Formula - Mensuration


MENSURATION

Mensuration is the science of measurement of the lenghts of lines, areas of surfaces and volumes of solids.

Perimeter : Perimeter is sum of all the sides. It is measured in cm,m, etc.

Area : The area of any figure is the amount of surface enclosed within its boundary lines. This is measured in square unit like cm2, m2, etc.

Volume : If an object is solid, then the space occupied by such an object is called its volume. This is measured in cubic unit like cm3, m3, etc.

Basic Conversions :

I. 1 km = 10 hm

1 hm = 10 dam

1 dam = 10 m

1 m = 10 dm

1 dm = 10 cm

1 cm = 10 mm

1 m = 100 cm = 1000 mm

1 km = 1000 m

II. 1 km = 58 miles

1 mile = 1.6 km

1 inch = 2.54 cm

1 mile = 1760 yd = 5280 ft.

1 nautical mile (knot) = 6080 ft

III. 100 kg = 1 quintal

10 quintal = 1 tonne

1 kg = 2.2 pounds (approx.)

IV. l litre = 1000 cc

1 acre = 100 m2

1 hectare = 10000 m2(100 acre)

Part I : Plane Figures

TRIANGLE

rnfj

Perimeter (P) = a + b + c

Area (A) = \(\sqrt{s\left ( s ? a \right )\left ( s ? b \right )\left ( s ? c \right )}\)

where s = \(\frac{a + b + c}{2}\) and a, b and c are three sides of the triangle.

Also, A = ½ × bh;

where b →  base
h → altitude

Equilateral triangle

 

23

Perimeter = 3a

A = √34 a2 ; where a → side

Right triangle

 

1

A = ½pb and h2 = p2 + b2 (Pythagoras triplet)

where,p → perpendicular
b → base
h → hypotenuse

Example 1

1.Find the area of a triangle whose sides are 50 m, 78 m, 112 m respectively and also find the perpendicular from the opposite angle on the side 112 m.

Sol. Here a = 50 m, b = 78 m, c = 112 m

s = ½(50 + 78 + 112) = 120 m

s – a = 120 – 50 = 70 m
s – b = 120 – 78 = 42 m
s – c = 120 – 112 = 8 m

∴ Area = \(\sqrt{120 \times 70 \times 42 \times 8} = 1680 \; sq.m.\)

∵ Area = ½ base × perpendicular

∴ perpendicular = \(\frac{2Area}{Base} = \frac{1680 \times 2}{112} = 30m.\)

Example 2

2. The base of a triangular field is 880 m and its height 550 m. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of 24.25 per sq. hectometre.

Sol. Area of the field = \(\frac{Base \times Height}{2}\)

= \(\frac{880 \times 550}{2} = 242000 \; sq.m. = 24.20 \; sq.hm.\)

Cost of supplying water to 1 sq. hm = 24.25

∴ Cost of supplying water to the whole field

= 24.20 × 24.25 = 586.85

SOME IMPORTANT RESULT

In a rectangle, (Perimeter)24 = (diagonal)2 + 2 × Area

In an isosceles right angled triangle,
Area = 23.3 × (perimeter)2

In a parallelogram,
Area = Diagonal × length of perpendicular on it

If area of circle is decreased by x%, then the radius of circle is decreased by (100 - 10√(100 - x))%

RECTANGLE

2

Perimeter = 2 (l + b)

Area = l × b; where l → length
b → breadth

SQUARE

3

Perimeter = 4 × side = 4a

Area = (side)2 = a2; where a → side

PARALLELOGRAM

4

Perimeter = 2 (a + b)

Area = b × h;

where a → breadth
b → base (or length)
h → altitude

RHOMBUS

5

Perimeter = 4a

Area = ½ d1 × d2

where a → side and
d1 and d2 are diagonals.

