TNTET Paper 2 - Area circumference of circles and surface area of solids. E-Book Study Material

Area, Circumference of Circles and Surface Area of Solids

Welcome to this in-depth session on the geometry of circles and solids. Understanding the area, circumference of circles, and the surface area of solids is fundamental in mathematics. These concepts have wide applications in real-world scenarios, from calculating the amount of material needed for a circular or spherical object to determining the space it occupies. We will break down each concept, providing clear explanations, formulas, and examples to ensure you master this topic for your exams.

1. Circles: Circumference and Area

A circle is a fundamental geometric shape defined as the set of all points in a plane that are at a fixed distance from a central point. The fixed distance is called the radius, and the central point is called the center of the circle.

1.1. Circumference of a Circle

The circumference of a circle is the distance around the circle, essentially its perimeter. It's the total length of the boundary of the circle.

The formula for the circumference (C) of a circle depends on its radius (r) or its diameter (d). The diameter is twice the radius (d = 2r). The constant pi (π) is a mathematical constant approximately equal to 3.14159 or 22/7. It represents the ratio of a circle's circumference to its diameter.

Formulas for Circumference:

Using Radius: C = 2πr
Using Diameter: C = πd

Example 1: Calculate the circumference of a circle with a radius of 7 cm.

Given: Radius (r) = 7 cm. Using the formula C = 2πr, and taking π ≈ 22/7: C = 2 * (22/7) * 7 C = 2 * 22 C = 44 cm. So, the circumference of the circle is 44 cm.

Example 2: A circular park has a diameter of 20 meters. Find its circumference.

Given: Diameter (d) = 20 meters. Using the formula C = πd, and taking π ≈ 3.14: C = 3.14 * 20 C = 62.8 meters. The circumference of the park is 62.8 meters.

Circumference Shortcut: Remember that 'C' for circumference is like 'C' for 'Complete circle' or 'around the circle'. The formula involves '2' (for diameter) and 'πr' (radius). If you know the radius, use 2πr. If you know the diameter, use πd.

1.2. Area of a Circle

The area of a circle is the amount of space enclosed within the boundary of the circle. It's the measure of the two-dimensional surface that the circle covers.

The formula for the area (A) of a circle is given by:

Formula for Area:

A = πr²

Here, 'r' is the radius of the circle. Notice that the area formula uses the radius squared, which means the area grows much faster than the circumference as the radius increases.

Example 3: Find the area of a circle with a radius of 10 cm.

Given: Radius (r) = 10 cm. Using the formula A = πr², and taking π ≈ 3.14: A = 3.14 * (10 cm)² A = 3.14 * 100 cm² A = 314 cm². The area of the circle is 314 square centimeters.

Example 4: A circular pizza has a diameter of 14 inches. What is its area?

Given: Diameter (d) = 14 inches. First, find the radius: r = d/2 = 14 inches / 2 = 7 inches. Using the formula A = πr², and taking π ≈ 22/7: A = (22/7) * (7 inches)² A = (22/7) * 49 inches² A = 22 * 7 inches² A = 154 inches². The area of the pizza is 154 square inches.

Area Shortcut: Think of 'A' for Area as 'amount of space'. The formula A = πr² involves squaring the radius. This means the area is proportional to the square of the radius. Always remember to square the radius before multiplying by π.

1.3. Area of a Semicircle and Quarter Circle

A semicircle is half of a circle, and a quarter circle is one-fourth of a circle. Their areas are simply fractions of the full circle's area.

Area of Semicircle = (1/2) * πr²
Area of Quarter Circle = (1/4) * πr²

Example 5: Calculate the area of a semicircle with a radius of 14 meters.

Given: Radius (r) = 14 meters, π ≈ 22/7. Area of Semicircle = (1/2) * (22/7) * (14 m)² Area = (1/2) * (22/7) * 196 m² Area = 11 * 28 m² Area = 308 m².

2. Surface Area of Solids

Surface area is the total area of the outer surfaces of a three-dimensional object. It's like finding the total area of all the faces or curved surfaces of a solid. We will focus on some common solids.

2.1. Cuboid

A cuboid is a solid figure bounded by six rectangular faces. It has length (l), breadth (b), and height (h).

The surface area of a cuboid is the sum of the areas of its six rectangular faces.

Formula for Surface Area of Cuboid:

Total Surface Area (TSA) = 2(lb + bh + hl)

Example 6: Find the total surface area of a cuboid with length 10 cm, breadth 5 cm, and height 3 cm.

Given: l = 10 cm, b = 5 cm, h = 3 cm. TSA = 2 * ( (10*5) + (5*3) + (3*10) ) cm² TSA = 2 * ( 50 + 15 + 30 ) cm² TSA = 2 * ( 95 ) cm² TSA = 190 cm².

2.2. Cube

A cube is a special type of cuboid where all sides are equal in length (l = b = h = a). It has six square faces.

Formula for Surface Area of Cube:

Total Surface Area (TSA) = 6a² (where 'a' is the length of one side)

This formula comes from the cuboid formula: 2(a*a + a*a + a*a) = 2(3a²) = 6a².

Example 7: Calculate the surface area of a cube with a side length of 6 cm.

Given: Side (a) = 6 cm. TSA = 6 * (6 cm)² TSA = 6 * 36 cm² TSA = 216 cm².

Cube vs. Cuboid Surface Area: A cube is a perfect box with equal sides. Its surface area is 6 times the area of one face (a²). A cuboid is a stretched box, so you need to calculate the area of each pair of different-sized faces (lb, bh, hl) and sum them up, then multiply by two.

2.3. Cylinder

A cylinder is a solid with two circular bases joined by a curved surface. It has a radius (r) and a height (h).

