krit.club logo

Mensuration - Area of Polygons

Grade 8ICSE

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

🔑Concepts

Polygon Definition and Classification: A polygon is a closed plane figure bounded by three or more line segments. A 'Regular Polygon' has all sides and all interior angles equal (like an equilateral triangle or a square), while an 'Irregular Polygon' has sides or angles of different measures. Visually, a regular polygon looks perfectly symmetrical around its center.

Area of a Trapezium: A trapezium is a quadrilateral with exactly one pair of parallel sides. To visualize its area, imagine a rectangle formed by the average length of the two parallel sides (aa and bb) and the perpendicular distance (hh) between them. The formula calculates the area as half the product of the sum of parallel sides and the height.

Area of a General Quadrilateral: Any general quadrilateral can be viewed as two triangles sharing a common diagonal. Visually, if you draw a diagonal dd and then draw perpendiculars (called offsets) h1h_1 and h2h_2 from the remaining two vertices to this diagonal, the area of the quadrilateral is the sum of the areas of these two triangles.

Area of a Rhombus: A rhombus is a special parallelogram where all sides are equal. Visually, its diagonals (d1d_1 and d2d_2) intersect at right angles (9090^{\circ}) and bisect each other. The area can be found using the lengths of these diagonals or by treating it as a parallelogram with a base and a height.

Area of a Parallelogram: A parallelogram has opposite sides that are parallel and equal. Visually, if you cut a right-angled triangle from one side and move it to the other, it forms a rectangle. Thus, the area is simply the product of its base length and its corresponding vertical height.

Area of Irregular Polygons (Field Book Method): To find the area of an irregular polygon or a field, we divide the entire shape into several smaller, non-overlapping known figures such as triangles and trapezia. Visually, this involves drawing a central baseline and measuring perpendicular offsets to various vertices, then summing the individual areas of the resulting sub-shapes.

Area of a Regular Hexagon: A regular hexagon consists of six congruent equilateral triangles meeting at a central point. Visually, if you draw lines from the center to each vertex, you see six identical triangles. The total area is 66 times the area of one equilateral triangle with side ss.

📐Formulae

Area of a Trapezium = 12×(a+b)×h\frac{1}{2} \times (a + b) \times h, where a,ba, b are parallel sides and hh is the height.

Area of a General Quadrilateral = 12×d×(h1+h2)\frac{1}{2} \times d \times (h_1 + h_2), where dd is the diagonal and h1,h2h_1, h_2 are the perpendicular offsets.

Area of a Rhombus = 12×d1×d2\frac{1}{2} \times d_1 \times d_2, where d1d_1 and d2d_2 are the lengths of the diagonals.

Area of a Rhombus = base×heightbase \times height

Area of a Parallelogram = base×heightbase \times height

Area of an Equilateral Triangle = 34s2\frac{\sqrt{3}}{4}s^2, where ss is the side length.

Area of a Regular Hexagon = 332s2\frac{3\sqrt{3}}{2}s^2, where ss is the side length.

💡Examples

Problem 1:

Find the area of a trapezium whose parallel sides are 12 cm12\text{ cm} and 20 cm20\text{ cm} long, and the distance between them is 8 cm8\text{ cm}.

Solution:

  1. Identify the given values: Parallel sides a=12 cma = 12\text{ cm}, b=20 cmb = 20\text{ cm}, and height h=8 cmh = 8\text{ cm}.
  2. Apply the formula for the area of a trapezium: Area=12×(a+b)×hArea = \frac{1}{2} \times (a + b) \times h.
  3. Substitute the values: Area=12×(12+20)×8Area = \frac{1}{2} \times (12 + 20) \times 8.
  4. Calculate the sum: Area=12×32×8Area = \frac{1}{2} \times 32 \times 8.
  5. Solve: Area=16×8=128 cm2Area = 16 \times 8 = 128\text{ cm}^2.

Explanation:

The area is calculated by taking the average of the two parallel sides and multiplying it by the perpendicular height.

Problem 2:

The area of a rhombus is 240 cm2240\text{ cm}^2 and one of the diagonals is 16 cm16\text{ cm}. Find the length of the other diagonal.

Solution:

  1. Given: Area=240 cm2Area = 240\text{ cm}^2 and d1=16 cmd_1 = 16\text{ cm}.
  2. Use the formula: Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2.
  3. Substitute the known values: 240=12×16×d2240 = \frac{1}{2} \times 16 \times d_2.
  4. Simplify: 240=8×d2240 = 8 \times d_2.
  5. Solve for d2d_2: d2=2408=30 cmd_2 = \frac{240}{8} = 30\text{ cm}.

Explanation:

Since the area and one diagonal of the rhombus are known, we use the diagonal-based area formula to isolate and solve for the unknown diagonal.