IRREGULAR QUADRILATERAL

6

Perimeter = p + q + r + s

Area = ½ × d × (h1 + h2)

TRAPEZIUM

7

Perimeter = a + b + m + n

Area = ½ (a + b)h;

where a and b are two parallel sides;
m and n are two non-parallel sides;
h → perpendicular distance between two parallel sides.

Area of pathways running across the middle of a rectangle

8

A = a (l + b) – a2;

where l → length
b → breadth,
a → width of the pathway.

PATHWAY SOUTSIDE

9

A = (l + 2a) (b + 2a) – lb;

where l → length
b → breadth
a → width of the pathway

PATHWAYS INSIDE

10

A = lb – (l – 2a) (b – 2a);

where l → length
b → breadth
a → width of the pathway

Example 3

3. A 5100 sq.cm trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40 m then find the length of the other parallel side.

Sol. Since, A = ½ (a + b)h;

=> 5100 = ½(40 + x) × 60

=> 170 = 40 + x

∴ other parallel side = 170 – 40 = 130 m

Example 4

4. A rectangular grassy plot is 112 m by 78 m. It has a gravel path 2.5 m wide all round it on the inside. Find the area of the path and the cost of constructing it at 2 per square metre?

Sol. A = lb – (l – 2a) (b – 2a)

= 112 × 78 – (112 – 5) (78 – 5)

= 112 × 78 – 107 × 73 = 8736 – 7811

= 925 sq.m

∴ Cost of construction = rate × area

= 2 × 925 = Rs. 1850

Example 5

5. The perimeter of a rhombus is 146 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus.

Sol. Let ABCD be the rhombus in which AC = 55 cm.

11

and AB = 1464 = 36.5 cm.

Also, AO = 552 = 27.5 cm.

∴ BO = \(\sqrt{\left ( 36.5 \right )^{2} ? \left ( 27.5 \right )^{2}} = 24 \; cm\)

Hence, the other diagonal BD = 48 cm

Now, Area of the rhombus = ½ AC × BD

= ½ × 55 × 48 = 1320 sq.cm.

Example 6

6. Find the area of a quadrilateral piece of ground,one of whose diagonals is 60 m long and the perpendicular from the other two vertices are 38 and 22 m respectively.

Sol. Area = ½ × d × (h1 + h2)

= ½ × 60 × (38 + 22) = 1800 sq.m.

CIRCLE

12

Perimeter (Circumference) = 2πr = πd

Area = πr2;

where r → radius
d → diameter
and π = 227 or 3.14

SEMICIRCLE

13

Perimeter = πr + 2r

Area = ½ × πr2

SECTOR OF A CIRCLE

14

Area of sector OAB = \(\frac{\theta}{360} \times \pi r^{2}\)

Length of an arc (l) = \(\frac{\theta}{360} \times 2\pi r\)

Area of segment = Area of sector – Area of triangle OAB

= \(\frac{\theta}{360^{\cdot}} \times \pi r^{2} - \frac{1}{2} r^{2} sin\theta\)

Perimeter of segment = length of the arc + length of segment

AB = \(\frac{\pi r\theta}{180} + 2r sin\frac{\theta}{2}\)

RING

15

Area of ring = π(R22 - R12)

Example 7

7. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of a side of the square.
(a) 44 cm
(b) 45 cm
(c) 46 cm
(d) 48 cm

Sol.(a) Length of the wire = Perimeter of the circle

= 2π × 28

= 176 cm2

Side of the square = 1764 = 44 cm

Example 8

8.The radius of a circular wheel is 134 m. How many revolutions will it make in travelling 11 km ?

Sol. Distance to be travelled = 11 km = 11000 m

Radius of the wheel = 134 = 74

∴ Circumference of the wheel = 2 × 227 × 74 = 11 m

∴ In travelling 11 m, wheel makes 1 revolution.

∴ In travelling 11000 m the wheel makes 111 × 11000 revolutions, i.e., 1000 revolutions.