A cylinder has three surfaces:

The top circular base.
The bottom circular base.
The curved lateral surface.

Formulas for Surface Area of Cylinder:

Area of each circular base = πr²
Curved Surface Area (CSA) or Lateral Surface Area (LSA) = 2πrh
Total Surface Area (TSA) = Area of two bases + CSA = 2πr² + 2πrh = 2πr(r + h)

Example 8: Find the curved surface area and total surface area of a cylinder with a radius of 7 meters and a height of 10 meters. (Take π = 22/7)

Given: r = 7 m, h = 10 m. CSA = 2πrh CSA = 2 * (22/7) * 7 m * 10 m CSA = 2 * 22 * 10 m² CSA = 440 m². TSA = 2πr(r + h) TSA = 2 * (22/7) * 7 m * (7 m + 10 m) TSA = 44 m * (17 m) TSA = 748 m².

Cylinder Surface Area Logic: Imagine unrolling the curved surface of a cylinder. It forms a rectangle. The height of the rectangle is the height of the cylinder (h). The length of the rectangle is the circumference of the base circle (2πr). So, the curved surface area is length × height = 2πrh. The total surface area is this plus the areas of the two circular bases (πr² each).

2.4. Cone

A cone is a solid figure with a circular base and a single vertex (apex) connected to the circumference of the base. It has a radius (r) and a slant height (l). The slant height is the distance from the apex to any point on the circumference of the base. The perpendicular height (h) is the distance from the apex to the center of the base. They are related by the Pythagorean theorem: l² = r² + h².

Formulas for Surface Area of Cone:

Curved Surface Area (CSA) = πrl
Total Surface Area (TSA) = Area of base + CSA = πr² + πrl = πr(r + l)

Example 9: A cone has a base radius of 3 cm and a slant height of 5 cm. Find its total surface area.

Given: r = 3 cm, l = 5 cm. TSA = πr(r + l) TSA = π * 3 cm * (3 cm + 5 cm) TSA = 3π * 8 cm² TSA = 24π cm². If we use π ≈ 3.14: TSA = 24 * 3.14 cm² TSA = 75.36 cm².

Example 10: Find the curved surface area of a cone with a radius of 6 meters and a perpendicular height of 8 meters.

Given: r = 6 m, h = 8 m. First, calculate the slant height (l) using l² = r² + h²: l² = (6 m)² + (8 m)² l² = 36 m² + 64 m² l² = 100 m² l = √100 m² = 10 m. Now, calculate CSA = πrl: CSA = π * 6 m * 10 m CSA = 60π m². If we use π ≈ 3.14: CSA = 60 * 3.14 m² CSA = 188.4 m².

Cone Formula Tip: The slant height 'l' is crucial for cone surface area. Always ensure you have 'l' or can calculate it using Pythagoras theorem (l² = r² + h²). The 'l' in πrl is like a 'lean' or 'slope' measurement.

2.5. Sphere

A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball. It is defined by its radius (r).

Formulas for Surface Area of Sphere:

Surface Area (SA) = 4πr²

Interestingly, the surface area of a sphere is exactly four times the area of a great circle of the sphere (a circle with the same radius).

Example 11: Calculate the surface area of a sphere with a radius of 14 cm. (Take π = 22/7)

Given: r = 14 cm. SA = 4πr² SA = 4 * (22/7) * (14 cm)² SA = 4 * (22/7) * 196 cm² SA = 4 * 22 * 28 cm² SA = 88 * 28 cm² SA = 2464 cm².

Sphere Surface Area Insight: The formula SA = 4πr² is quite elegant. It means a sphere's surface area is equivalent to the area of four circles with the same radius. This is a key fact to remember for quick recall.

2.6. Hemisphere

A hemisphere is half of a sphere. It has a curved surface and a flat circular base.

Formulas for Surface Area of Hemisphere:

Curved Surface Area (CSA) = (1/2) * Surface Area of Sphere = (1/2) * 4πr² = 2πr²
Total Surface Area (TSA) = CSA + Area of the base = 2πr² + πr² = 3πr²

Example 12: Find the total surface area of a hemispherical bowl with a radius of 7 cm. (Take π = 22/7)

Given: r = 7 cm. TSA = 3πr² TSA = 3 * (22/7) * (7 cm)² TSA = 3 * (22/7) * 49 cm² TSA = 3 * 22 * 7 cm² TSA = 66 * 7 cm² TSA = 462 cm².

3. Applications and Key Takeaways

Understanding these formulas allows us to solve practical problems. For instance, calculating the amount of paint needed for a cylindrical tank involves its curved surface area, while determining the fabric required for a conical hat uses its CSA. The area of a circular garden helps in landscaping, and the surface area of a spherical ball is relevant in physics and engineering.

Key Formulas Summary:

Shape	Circumference/Perimeter	Area	Curved Surface Area (CSA)	Total Surface Area (TSA)
Circle	2πr or πd	πr²	N/A	N/A
Semicircle	πr + 2r	(1/2)πr²	N/A	N/A
Cuboid	N/A	N/A	N/A	2(lb + bh + hl)
Cube	N/A	N/A	N/A	6a²
Cylinder	N/A	N/A	2πrh	2πr(r + h)
Cone	N/A	N/A	πrl	πr(r + l)
Sphere	N/A	N/A	N/A	4πr²
Hemisphere	N/A	N/A	2πr²	3πr²

Mastering these formulas and understanding their derivation will equip you to tackle any problem related to circles and solids in your examination. Practice consistently with various examples, and don't hesitate to draw diagrams to visualize the shapes and dimensions.