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Geometry and Measurement - Perimeter and Area of 2D Shapes including Circles

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Perimeter is the total linear distance around the outside boundary of a 2D shape. For any polygon, it is the sum of all its side lengths. Visually, if you were to place a piece of string along the edge of a shape and then straighten it out, the length of that string is the perimeter.

Area is the measure of the total surface space contained within the boundaries of a 2D shape, expressed in square units like cm2cm^2 or m2m^2. You can visualize area as the number of 1×11 \times 1 unit squares required to perfectly tile the inside of the shape.

A parallelogram is a quadrilateral with two pairs of parallel sides. Its area is calculated by multiplying the base by the perpendicular height. Visually, if you cut a right-angled triangle from one side of a parallelogram and shift it to the other side, it forms a rectangle, which explains why the area formula A=b×hA = b \times h is used.

A triangle's area is always exactly half the area of a parallelogram with the same base and height. When calculating the area, the height (hh) must be the perpendicular distance from the base to the opposite vertex. Visually, any triangle can be seen as half of a rectangle or parallelogram that encloses it.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides (labeled aa and bb). The area is the product of the average of the parallel sides and the perpendicular height. Imagine the trapezium being transformed into a rectangle by 'leveling out' the slanted sides to the average width.

The circle is defined by its radius (rr), the distance from the center to the edge, and its diameter (dd), which is the distance across the circle through the center (d=2rd = 2r). The perimeter of a circle is specifically called the circumference. The constant π\pi (approximately 3.141593.14159 or 227\frac{22}{7}) represents the fixed ratio between any circle's circumference and its diameter.

The area of a circle represents the space inside the curved boundary. Visually, if a circle is sliced into many thin sectors (like pizza slices) and rearranged, they form a shape resembling a rectangle with a width of πr\pi r and a height of rr, leading to the formula A=πr2A = \pi r^2.

Composite shapes are complex figures made by combining two or more simple 2D shapes (such as a rectangle joined to a semi-circle). To find the total area, calculate the areas of the individual parts and add them. To find the perimeter, sum only the exterior lengths that form the outer boundary.

📐Formulae

Perimeter of a Square: P=4sP = 4s

Area of a Square: A=s2A = s^2

Perimeter of a Rectangle: P=2(l+w)P = 2(l + w)

Area of a Rectangle: A=l×wA = l \times w

Area of a Triangle: A=12×b×hA = \frac{1}{2} \times b \times h

Area of a Parallelogram: A=b×hA = b \times h

Area of a Trapezium: A=12(a+b)hA = \frac{1}{2}(a + b)h

Circumference of a Circle: C=2πrC = 2\pi r or C=πdC = \pi d

Area of a Circle: A=πr2A = \pi r^2

💡Examples

Problem 1:

Calculate the area and circumference of a circular garden with a radius of 1414 m. (Take π227\pi \approx \frac{22}{7})

Solution:

  1. Identify the given radius: r=14r = 14 m.
  2. Calculate circumference using C=2πrC = 2\pi r: C=2×227×14C = 2 \times \frac{22}{7} \times 14 C=2×22×2=88C = 2 \times 22 \times 2 = 88 m.
  3. Calculate area using A=πr2A = \pi r^2: A=227×142A = \frac{22}{7} \times 14^2 A=227×196A = \frac{22}{7} \times 196 A=22×28=616A = 22 \times 28 = 616 m2m^2.

Explanation:

We substitute the radius into the standard circle formulas. Using the fraction 227\frac{22}{7} for π\pi is helpful here because 1414 is a multiple of 77, allowing for easy simplification.

Problem 2:

A trapezium has parallel sides of length 1010 cm and 1616 cm. If the perpendicular height between them is 55 cm, find its area.

Solution:

  1. Identify the parallel sides: a=10a = 10 cm, b=16b = 16 cm.
  2. Identify the height: h=5h = 5 cm.
  3. Apply the trapezium area formula: A=12(a+b)hA = \frac{1}{2}(a + b)h
  4. Substitute the values: A=12(10+16)×5A = \frac{1}{2}(10 + 16) \times 5
  5. Simplify the sum in parentheses: A=12(26)×5A = \frac{1}{2}(26) \times 5
  6. Calculate the final result: A=13×5=65A = 13 \times 5 = 65 cm2cm^2.

Explanation:

The area is found by taking the sum of the parallel bases, dividing by 22 to find the average length, and then multiplying by the vertical height.