Here, we have provided the links which contains the study materials which will help you in studying and preparing for your examinations of the Mensuration Exercise. Referring to the links we’ve provided below and the links which contains the study materials of Mensuration Exercise. PDF Format along with the list of recommended books which we’ve provided below, you will be able to ace your examinations. We have also provided you the further details which will allow you to do well in your exams and learn more. These study materials help you understand the concepts and everything easily and creates a better space for you to work on. These study materials give you the best resources to study from.
Free Download Study Materials
Every year, they release notifications for several vacant posts, so required people interested in the job can apply for the job. To ace the exam, every candidate needs to prepare for the exam as per the syllabus and information provided in the study materials exclusively for this purpose. To get the study materials that are required, you must download the files provided below. You can click on the links given below to download the files and then study them. We hope you find our resources helpful. Every student who has studied using our study materials finds the exam easy to ace since question banks and solution keys are attached along. Once you’ve mastered the content provided in the study materials we’ve provided for you, you have nothing to worry about when it comes to acing the exam. Use our resources to study well for your exam and get your dream job.
- 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 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 liter = 1000cc
1 acre = 100m2
1 hectare =10000 m2(100 acre)
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.
Here a = 50 m, b = 78 m, c = 112m
∴ Area = 1⁄2 base × perpendicular
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
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
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.
Since, A =1/2 (a+b)h
∴ 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?
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
Area =1/2×d×(h_1+h_2 )
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.
(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?
Distance travelled in one revolution = Circumference of
∴ No. of revolutions required to travel 26.4 km
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
- Area of sector
=πr2.θ/(360°)=22/7×6×6×35/360 〖cm〗^2=11 sq.cm
- Here length of arc l = 22 cm.
Area of sector =πr^2.θ/(360°)=1/2 r.2πr θ/(360°)
Example 9: The radius of a circular wheel is m. How many revolutions will it make in travelling 11 km?
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.
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.
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
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
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
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
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.
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
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?
Area covered in one revolution = curved surface area
Number of revolutions=(Total area to be pressed)/(Curved surface area)
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.
Area of land = 10000 sqm
Volume of rainfall =(10000×43)/100=4300m^3
Weight of water = 4300 × 1 m tonnes = 4300 m tonnes