Mensuration PDF

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Mensuration PDF

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and volumes in the broadest sense, it is all about the process of measurement.

These are two types of geometrical shapes (1) 2D (2) 3D

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.

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Basic Conversions:

  1. 1 km = 10 hm

1 hm = 10 dam

1 dam = 10m

1 m = 10dm

1 dm = 10 cm

1 cm = 10mm

1 m = 100 cm = 1000 mm

1 km = 1000m

  1. 1 km = miles

1 mile = 1.6 km

1 inch = 2.54 cm

1 mile = 1760 yd = 5280 ft.

1 nautical mile (knot) = 6080 ft

  • 100 kg = 1 quintal

10 quintal = 1 tonne

1 kg = 2.2 pounds (approx.)

  1. 1 liter = 1000cc

1 acre = 100m2

1 hectare =10000 m2(100 acre) 

PART I: PLANE FIGURES

Perimeter (P) = a+b+c

Area (A) =√(s(s-a)(s-b)(s-c) )

Where s=(a+b+c)/2and a, band care three sides of the triangle.

Also A = 1/2×b×h: Where b →base h→ altitude

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

Solution:
Here a = 50 m, b = 78 m, c = 112m

s=1/2 (50+78+112)=120m

s-a=120-50=70m

s-b=120-78=42m

s-c=120-112=8m

∵Area=√(120×70×42×8)=1680 sq.m.

∴ Area = 1⁄2 base × perpendicular

∴Perpendicular=(2 Area)/Base=(1680×2)/112=30m.

Example 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. Hectometer

Solution:
Area of the field =(Base×Height)/2

=(880×550)/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

Note:

In a rectangle, (Perimeter)^2/4 = (diagonal)2 + 2 × Area

In an isosceles right angled triangle,

Area =b/4 √(4a^2-b^2 )

where a is two equal side and b is different side Ina 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))%

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

Solution:

Since, A =1/2 (a+b)h

⇒5100=1/2 (40+x)×60

⇒170=40+x

∴ other parallel side = 170 – 40 = 130m

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

Solution:

A = lb – (l – 2a) (b – 2a)

= 112 × 78 – (112 – 5) (71 – 5)

=112 × 78 – 107 × 73 – 8736 – 7811 = 925 sq.m

∴ Cost of construction = rate × area = 2 × 925 = `1850

Example 5: 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 22m respectively

Solution:

Area =1/2×d×(h_1+h_2 )

=1/2×60(38+22)=1800 sq.m

Example 6: A wire is looped in die 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.

Solution:

(a) Length of the wire – Perimeter of the circle

= 2? × 28

= 176 cm2

Side of the square =176/4=44cm

Example 7: The radius of a wheel is 42 cm. How many revolutions will it make in going 26.4 km?

Solution:

Distance travelled in one revolution = Circumference of

the whee=2πr=2×22/7×42cm=264cm

∴ No. of revolutions required to travel 26.4 km

=(26.4×1000×100)/264=10000

Example 8: Find the area of sector of a circle whose radius is 6 cm when

  • the angle at the centre is 35°
  • when the length of arc is 22 cm

Solution:

  • Area of sector

=πr2.θ/(360°)=22/7×6×6×35/360 〖cm〗^2=11 sq.cm

  • Here length of arc l = 22 cm.

∴2πr θ/(360°)=22cm

Area of sector =πr^2.θ/(360°)=1/2 r.2πr θ/(360°)

=1/2 rl=1/2×6×22sq.cm=66sq.cm.

Example 9: The radius of a circular wheel is  m. How   many revolutions will it make in travelling 11 km?

Solution:

Distance to be travelled =11km=11000m

Radius of the wheel =13/4 m=7/4 m

∴ Circumference of the wheel=2×22/7×7/4=11m

∴ In travelling 11 m, wheel makes 1 revolution.

∴In travelling 11000 m the wheel makes 1/11×11000 revolutions. i.e. 1000 revolutions.

CUBOID

A cuboid is a three dimensional box.

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

Volume of the cuboid = lbh

Length of diagonal √(l^2+b^2+h^2 )

(Area of four walls = 2(l + b) × h)

Rectangular Parallel piped box. It is same as cuboid. Formally a polyhedron for which all faces are rectangles.

CUBE

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

Total surface area of a cube = 6a2

Volume of longest the cube = a3

Length of longest diagonal = √3 a

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

RIGHT CIRCULAR CYLINDER

It is a solid which has both its ends in the form of a circle, lateral surface area = 2?rh

Total surface area = 2?r (r + h)

Volume = ?r2h; where r is radius of the base and h is height

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

=1/2 (perimeter of base)×slant height=1/2 pl

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

Volume 1/3 (area of the base) × height

RIGHTCIRCULAR 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 = ?rl

Total surface area = ?r (l + r)

Volume ==1/3 πr^2 h; where r: radius of the base

h: height

l: slant height

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

Solution:

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

l + b + h = 19 … (i)

and √(l^2+b^2+h^2 )=11

⇒l^2+b^2+h^2=121 … (ii)

Area of the surface to be painted

=2(lb+bh+hl)

(l+b+h)^2=l^2+b^2+h^2+2(lb+bh+hl)

⇒2(lb+bh+hl)=(19)^2-121=361-121=240

Surface area of the room = 240m2

Cost of painting the required area = 10 × 240 = `2400

Example 11: A road roller of diameter 1.75 m and length 1m has to press a ground of area 1100 sqm. How many revolutions does it make?

Solution:

Area covered in one revolution = curved surface area

Number of revolutions=(Total area to be pressed)/(Curved surface area)

=1100/2πrh=1100/(2×22/7×1.75/2×1)

= 200

Example 12: 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.

Solution:

Area of land = 10000 sqm

Volume of rainfall =(10000×43)/100=4300m^3

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

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