Shapes & Spaces3D Shapes

Geometry is one of the most dynamic and practical sections of Mathematics that primarily involves a wide range of sizes and shapes of numerous figures and their respective properties. The entire concept of geometry can be divided into two different types, i.e. Plane and Solid Geometry. Plane Geometry primarily deals with different flat shapes, such as polygons, lines, curves, etc.

On the other hand, Solid Geometry mainly involves Three-Dimensional (3D) objects such as spheres, cylinders, and cubes, etc. We will focus our discussion on 3D-objects along with various fundamental aspects. A detailed discussion regarding the Surface Area and Volumes of different figures have also been provided along with appropriate formulas for the same.

# What are Three-Dimensional Shapes?

Shapes that can be measured in three different directions such as length, width, and height could be defined as 3D-shapes. These are different from 2D shapes primarily because of the fact because they have some thickness. Some of the most common 3D shapes have been illustrated below.

# 3D Shapes

Two different measures can be used to define 3D shapes, namely – Surface Area and Volume.

The surface area could be understood as a total area of the surface of any 3D-object. It is primarily denoted by the term – *SA. *Also, the square unit is used to denote surface areas. Further, the surface area can be classified into three different categories as given below –

**CSA**– Curved Surface Area – Area of all curved regions**LSA**– Lateral Surface Area – Area of all curved regions including flat surfaces but excluding base areas**TSA –**Total Surface Area –Area of all surfaces of the object including a base of the object.

Further, Volume could be understood as the total space the three-dimensional shape or solid object occupies. “V” is most commonly used for denoting Volume and it is measured in cubic units.

A three-dimensional shape could have attributes of edges, vertices, and faces. Any flat surface in 3D shape could be termed as a face. Any line where two different faces of the object meets could be termed as an edge. The point where 3 edges meet is termed as a vertex.

** **

# Cube

A cube can be defined as 3D shapes six square faces or sides. Properties of a cube include 8 vertices, 6 faces, and 12 edges. Considering “a” as the length of each side of the cube,

**SA of Cube** = 6a^{2 }square unit

**The Volume of Cube** = a^{3 }cubic units

# Cuboid

A cuboid or rectangular prism has rectangular faces where all the angles have a measure of 90 degrees. A cuboid comprises of 8 vertices, 12 edges, and 6 faces.

SA of Cuboid = 2 (lb + bh + hl) Sq. units.

The Volume of Cuboid = lbh cubic units

# Pyramid

A pyramid could be defined as a structure that has a solid shape where outer faces are triangular that converge to meet a singular point at the top. The base of the pyramid could be of any shape i.e. quadrilateral, square, triangular, etc. A square pyramid would have the properties of 5 vertices, 5 faces, and 8 edges.

SA of Pyramid = Slant Height x (1/2) (Perimeter of Triangle) + Base Area Square Units

Volume of Pyramid = (Base Area) x Height x (1/3) Cubic Units

# Cylinder

A cylinder could be understood as a 3D figure that has two circular bases that are connected by a closed curved surface. A cylinder does not have any vertex but comprises of 2 edges, 1 curved face, and 2 flat faces or circles.

SA of Cylinder = 2πr (h +r) sq. units

The volume of Cylinder = πr^{2} h cubic units

# Cone

A 3D object or solid that has a circular base along with a single vertex.The cone decreases gradually in size from circular flat to the top point, which is known as apex. A cone primarily comprises of 1 curved face, 1 edge, 1 vertex, and 1 flat surface, i.e. circle.

SA of Cone = πr (r +√(r^{2}+h^{2}) Sq. units

The volume of Cone = ⅓ πr^{2}h Cubic units

# Sphere

A sphere could be defined as a solid-figure that has a perfectly round shape and every point on the surface has an equal distance from the point which is known as a centre.A fixed distance from the centre of the sphere is termed as a radius. Note that a sphere does neither has a vertex or an edge and only has a curved face.

CSA of Sphere = 2πr² sq. units

TSA of Sphere = 4πr² sq. units

Volume of Sphere = 4/3 πr² cubic units

QUIZ

By utilizing the discussion that has been made above, try to find out appropriate answers to the below-given questions. This will help in enhancing clarity of different concepts that have been discussed above.

**Q1. **Two cubes having edges of 14 cm are joined together to form a cuboid. Find out Volume and TSA of the resultant cuboid.

- 5488 cm
^{3}, 1960 cm^{2} - 1488 cm
^{3}, 1960 cm^{2} - 5488 cm
^{3}, 2060 cm^{2} - 3488 cm
^{3}, 2160 cm^{2}

**Answer – **(a)

**Solution – **We can do the question by finding TSA and Volume of two cubes and add them together or find out TSA and Volume of the resultant cuboid.

**Volume** of Cuboid = 2 (Volume of Cube) = 2 a^{3} cm ^{3} = 2 x 14^{3}cm^{3}** = 5488 cm ^{3}**

**TSA **of Cuboid = 2 (lb + bh + hl) = 2 (28.14 + 14.14 + 14.28) cm^{2}** = 1960 cm ^{2}**

**Q2. **A cuboid has dimensions of length, breadth, and height – 20 cm, 15 cm, and 10 cm respectively. Find out volume of the cuboid.