Example 9

9. The circumference of a circular garden is 1012 m. Find the area of outsider road of 3.5 m width runs around it. Calculate the area of this road and find the cost of travelling the road at 32 per 100 sqm.

Solution : A = πr2, C = 2πr = 1012

16

=> r = 1012 × 12 × 722 = 161 m

Area of garden = 227 × 161 × 161 = 81466 sqm

Area of the road = area of bigger circle – area of the garden

Now, radius of bigger circle = 161 + 3.5 = 3292 m

∴ Area of bigger circle = 227 × 3292 × 3292 = 85046½ sq.m.

Thus, area of the road = 85046½ – 81466 = 3580½ sqm.

Hence, cost = 71612 × 32100 = 1145.76

Example 10

10. There is an equilateral triangle of which each side is 2m. With all the three corners as centres, circles each of radius 1 m are described.

(i) Calculate the area common to all the circles and the triangle.
(ii) Find the area of the remaining portion of the triangle.

Sol.

17

Area of each sector = \(\frac{\theta}{360} \times \pi r^{2} = \frac{60}{360} \times \frac{22}{7} \times 1 \times 1\)

= \(\frac{1}{6} \times \frac{22}{7} = \frac{11}{21} m^{2}\)

Area of equilateral triangle = \(\frac{\sqrt{3}}{4}a^{2}\)

= \(\frac{\sqrt{3}}{4} \times 2 \times 2 = \sqrt{3} m^{2}\)

(i) Common area = 3 × Area of each sector

= 3 × 1121 = 117 = 1.57 m2

(ii) Area of the remaining portion of the triangle = Ar. of equilateral triangle – 3(Ar. of each sector)

√3 – 1.57 = 1.73 – 1.57 = 0.16 m2

PART-II SOLID FIGURE

CUBOID

A cuboid is a three dimensional box.

Total surface area of a cuboid = 2 (lb + bh + lh)

Volume of the cuboid = lbh

18

Area of four walls = 2(l + b) × h

CUBE

A cube is a cuboid which has all its edges equal.

Total surface area of a cube = 6a2

Volume of the cube = a3

19

RIGHT PRISM

A prism is a solid which can have any polygon at both its ends.

Lateral or curved surface area = Perimeter of base × height

Total surface area = Lateral surface area + 2 (area of the end)

Volume = Area of base × height

20

RIGHT CIRCULAR CYLINDER

It is a solid which has both its ends in the form of a circle.

Lateral surface area = 2πrh

Total surface area = 2πr (r + h)

Volume = πr2h; where r is radius of the base h is height

21

PYRAMID

A pyramid is a solid which can have any polygon at its base and its edges converge to single apex.

Lateral or curved surface area

= ½ (perimeter of base) × slant height

Total surface area = lateral surface area + area of the base

22

Volume = ⅓(area of the base) × height

RIGHT CIRCULAR CONE

It is a solid which has a circle as its base and a slanting lateral surface that converges at the apex.

Lateral surface area = πrl

Total surface area = πr (l + r)

Volume = ⅓ πr2h;

where r : radius of the base
h : height
l : slant height

23

SPHERE

It is a solid in the form of a ball with radius r.

Lateral surface area = Total surface area = 4πr2

Volume = 43πr3

where r is radius.

24

HEMISPHERE

It is a solid half of the sphere.

Lateral surface area = 2πr2

Total surface area = 3πr2

Volume = ⅔πr3

where r is radius

25

FRUSTUM OF A CONE

When a cone cut the left over part is called the frustum of the cone.

