Mensuration in Maths: Definition, Formulas and Solved Questions
Curious about what mensuration is and its significance? Get ready to enhance your knowledge and grasp the fundamental formulas and principles that govern shapes, sizes, and measurements in various geometrical contexts.
What is Mensuration in Maths?
In mathematics, mensuration refers to the branch that measures geometric figures and shapes. It involves calculating various properties of 2D and 3D shapes, such as:
- Area: The measure of the space enclosed within a shape.
- Perimeter: The total length around a 2D shape.
- Volume: The amount of space occupied by a 3D shape.
- Surface Area: The total area of all the surfaces of a 3D shape.
Mensuration uses formulas and techniques to determine these measurements, which are essential for solving problems related to geometry and practical applications like construction, design, and spatial analysis.
What are 2D figures in Mensuration?
In mensuration, 2D figures (two-dimensional shapes) refer to flat geometric figures that have only two dimensions: length and width (or height). These shapes do not have thickness or depth, and they lie entirely in a single plane. Studying 2D shapes in mensuration involves calculating various properties such as area, perimeter, and sometimes the length of diagonals.
Mensuration deals with finding the area (the amount of space inside the shape) and the perimeter (the total length around the shape) of 2D shapes. Examples of 2D shapes in mensuration include square, rectangle, triangle, circle, parallelogram, and trapezium.
What are 3D figures in Mensuartion?
In mensuration, 3D figures (three-dimensional shapes) refer to solid geometric figures that have three dimensions: length, width, and height (or depth). Unlike 2D shapes, which are flat and only have area and perimeter, 3D shapes have volume and surface area.
Examples of 3D shapes include cubes, cuboids (Rectangular Prisms), spheres, cylinders, cones, and pyramids. Measuring 3D shapes involves calculating their volume (the amount of space they occupy) and surface area (the total area of all their surfaces).
Basic Terminologies In Mensuration
Let us study the essential terminologies in mensuration:
Terminology |
Definition |
---|---|
Area | The amount of space inside a shape or figure |
Perimeter | The total distance around the outside of a shape |
Volume | The amount of space occupied by a 3D object |
Circumference | The distance around the edge of a circle |
Surface Area | The total area of all the faces of a 3D object |
Diagonal | A line segment connecting two non-adjacent vertices of a polygon |
Height | The vertical measurement of an object or figure |
Base | The bottom side of a shape or figure |
Radius | The distance from the centre to the edge of a circle |
Diameter | The distance across a circle through its centre |
Basic 2D Formulas in Mensuration
For 2D shapes in mensuration, the area formulas are essential for calculating the space enclosed within the shape. Here are some common area formulas to find the area of 2D shapes:
Shape |
Formula |
---|---|
Square | Area = side length×side length |
Rectangle | Area = length×width |
Triangle | Area = 1/2×base×height |
Circle | Area = πr2 |
Basic 3D Formulas in Mensuration
Calculating the volume of 3D objects involves different formulas based on the shape. Here are some of the important formulas to find the volume of 3D shapes in mensuration:
Shape |
Formula |
---|---|
Cube | V = s3 |
Rectangular Prism | V = l×w×h |
Sphere | V = (4/3)πr3 |
Cylinder | V = πr2h |
Cone | V = (1/3)πr2h |
Pyramid | V = (1/3)Bh |
2D vs 3D in Mensuration
Let us study some of the key features and the differences between 2D & 3D:
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A comparative table of 2D and 3D figures, including their definitions and formulas for area and volume:
Figure | Definition | 2D Formulas | 3D Formulas |
---|---|---|---|
Square | A quadrilateral with four equal sides and four right angles. | Area: side2 Perimeter: 4×side |
|
Rectangle | A quadrilateral with opposite sides equal and four right angles. | Area: length×width Perimeter: 2×(length+width) |
|
Triangle | A polygon with three sides and three angles. | Area: 1/2×base×height Perimeter: Sum of all sides |
|
Circle | A round shape where all points are equidistant from the centre. | Area: πr2 Circumference: 2πr2 |
|
Cube | A solid with six equal square faces. | Volume: side3 Surface Area: 6×side2 |
|
Cuboid | A solid with six rectangular faces. | Volume: length×width×height Surface Area: 2×(length×width+width×height+height×length) |
|
Sphere | A perfectly round solid where all points on the surface are equidistant from the centre. | Volume: 4/3πr3 Surface Area: 4πr2 |
|
Cylinder | A solid with two parallel circular bases connected by a curved surface. | Volume: πr2h Surface Area: 2πr(r+h) |
|
Cone | A solid with a circular base and a single apex. | Volume: 1/3πr2h Surface Area: πr(r+slant height) |
|
Pyramid | A solid with a polygonal base and triangular faces meeting at an apex. | Volume: 1/3×Base Area×Height Surface Area: Base Area + Area of triangular faces |
Solved Questions With Solutions (MCQs)
Provided below are some of the top selected questions with detailed answers for practice:
Question 1. When three metal cubes measuring 6 cm, 8 cm, and 10 cm on each side are combined and reshaped into a single cube, determine the length of the new cube's side.
(a) 13 cm
(b) 11 cm
(c) 12 cm
(d) 24 cm
Solution: c) 12 cm
Explanation: The total volume will remain the same. Let the side of the resulting cube be = 'a'. Then, 63 + 83 + 103 = a3 =3√1728 = 12 cm
Question 2. Take the volume of a cube is as 216 cm3. Part of this cube is then melted to form a cylinder of length 8 cm. Find the volume of the cylinder.
