter of sphere = 12 cmDiameter of wire = 2 cmRadius of sphere = 6 cmRadius of sphere = 1 cm.Now,

4

3x x 6 x 6 x 6 x = 1 x 1 x h

288 cm = h.

12.a) 1

2

SoA 1

SoA 5

2122

6S 1

56S

2122

S 1

5S

15

2

S 1

S

Now,

3

1 13

2 2

V S

V S

3

31

( 5)

1

5 5

13.b) Here

1 1,2 2

R H2 5

R 3 H 3

2 2

2 211

2

2H1 1 12H2 2 2

V R H R 2 5

V R H 3 3R

20

27

14.c) Area of ring = [R²]-[r²]

=

252

-

232

=

25 94 4

= 164 = 4

Volume = 4 x 100 W = 4 x 22 x 100 x 21

= 26400g= 26.4 kg

15.b) H =116 radius

r = 16 HVolume of cone = Volume of sphere.

1

3r²x

1

15r =

4

3r³

r³

64= R³

R³

r³=

1

64

R

r=

1

4= 1 : 4