# Area, Shapes and Perimeter

Area, Shapes and Perimeter

In the Quantitative Ability section of the TCS NQT (National Qualifier Test), questions related to area, shapes, and perimeter are commonly included. Here are some key concepts and examples you may encounter:

1. Area and Perimeter of Basic Shapes:
• Rectangle: The area of a rectangle is given by length × width, and the perimeter is given by 2 × (length + width).

Example 1: Find the area and perimeter of a rectangle with a length of 8 metres and a width of 5 metres.

Solution: Area of the rectangle = length × width = 8 metres × 5 metres = 40 square metres Perimeter of the rectangle = 2 × (length + width) = 2 × (8 metres + 5 metres) = 2 × 13 metres = 26 metres

So, the area of the rectangle is 40 square metres, and the perimeter is 26 metres.

Example 2: The length of a rectangle is 12 centimetres, and its perimeter is 38 centimetres. Calculate the width and area of the rectangle.

Solution: Let the width of the rectangle be 'w' centimetres.

The perimeter of the rectangle = 2 × (length + width) = 38 centimetres 38 = 2 × (12 + w) 38 = 24 + 2w 2w = 38 - 24 2w = 14 w = 14 / 2 w = 7 centimetres

Now, we have the width of the rectangle (w = 7 centimetres), we can find the area.

Area of the rectangle = length × width = 12 centimetres × 7 centimetres = 84 square centimetres

So, the width of the rectangle is 7 centimetres, and the area is 84 square centimetres.

Example 3: The perimeter of a rectangle is 30 metres, and its width is 6 metres. Calculate the length and area of the rectangle.

Solution: Let the length of the rectangle be 'l' metres.

The perimeter of the rectangle = 2 × (length + width) = 30 metres 30 = 2 × (l + 6) 30 = 2l + 12 2l = 30 - 12 2l = 18 l = 18 / 2 l = 9 metres

Now, we have the length of the rectangle (l = 9 metres), we can find the area.

Area of the rectangle = length × width = 9 metres × 6 metres = 54 square metres

So, the length of the rectangle is 9 metres, and the area is 54 square metres.

• Square: The area of a square is given by side × side (or side^2), and the perimeter is given by 4 × side.

Example 1: Find the area and perimeter of a square with a side length of 6 centimetres.

Solution: Area of the square = side × side = 6 centimetres × 6 centimetres = 36 square centimetres Perimeter of the square = 4 × side = 4 × 6 centimetres = 24 centimetres

So, the area of the square is 36 square centimetres, and the perimeter is 24 centimetres.

Example 2: The perimeter of a square is 40 metres. Calculate the side length and area of the square.

Solution: Let the side length of the square be 's' metres.

Perimeter of the square = 4 × side = 40 metres 40 = 4 × s s = 40 / 4 s = 10 metres

Now, we have the side length of the square (s = 10 metres), we can find the area.

Area of the square = side × side = 10 metres × 10 metres = 100 square metres

So, the side length of the square is 10 metres, and the area is 100 square metres.

• Circle: The area of a circle is given by π × radius^2, and the perimeter (also known as the circumference) is given by 2 × π × radius.

Example 1: Find the area and perimeter (circumference) of a circle with a radius of 5 centimetres. (Take π = 3.14)

Solution: Area of the circle = π × radius^2 = 3.14 × 5 centimetres × 5 centimetres ≈ 78.5 square centimetres Perimeter (circumference) of the circle = 2 × π × radius = 2 × 3.14 × 5 centimetres ≈ 31.4 centimetres

So, the area of the circle is approximately 78.5 square centimetres, and the perimeter (circumference) is approximately 31.4 centimetres.

Example 2: The area of a circular garden is 314 square metres. Calculate the radius and circumference of the circle. (Take π = 3.14)

Solution: Let the radius of the circle be 'r' metres.

Area of the circle = π × radius^2 = 3.14 × r^2 = 314 square metres r^2 = 314 / 3.14 r^2 ≈ 100 r ≈ √100 r ≈ 10 metres

Now, we have the radius of the circle (r ≈ 10 metres), we can find the circumference.

Circumference of the circle = 2 × π × radius = 2 × 3.14 × 10 metres ≈ 62.8 metres

So, the radius of the circle is approximately 10 metres, and the circumference is approximately 62.8 metres.

1. Area and Perimeter of Triangles:
• Right Triangle: The area of a right triangle is given by (base × height) / 2, and the perimeter is given by the sum of the lengths of all three sides.

Example 1: Find the area and perimeter of a right triangle with a base of 6 centimetres and a height of 8 centimetres.

