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Mensuration - Area of Polygons

Grade 8CBSE

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 lengths of parallel sides multiplied by the perpendicular distance (height) between them. Formula: Area = 12×(a+b)×h\frac{1}{2} \times (a + b) \times h.

Diagram of a trapezium with parallel sides a and b, and height h.
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A general quadrilateral can be split into two triangles by drawing a diagonal. Its area is the sum of the areas of these two triangles: Area = 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 perpendiculars from opposite vertices.

General quadrilateral divided by a diagonal into two triangles with heights h1 and h2.
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The area of a rhombus is half the product of its diagonals. Since diagonals of a rhombus bisect each other at right angles, Area = 12×d1×d2\frac{1}{2} \times d_1 \times d_2.

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Area of any polygon can be found by dividing it into known shapes like triangles, rectangles, or trapeziums and summing their individual areas.

📐Formulae

Area of a Trapezium = 12×(a+b)×h\frac{1}{2} \times (a + b) \times h

Area of a General Quadrilateral = 12×d×(h1+h2)\frac{1}{2} \times d \times (h_1 + h_2)

Area of a Rhombus = 12×d1×d2\frac{1}{2} \times d_1 \times d_2

Area of a Parallelogram = base×heightbase \times height

Area of a Triangle = 12×base×height\frac{1}{2} \times base \times height

Area of a Square = side×side=s2side \times side = s^{2}

Area of a Rectangle = length×breadthlength \times breadth

💡Examples

Problem 1:

Find the area of a trapezium whose parallel sides are 1515 cmcm and 2525 cmcm, and the perpendicular distance between them is 1010 cmcm.

Solution:

Given: Parallel side a=15a = 15 cmcm Parallel side b=25b = 25 cmcm Height h=10h = 10 cmcm

Using the formula: Area = 12×(a+b)×h\frac{1}{2} \times (a + b) \times h Area = 12×(15+25)×10\frac{1}{2} \times (15 + 25) \times 10 Area = 12×40×10\frac{1}{2} \times 40 \times 10 Area = 20×1020 \times 10 Area = 200200 cm2cm^{2}

Explanation:

First, identify the two parallel sides (bases) and the height. Substitute these values into the trapezium area formula and perform the arithmetic operations.

Problem 2:

The area of a rhombus is 240240 cm2cm^{2}. If one of its diagonals is 1616 cmcm, find the length of the other diagonal.

Solution:

Given: Area of rhombus = 240240 cm2cm^{2} Diagonal d1=16d_1 = 16 cmcm

Using the formula: Area = 12×d1×d2\frac{1}{2} \times d_1 \times d_2 240=12×16×d2240 = \frac{1}{2} \times 16 \times d_2 240=8×d2240 = 8 \times d_2 d2=2408d_2 = \frac{240}{8} d2=30d_2 = 30 cmcm

Explanation:

Apply the area formula for a rhombus. Since the area and one diagonal are known, rearrange the equation to solve for the unknown diagonal d2d_2 by dividing the area by half of the known diagonal.

Problem 3:

Calculate the area of a quadrilateral where the length of one diagonal is 2020 cmcm and the perpendiculars dropped on it from the remaining vertices are 8.58.5 cmcm and 11.511.5 cmcm.

A quadrilateral with diagonal 20 cm and heights 8.5 cm and 11.5 cm.

Solution:

Area=12×d×(h1+h2)Area = \frac{1}{2} \times d \times (h_1 + h_2) Area=12×20×(8.5+11.5)Area = \frac{1}{2} \times 20 \times (8.5 + 11.5) Area=10×20Area = 10 \times 20 Area=200 cm2Area = 200 \text{ cm}^2

Explanation:

To find the area of a general quadrilateral, we identify the diagonal (d=20d = 20 cmcm) and the two perpendicular heights (h1=8.5h_1 = 8.5 cmcm, h2=11.5h_2 = 11.5 cmcm) and apply the formula.

Problem 4:

The area of a trapezium is 3434 cm2cm^2 and the length of one of the parallel sides is 1010 cmcm and its height is 44 cmcm. Find the length of the other parallel side.

Trapezium with height 4 cm, top side 10 cm, and unknown bottom side b.

Solution:

Area=12×(a+b)×hArea = \frac{1}{2} \times (a + b) \times h 34=12×(10+b)×434 = \frac{1}{2} \times (10 + b) \times 4 34=(10+b)×234 = (10 + b) \times 2 342=10+b\frac{34}{2} = 10 + b 17=10+b17 = 10 + b b=17−10=7 cmb = 17 - 10 = 7 \text{ cm}

Explanation:

Given the Area (3434), height (44), and one parallel side (1010), we substitute these into the trapezium area formula and solve for the unknown side 'bb'.