Curved surface area = πl(r1 + r2)

Total surface area = πl(r1 + r2) + πr12 + πr22

where \(l = \sqrt{h^{2} + \left ( r_{1} - r_{2} \right )^{2}}\)

24

Volume = \(\frac{1}{3}\pi h\left ( r_{1}^{2} + r_{1}r_{2} + r_{2}^{2} \right )\)

Example 11

11. The sum of length, breadth and height of a room is 19 m. The length of the diagonal is 11 m.The cost of painting the total surface area of the room at the rate of 10 per m2 is :

(a)240
(b)2400
(c)420
(d)4200

Sol.(b) Let length, breadth and height of the room be l, b and h, respectively. Then,

l + b + h = 19

and \(\sqrt{l^{2} + b^{2} + h^{2}} = 11\)

=> \(l^{2} + b^{2} + h^{2} = 121\)

Area of the surface to be painted

= 2(lb + bh + hl)

(l + b + h)2 = l2 + b2 + h2 + 2 (lb + lh + hl)

=> 2(lb + bh + hl) = (19)2 – 121 = 341 – 121 = 240

Surface area of the room = 240 m2.

Cost of painting the required area = 12 × 240 = 2400

Example 12

12. ABCD is a parallelogram. P,Q, R and S are points on sides AB, BC, CD and DA, respectively such that AP = DR. If the area of the rectangle ABCD is 16 cm2, then the area of the quadrilateral PQRS is:

26

(a) 6 cm2
(b) 6.4 cm2
(c) 4 cm2
(d) 8 cm2

Sol.(d) Area of the quadrilateral PQRS

= Area of ΔSPR + Area of ΔPQR

= ½ × PR × AP + ½ × PR × PB

= ½ × PR(AP + PB) = ½ × AD × AB

(PR = AD and AP + PB = AB)

= ½ × Area of rectangle ABCD = ½ × 16 = 8 cm2

Example 13

13. A road roller of diameter 1.75 m and length 1 m has to press a ground of area 1100 sq m. How many revolutions does it make ?

Sol. Area covered in one revolution = curved surface area

∴ Number of revolutions = \(\frac{Total \; area \; to \; be \; pressed}{Curved \; surface \; area}\)

= \(\frac{1100}{2\pi rh} = \frac{1100}{2 \times \frac{22}{7} \times \frac{1.75}{2} \times 1}\)

= 200

Example 14

14. The annual rainfall at a place is 43 cm. Find the weight in metric tonnes of the annual rain falling there on a hectare of land, taking the weight of water to be 1 metric tonne to the cubic metre.

Sol. Area of land = 10000 sqm

Volume of rainfall = \(\frac{10000 \times 43}{100} = 4300 \, m^{3}\)

Weight of water = 4300 × 1 m tonnes = 4300 m tonnes

Example 15

15. The height of a bucket is 45 cm. The radii of the two circular ends are 28 cm and 7 cm, respectively. The volume of the bucket is :
(a) 38610 cm3
(b) 48600 cm3
(c) 48510 cm3
(d) None of these

Sol.(c) Here r1 = 7 cm, r2 = 28 cm and h = 45 cm

27

Volume of the frustum of a cone

Volume of the bucket = \(\frac{1}{3} \pi h\left ( r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}\right )\)

Hence, the required volume

= \(\frac{1}{3} \times \frac{22}{7} \times 45\left ( 28^{2} + 7^{2} + 28 \times 7 \right ) = 48510 \, cm^{3}\)

Example 16

16. A hollow cylindrical tube open at both ends is made of iron 2 cm thick. If the external diameter be 50 cm and the length of the tube be 140 cm,find the number of cubic cm of iron in it.

Sol. Height = 140 cm
External diameter = 50 cm
∴ External radius = 25 cm

29

Also, internal radius OA = OB – AB = 25 – 2 = 23 cm

∴ Volume of iron = Vexternal – Vinternal

= \(\frac{22}{7} \times 140\left ( 25^{2} ? 23^{2}\right) = 42240 \, cu.cm.\)

Example 17

17. A cylindrical bath tub of radius 12 contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

(a) 80 cm2
(b) 84 cm2
(c) 104 cm2
(d) 76 cm2

Sol.(b) Volume of the spherical ball = volume of the water displaced.

=> 43 πr3 = π(12)3 × 6.75

=> r3 = \(\frac{144 \times 6.75 \times 3}{4} = 729\)

or r = 9 cm.