- 2000 cm
^{3} - 4000 cm
^{3} - 5000 cm
^{3} - 3000 cm
^{3}

**Answer –**(d)

**Solution –**As per the given dimensions, length, breadth, and height of cuboid are 20 cm, 15 cm, and 10 cm.

For calculating volume = (l x b x h) cm^{3} = (20 x 15 x 10) cm^{3} = 3000 cm^{3}

Hence, correct answer to the given problem is 3000 cm^{3}. ** Q3. **Faces of square pyramid primarily comprise of –

- Six Squares
- Four Rectangles and Two Squares
- Three Triangles and One Square
- Four Triangles and One Square

**Answer – **(D)

**Solution – **A square primarily comprises of four triangles and one square. The base of the pyramid is in square, and other faces are triangular.

**Q4**. As per the given diagram, the pyramid has been given that has rectangular pentagon of area 42 cm^{2} with a height of 7 cm. Find out the Volume of the pyramid.

- 198 cm
^{3} - 98 cm
^{3} - 89 cm
^{3} - 147 cm
^{3}

**Answer – **(b)

**Solution –**Volume of Pyramid = 1/3 x Height x Base Area = 1/3 x 7 cm x 42 cm^{2 }= 98 cm^{3}. Hence, the answer to the problem is (b) i.e. 98 cm^{3}.

**Q5. **Find out volume of a cylinder if the radius is 7 cm and height is 10 cm.

- 1440 cm
^{3} - 2440 cm
^{3} - 1480 cm
^{3} - 1539 cm
^{3}

**Answer – **(d)

**Solution – **We can find volume of cylinder by using rudimentary formula i.e. πr²h. We can take π = 22/7, r = 7 cm, and h = 10 cm. Thus,

V = πr²h = (22/7) x 7 x 7 x 10 = 1539 cm^{3}.

Therefore, the Volume of the cylinder is 1539 cm^{3}. ∴ option (d)

**Q6. **The Volume of a cone is 150.72 cm^{3}. If the height of the cone is 9 cm, what will be the radius?

- 9 cm
- 16 cm
- 4 cm
- 10 cm

**Solution –**It has been given that Volume (V) = 150.72 cm^{3}, h = 9 cm. We have to find out the radius (r) of the cone. We can take pi = 3.14.

By using the formula for volume of cone i.e. V = 1/3 πr^{2}h

- r
^{2}= (3V)/ πh - r
^{2}== = 16 - r = ±4 cm.

The height of the cone cannot be negative. Hence, the height of the cone is 4 cm.

**Q7. **Which of the following formula can be used to find Volume of an ice cream cone?

- V = 1/3 πr
^{2}h - V = 3 πr
^{2}h - V = 2/3 πr
^{2}h - V = 1/3 πrh

**Answer – **(a)

**Solution – **We can find out the Volume of a cone by using a simple formula i.e. V = 1/3 πr^{2}h. This primarily regards option (a) as correct and other options are incorrect. Hence, the correct answer to the problem is (a).

**Q8. **The surface area for a cylinder is 528 cm^{2, }and the diameter for the base is 12 cm. Find out the height of the cylindrical object.

- 4 cm
- 2 cm
- 8 cm
- 6 cm

**Answer – **(c)

**Solution – **Diameter of the cylinder = 12 cm.

Radius of the cylinder = (Diameter/2) = 6 cm.

Surface area = 528 cm^{2}

As surface area of the cylinder = 2πr (r + h)

- 2πr (r + h) = 528 cm
^{2} - 2π6
^{2}+ 12πh = 528 cm^{2} - 2π36 + 12πh = 528 cm
^{2} - 2 x (22/7) x 36 + 12 x(22/7) xh = 528 cm
^{2} - h = 8 cm.

By solving the above-given equation, we can find out the height of the cylinder. Hence, the height of the cylinder is 8 cm.

**Q9. **The surface area of a cube is 384 cm^{2}. Find out the length of each side of the cube.

- 4 cm
- 2 cm
- 8 cm
- 6 cm

**Answer – **(c)

**Solution –**Surface area of the cube = 6a^{2}

- 6a
^{2 }=384 - a
^{2}= 64 - a = ±8 cm

As the side of the cube cannot be negative, we can eliminate negative value. Hence, the length of each side of the cube is 8 cm.

**Q10. **Find out the height of square pyramid if volume = 9 cm^{3}and base area = 3 cm^{2}.

- 3 cm
- 6 cm
- 9 cm
- 12 cm

**Answer – **(c)

**Solution – **For finding out volume of pyramid, we can use the below-mentioned formula –

Volume of Pyramid = (Base Area) x Height x (1/3) Cubic Units

9 cm^{3}= (3 cm^{2}) x height x 1/3

- height = (9 x 3)/3 = 9 cm.

Thus, by solving the equation in a linear format, we can find out the height of the square pyramid which equates to 9 cm.