(a) 342 cm3
(b) 216 cm3
(c) 36 cm3
(d) Data inadequate
Solution: d) Data inadequate
Explanation: Data is inadequate as it’s not mentioned what part of the cube is melted to form a cylinder.
Question 3. The diameters of the two cones are equal. If their slant height is in the ratio 5: 7, find the ratio of their curved surface areas.
(a) 25:7
(b) 25:49
(c) 5:49
(d) 5:7
Solution: d) 5:7
Explanation: Let the radius of the two cones be = x cm
Let the slant height of 1st cone = 5 cm and
the slant height of 2nd cone = 7 cm
Then the ratio of covered surface area = πx5/πx7 =5:7
Question 4. The curved surface area of a cone is 2376 square cm, and its slant height is 18 cm. Find the diameter.
(a) 6 cm
(b) 18 cm
(c) 84 cm
(d) 12 cm
Solution: c) 84 cm
Explanation: Radius = πrl/πl = 2376/3.14x18 = 42 cm
Diameter = 2 x Radius = 2 x 42 = 84 cm
Question 5. The ratio of the radii of a cylinder to that of a cone is 1: 2. If their heights are equal, find the ratio of their volumes.
(a) 1:3
(b) 2:3
(c) 3:4
(d) 3:1
Solution: c) 3:4
Explanation: Let the radius of the cylinder = 1(r)
Then the radius of cone be = 2(R)
Then as per question = πr2h/R2h/3= 3πr2h/πR2h=3r2/R2= 3: 4
Question 6. The circumference of a circle is more than its diameter by 16.8 cm. Find the circumference of the circle.
(a) 12.32 cm
(b) 49.28 cm
(c) 58.64 cm
(d) 24.64 cm
Solution: d) 24.64 cm
Explanation: Let the radius of the circle be = p
Then 2pr – 2r = 16.8
r = 3.92 cm
Then, 2πr = 24.64 cm
Question 7. A spherical shell's outer and inner diameters are 10 cm and 9 cm, respectively. Find the volume of the metal contained in the shell.
(a) 6956 cm3
(b) 141.95 cm3
(c) 283.9 cm3
(d) 478.3 cm3
Solution: b) 141.95 cm3
Explanation: Inner radius(p) = 9 2 = 4.5 cm
Outer radius (R) = 10 2 = 5 cm
Now, the volume of metal contained in the shell = 4πR3- 4πr3 = 4π/3 (R3-r3) = 141.95 cm3
Question 8. The radii of two spheres are in the ratio of 1: 2. Find the ratio of their surface areas.
(a) 1:3
(b) 2:3
(c) 1:4
(d) 3:4
Solution: c) 1:4
Explanation: Let smaller radius (r) = 1
The bigger radius (R) = 2
Then, as per the question
4πr2/4πR2 = (r/R)2= (1/2)2=1:4
Question 9. A 7 m wide road surrounds a circular path whose circumference is 352 m. What will be the area of the road?
(a) 2618 m2
(b) 654.5 m2
(c) 1309 m2
(d) 5236 m2
Solution: a) 2618 m2
Explanation: Let the inner radius = r
Then 2πr = 352 m. Then r = 56
The outer radius = r + 7 = 63 = R
Now, πR2 – πr2 = Area of road
π(R2 – r2) = 2618 m2
Question 10. Seven equal cubes, each side 5 cm, are joined end to end. Find the surface area of the resulting cuboid.
(a) 750 cm2
(b) 1500 cm2
(c) 2250 cm2
(d) 700 cm2
Solution: a) 750 cm2
Explanation: Total surface area of 7 cubes = 7 X 6a2 = 1050
However, when joining end to end, 12 sides will be covered.
So their area = 12 x a2 = 12 x 25 = 300
So the surface area of the resulting figure = 1050 – 300 = 750 cm2
Conclusion
The exploration of mensuration has unveiled a world where mathematical concepts intertwine with practical applications. Understanding the distinctions between 2D and 3D shapes, mastering essential terminologies, and delving into shape formulas are crucial steps in grasping the complexities of mensuration.
One can appreciate how mensuration extends beyond theoretical realms into real-world scenarios by uncovering key concepts and tools necessary for calculations.
Frequently Asked Questions (FAQs)
1. What is the significance of mensuration in mathematics?
Mensuration in mathematics deals with the measurement of geometric shapes and figures, aiding in calculating areas, volumes, and perimeters involving spatial concepts and measurements.
2. How do 2D and 3D shapes differ in mensuration?
2D shapes are flat figures with length and width, while 3D shapes have depth as well. Mensuration involves different formulas for calculating the areas and volumes of these shapes based on their dimensions.
3. What are some essential terminologies used in mensuration?
Common terminologies in mensuration include area, perimeter, volume, base, height, radius, diameter, circumference, surface area, and lateral area. Understanding these terms is crucial for accurate calculations in geometry.
4. Why is it important to know key mensuration concepts?
Understanding key mensuration concepts enables individuals to solve complex mathematical problems involving measurement and spatial relationships. It provides a foundation for various fields such as engineering, architecture, physics, and more.
5. What practical applications does mensuration have in real life?
Mensuration finds applications in various real-life scenarios such as construction (calculating materials needed), landscaping (determining area for gardening), packaging (estimating box sizes), and sports (measuring playing fields).
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