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Mensuration - Area of Polygons

Grade 8ICSE

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

🔑Concepts

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The area of a trapezium is calculated as half the sum of the parallel sides multiplied by the perpendicular distance (height) between them: Area=12(a+b)hArea = \frac{1}{2}(a+b)h.

Trapezium with parallel sides a and b and height h.
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A general quadrilateral can be split into two triangles by a diagonal. Its area is the sum of the areas of these triangles: Area=12×d×(h1+h2)Area = \frac{1}{2} \times d \times (h_1 + h_2), where dd is the diagonal and h1,h2h_1, h_2 are perpendiculars from the opposite vertices.

Quadrilateral split by diagonal d into two triangles with heights h1 and h2.
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The area of a rhombus is half the product of its diagonals: Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2. The diagonals of a rhombus bisect each other at right angles (90∘90^\circ).

Rhombus showing diagonals d1 and d2.
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Regular polygons like hexagons can be divided into congruent triangles. A regular hexagon consists of 6 equilateral triangles. Total Area = 6×34s2=332s26 \times \frac{\sqrt{3}}{4}s^2 = \frac{3\sqrt{3}}{2}s^2.

📐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.

Problem 3:

Find the area of a quadrilateral ABCDABCD where the diagonal AC=15 cmAC = 15\text{ cm} and the lengths of the perpendiculars from BB and DD to ACAC are 5 cm5\text{ cm} and 7 cm7\text{ cm} respectively.

Quadrilateral ABCD with diagonal AC=15cm and offsets 5cm and 7cm.

Solution:

Given: Diagonal d=15 cmd = 15\text{ cm} Height h1=5 cmh_1 = 5\text{ cm} Height h2=7 cmh_2 = 7\text{ cm}

Area of quadrilateral ABCD=12×d×(h1+h2)ABCD = \frac{1}{2} \times d \times (h_1 + h_2) Area=12×15×(5+7)Area = \frac{1}{2} \times 15 \times (5 + 7) Area=12×15×12Area = \frac{1}{2} \times 15 \times 12 Area=15×6Area = 15 \times 6 Area=90 cm2Area = 90\text{ cm}^2

Explanation:

We use the formula for a general quadrilateral by treating it as two triangles sharing a common base (the diagonal).

Problem 4:

Calculate the area of a regular hexagon with each side measuring 4 cm4\text{ cm}. (Take 3≈1.732\sqrt{3} \approx 1.732)

Regular hexagon divided into equilateral triangles with side 4cm.

Solution:

Side s=4 cms = 4\text{ cm} Area of a regular hexagon = 332s2\frac{3\sqrt{3}}{2}s^2 Area=332×(4)2Area = \frac{3\sqrt{3}}{2} \times (4)^2 Area=332×16Area = \frac{3\sqrt{3}}{2} \times 16 Area=33×8Area = 3\sqrt{3} \times 8 Area=243Area = 24\sqrt{3} Substituting 3=1.732\sqrt{3} = 1.732: Area=24×1.732Area = 24 \times 1.732 Area=41.568 cm2Area = 41.568\text{ cm}^2

Explanation:

A regular hexagon is composed of 6 equilateral triangles. We find the area of one triangle and multiply by 6.