Example 18

18. A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy. (Use π = 3.14).

Sol. The radius of the hemisphere = ½ × 6 = 3 cm

Now, slant height of cone = \(\sqrt{3^{2} + 4^{2}} = 5 \, cm\)

30

The surface area of the toy

= Curved surface of the conical portion + curved surface of the hemisphere

= (π × 3 × 5 + 2π × 32) cm2 = 3.14 × 3 (5 + 6) cm2 = 103.62 cm2.

Example 19

19. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Re. 1 per dm2.

Sol.

30

Let the height of the cylinder be h cm.

Then h + 7 + 7 = 104

=> h = 90

Surface area of the solid

= 2 × curved surface area of hemisphere + curved surface area of the cylinder

= (2 × 2 × 227 × 7 × 7 + 2 × 227 × 7 × 90) cm2

= 616 + 3960 cm2 = 4576 cm2

Cost of polishing the surface of the solid

= \(\frac{4576 \times 1}{100} = Rs. 45.76\)

Example 20

20. A regular hexagonal prism has perimeter of its base as 600 cm and height equal to 200 cm. How many litres of petrol can it hold ?Find the weight of petrol if density is 0.8 gm/cc.

Sol. Side of hexagon = \(\frac{Perimeter}{Number \; of \; sides} = \frac{600}{6} = 100 \, cm\)

Area of regular hexagon = 3√32 × 100 × 100 = 25950 sq.cm.

Volume = Base area × height

= 25950 × 200 = 5190000 cu.cm. = 5.19 cu.m.

Weight of petrol = Volume × Density

= 5190000 × 0.8 gm/cc

= 4152000 gm = 4152 kg.

Example 21

21. A right pyramid, 12 cm high, has a square base each side of which is 10 cm. Find the volume of the pyramid.

Sol. Area of the base = 10 × 10 = 100 sq.cm.

Height = 12 cm

∴ Volume of the pyramid = 13 × 100 × 12 = 400 cu.cm.

Example 22

22. Semi-circular lawns are attached to both the edges of a rectangular field measuring 42 m × 35 m. The area of the total field is :

(a) 3818.5 m2
(b) 8318 m2
(c) 5813 m2
(d) 1358 m2

Sol.(a) Area of the field

= 42 × 35 + 2 × 12 × 227 × (21)2 + 2 × 12 × 227 × (17.5)2

= 1470 + 1386 + 962.5 = 3818.5 m2

Example 23

23. A frustum of a right circular cone has a diameter of base 10 cm, of top 6 cm, and a height of 5 cm; find the area of its whole surface and volume.

Sol. Here r1 = 5 cm, r2 = 3 cm and h = 5 cm.

∴ l = \(\sqrt{h^{2} + \left ( r_{1} ? r_{2} \right )^{2}}\)

= \(\sqrt{5^{2} + \left ( 5 ? 3 \right )^{2}} = \sqrt{29} = 5.385 \, cm\)

∴ Whole surface of the frustum

= πl(r1 + r2) + πr12 + πr22

= 227 × 5.385(5 + 3) + 227 × 52 + 227 × 32 = 242. 25 sq.cm.

Volume = πh3(r12 + r1r2 + r22)

= 227 × 53[52 + 5 × 3 + 32] = 256.67 cu.cm.

Example 24

24. A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end, and the other end as its base.The volume of the cylinder, hemisphere and the cone are,respectively in the ratio :

(a) 2: 3 : 2
(b) 3 : 2 : 1
(c) 3 : 1 : 2
(d) 1 : 2 : 3

Sol.(b) We have,

radius of the hemisphere = radius of the cone

= height of the cone

= height of the cylinder = r (say)

Then, ratio of the volumes of cylinder, hemisphere and cone

121

= πr3 : 23 πr3 : 13 πr3 = 1 : 23 : 13 = 3 : 2 : 1