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Mensuration - Area of Trapezium and General Quadrilateral

Grade 8ICSE

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

🔑Concepts

A Trapezium is a quadrilateral with at least one pair of parallel sides, known as the bases. Visually, it appears as a four-sided figure where two sides run in the same direction but have different lengths, like the cross-section of a gold bar.

The height (h) or altitude of a trapezium is the perpendicular distance between its two parallel sides. In a diagram, this is often represented as a dashed line forming a 9090^{\circ} angle with both parallel bases.

An Isosceles Trapezium is a special type where the non-parallel sides are equal in length. Visually, this shape is perfectly symmetrical along a vertical axis passing through the midpoints of the parallel sides.

A general quadrilateral can be divided into two triangles by drawing one of its diagonals. Visually, if you take any four-sided polygon and draw a line from one corner to the opposite corner (the diagonal dd), you create two triangles sharing that diagonal as a common base.

Offsets in a general quadrilateral are the perpendiculars dropped from the remaining two vertices to the chosen diagonal. If the diagonal is dd, the lengths of these two perpendicular lines are denoted as h1h_{1} and h2h_{2}, effectively showing the 'height' of each constituent triangle.

The area of a general quadrilateral is the sum of the areas of the two triangles it is composed of. Visually, the total space inside the quadrilateral is filled by 1\triangle 1 with height h1h_{1} and 2\triangle 2 with height h2h_{2}, both sharing base dd.

A Rhombus is a quadrilateral where all sides are equal and diagonals bisect each other at right angles (9090^{\circ}). Visually, it looks like a diamond shape where the two diagonals divide it into four identical right-angled triangles.

Area calculation often involves unit consistency. Ensure all measurements (cm, m, mm) are converted to the same unit before applying formulas to avoid calculation errors.

📐Formulae

Area of a Trapezium=12×(sum of parallel sides)×heightArea\ of\ a\ Trapezium = \frac{1}{2} \times (sum\ of\ parallel\ sides) \times height

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

Area of a General Quadrilateral=12×diagonal×(sum of perpendicular offsets)Area\ of\ a\ General\ Quadrilateral = \frac{1}{2} \times diagonal \times (sum\ of\ perpendicular\ offsets)

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

Area of a Rhombus=12×(product of diagonals)Area\ of\ a\ Rhombus = \frac{1}{2} \times (product\ of\ diagonals)

Area of a Rhombus=12×d1×d2Area\ of\ a\ Rhombus = \frac{1}{2} \times d_{1} \times d_{2}

💡Examples

Problem 1:

Find the area of a trapezium whose parallel sides are 14 cm14\text{ cm} and 10 cm10\text{ cm} and the distance between them is 6 cm6\text{ cm}.

Solution:

  1. Identify the given values: Parallel sides a=14 cma = 14\text{ cm}, b=10 cmb = 10\text{ cm}, and height h=6 cmh = 6\text{ cm}.\2. Use the formula: Area=12×(a+b)×hArea = \frac{1}{2} \times (a + b) \times h\3. Substitute the values: Area=12×(14+10)×6Area = \frac{1}{2} \times (14 + 10) \times 6\4. Calculate the sum: Area=12×24×6Area = \frac{1}{2} \times 24 \times 6\5. Simplify: Area=12×6=72 cm2Area = 12 \times 6 = 72\text{ cm}^{2}.

Explanation:

We use the standard trapezium formula by taking the average of the two parallel bases and multiplying it by the vertical height.

Problem 2:

Calculate the area of a quadrilateral ABCDABCD where the diagonal AC=12 cmAC = 12\text{ cm} and the perpendiculars (offsets) from vertices BB and DD on diagonal ACAC are 5 cm5\text{ cm} and 3 cm3\text{ cm} respectively.

Solution:

  1. Identify the given values: Diagonal d=12 cmd = 12\text{ cm}, offset h1=5 cmh_{1} = 5\text{ cm}, and offset h2=3 cmh_{2} = 3\text{ cm}.\2. Use the formula for a general quadrilateral: Area=12×d×(h1+h2)Area = \frac{1}{2} \times d \times (h_{1} + h_{2})\3. Substitute the values: Area=12×12×(5+3)Area = \frac{1}{2} \times 12 \times (5 + 3)\4. Calculate the sum: Area=12×12×8Area = \frac{1}{2} \times 12 \times 8\5. Simplify: Area=6×8=48 cm2Area = 6 \times 8 = 48\text{ cm}^{2}.

Explanation:

The quadrilateral is treated as two triangles with a common base (the diagonal). We sum the heights of these triangles and multiply by half the base length.