Solution: Area of the right triangle = (base × height) / 2 = (6 centimetres × 8 centimetres) / 2 = 48 square centimetres

To find the hypotenuse, we can use the Pythagorean theorem: c^2 = a^2 + b^2, where 'c' is the hypotenuse, and 'a' and 'b' are the other two sides (base and height).

Hypotenuse (c) = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 centimetres

Perimeter of the right triangle = base + height + hypotenuse = 6 centimetres + 8 centimetres + 10 centimetres = 24 centimetres

So, the area of the right triangle is 48 square centimetres, and the perimeter is 24 centimetres.

Example 2: The perimeter of a right triangle is 30 metres, and its base is 12 metres. Find the height and area of the right triangle.

Solution: Let the height of the right triangle be 'h' metres.

The perimeter of the right triangle = base + height + hypotenuse = 30 metres 30 = 12 metres + h + hypotenuse

Since we do not have the hypotenuse value, we cannot find the exact value of 'h' in this example without additional information or additional equations.

However, we can calculate the area of the right triangle using the given base and height: Area of the right triangle = (base × height) / 2 = (12 metres × h) / 2 = 6h square metres

So, the area of the right triangle is 6h square metres.

Example 3: The base and height of a right triangle are both 5 metres. Find the area and perimeter of the right triangle.

Solution: Area of the right triangle = (base × height) / 2 = (5 metres × 5 metres) / 2 = 12.5 square metres

To find the hypotenuse, we can use the Pythagorean theorem: Hypotenuse (c) = √(5^2 + 5^2) = √(25 + 25) = √50 ≈ 7.07 metres

Perimeter of the right triangle = base + height + hypotenuse = 5 metres + 5 metres + 7.07 metres ≈ 17.07 metres

So, the area of the right triangle is 12.5 square metres, and the perimeter is approximately 17.07 metres.

• Equilateral Triangle: The area of an equilateral triangle is given by (side^2 × √3) / 4, and the perimeter is given by 3 × side.

Example 1: Find the area and perimeter of an equilateral triangle with a side length of 6 centimetres.

Solution: Area of the equilateral triangle = (side^2 × √3) / 4 = (6 centimetres)^2 × √3 / 4 = 36√3 / 4 ≈ 9√3 square centimetres

Perimeter of the equilateral triangle = 3 × side = 3 × 6 centimetres = 18 centimetres

So, the area of the equilateral triangle is approximately 9√3 square centimetres, and the perimeter is 18 centimetres.

Example 2: The perimeter of an equilateral triangle is 45 metres. Find the side length and area of the equilateral triangle.

Solution: Let the side length of the equilateral triangle be 's' metres.

Perimeter of the equilateral triangle = 3 × side = 45 metres 45 = 3 × s s = 45 / 3 s = 15 metres

Now, we have the side length of the equilateral triangle (s = 15 metres), we can find the area.

Area of the equilateral triangle = (side^2 × √3) / 4 = (15 meters)^2 × √3 / 4 = 225√3 / 4 ≈ 56.25√3 square metres

So, the side length of the equilateral triangle is 15 metres, and the area is approximately 56.25√3 square metres.

Example 3: The area of an equilateral triangle is 36√3 square centimetres. Find the side length and perimeter of the equilateral triangle.

Solution: Let the side length of the equilateral triangle be 's' centimetres.

Area of the equilateral triangle = (side^2 × √3) / 4 = 36√3 square centimetres 36√3 = (s^2 × √3) / 4

To find the side length 's', we can simplify the equation: s^2 = 36√3 × 4 / √3 s^2 = 36 × 4 s^2 = 144 s = √144 s = 12 centimetres

Now, we have the side length of the equilateral triangle (s = 12 centimetres), we can find the perimeter.

Perimeter of the equilateral triangle = 3 × side = 3 × 12 centimetres = 36 centimetres

So, the side length of the equilateral triangle is 12 centimetres, and the perimeter is 36 centimetres.

1. Area and Perimeter of Composite Shapes:
• Composite shapes are made up of multiple basic shapes.
• The area of a composite shape can be found by finding the sum of the areas of its individual components.
• The perimeter of a composite shape can be found by finding the sum of the lengths of its sides.

Example: Find the area and perimeter of a composite shape made up of a rectangle with a length of 6 cm and a width of 4 cm, and a square with a side length of 3 cm.

These are just a few examples of the concepts related to area, shapes, and perimeter that you may encounter in the TCS NQT Quantitative Ability section. It is recommended to practise solving problems involving different shapes and composite shapes to strengthen your understanding of these concepts.