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CIE IGCSE Maths Study Notes

5.2.1 Calculations for Common Shapes

This section is dedicated to mastering the calculations involved in determining the perimeters and areas of fundamental geometric shapes: rectangles, triangles, parallelograms, and trapeziums. By engaging with the formulas and examples provided, students will develop a strong foundation in handling a variety of geometrical problems.

Perimeter and Area of Rectangles

Perimeter and Area of Rectangles

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Perimeter of a Rectangle

The perimeter PP of a rectangle is twice the sum of its length ll and width ww:

P=2(l+w)P = 2(l + w)

Example:

Calculate the perimeter of a rectangle where l=6cml = 6 \, \text{cm} and w=4cmw = 4 \, \text{cm}.

P=2(6+4)=20cmP = 2(6 + 4) = 20 \, \text{cm}

Area of a Rectangle

The area AA is the product of its length and width:

A=l×wA = l \times w

Example:

Determine the area for l=5cml = 5 \, \text{cm} and w=3cmw = 3 \, \text{cm}.

A=5×3=15cm2A = 5 \times 3 = 15 \, \text{cm}^2

Perimeter and Area of Triangles

Perimeter and Area of Triangles

Area of a Triangle

A triangle's area is half the product of its base bb and height hh:

A=12×b×hA = \frac{1}{2} \times b \times h

Example:

For a base of 10cm10 \, \text{cm} and a height of 4cm4 \, \text{cm}, the area is:

A=12×10×4=20cm2A = \frac{1}{2} \times 10 \times 4 = 20 \, \text{cm}^2

Perimeter and Area of Parallelograms

Perimeter and Area of Parallelograms

Perimeter of a Parallelogram

The perimeter PP involves adding the lengths of all sides, or twice the sum of adjacent sides aa and bb:

P=2(a+b)P = 2(a + b)

Example:

With a=5cma = 5 \, \text{cm} and b=3cmb = 3 \, \text{cm}:

P=2(5+3)=16cmP = 2(5 + 3) = 16 \, \text{cm}

Area of a Parallelogram

The area AA is found by multiplying the base bb by the height hh:

A=b×hA = b \times h

Example:

For b=8cmb = 8 \, \text{cm} and h=5cmh = 5 \, \text{cm}:

A=8×5=40cm2A = 8 \times 5 = 40 \, \text{cm}^2

Perimeter and Area of Trapeziums

Area of a Trapezium

The area AA is half the sum of the lengths of the parallel sides (a and b)(a \text{ and } b) times the height (h)(h):

A=12(a+b)×hA = \frac{1}{2} (a + b) \times hArea of a Trapezium

Example:

For parallel sides a=6cma = 6 \, \text{cm}, b=4cmb = 4 \, \text{cm}, and height h=5cmh = 5 \, \text{cm}:

A=12(6+4)×5=25cm2A = \frac{1}{2} (6 + 4) \times 5 = 25 \, \text{cm}^2

Perimeter of a Trapezium

To find the perimeter PP, add the lengths of all four sides:

Perimeter of Trapezium

Example:

Given sides of lengths 4cm4 \, \text{cm}, 3cm3 \, \text{cm}, 5cm5 \, \text{cm}, and 6cm6 \, \text{cm}:

P=4+3+5+6=18cmP = 4 + 3 + 5 + 6 = 18 \, \text{cm}

Practice Problems

Problem 1: Rectangle

Given: l=7cm,w=2cml = 7 \, \text{cm}, w = 2 \, \text{cm}

Find the area and perimeter.

Solution:

Area:

A=l×w=7×2=14cm2A = l \times w = 7 \times 2 = 14 \, \text{cm}^2

Perimeter:

P=2(l+w)=2(7+2)=18cmP = 2(l + w) = 2(7 + 2) = 18 \, \text{cm}

Problem 2: Triangle

Given: b=9cm,h=3cmb = 9 \, \text{cm}, h = 3 \, \text{cm}

Calculate the area.

Solution:

A=12×b×h=12×9×3=13.5cm2A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 9 \times 3 = 13.5 \, \text{cm}^2

Problem 3: Parallelogram

Given: a=6cm,b=4cm,h=3cma = 6 \, \text{cm}, b = 4 \, \text{cm}, h = 3 \, \text{cm}

Determine the perimeter and area.

Solution:

Perimeter:

P=2(a+b)=2(6+4)=20cmP = 2(a + b) = 2(6 + 4) = 20 \, \text{cm}

Area:

A=b×h=6×3=18cm2A = b \times h = 6 \times 3 = 18 \, \text{cm}^2

Problem 4: Trapezium

Given: a=8cm,b=5cm,h=4cma = 8 \, \text{cm}, b = 5 \, \text{cm}, h = 4 \, \text{cm}

Find the area.

Solution:

A=12(a+b)×h=12(8+5)×4=26cm2A = \frac{1}{2} (a + b) \times h = \frac{1}{2} (8 + 5) \times 4 = 26 \, \text{cm}^2


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