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Geometry - Theorems on Area

Grade 9ICSE

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

🔑Concepts

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Parallelograms on the same base and between the same parallels are equal in area. Visualise two parallelograms ABCDABCD and ABEFABEF sharing the common base ABAB, with their opposite sides CDCD and EFEF lying on the same horizontal line parallel to ABAB. Because they share the same vertical height hh and base bb, their areas, calculated as Area=b×hArea = b \times h, are identical.

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Triangles on the same base (or equal bases) and between the same parallels are equal in area. For example, consider triangles ABCABC and ABDABD where the base ABAB is fixed on one parallel line and the vertices CC and DD lie anywhere on a second line parallel to ABAB. Regardless of where the vertices CC and DD are moved along that parallel line, the area of the triangles remains constant.

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If a triangle and a parallelogram stand on the same base and lie between the same parallels, the area of the triangle is equal to half the area of the parallelogram. Imagine a parallelogram ABCDABCD on base ABAB and a triangle PABPAB where point PP is any point on the opposite side CDCD. The triangle occupies exactly half the area of the parallelogram: Area(△PAB)=12Area(ABCD)Area(\triangle PAB) = \frac{1}{2} Area(ABCD).

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A median of a triangle divides it into two triangles of equal area. In △ABC\triangle ABC, if you draw a line from vertex AA to the midpoint MM of the opposite side BCBC, the median AMAM creates triangles △ABM\triangle ABM and △ACM\triangle ACM. Both have equal base lengths (BM=MCBM = MC) and share the same altitude from AA, hence Area(△ABM)=Area(△ACM)Area(\triangle ABM) = Area(\triangle ACM).

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The converse theorem states that triangles having equal areas and standing on the same base (or equal bases) must lie between the same parallels. If two triangles ABCABC and ABDABD share the base ABAB and have the same area, their altitudes relative to ABAB must be equal. This means vertices CC and DD are at the same distance from the base, making CD∥ABCD \parallel AB.

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A diagonal of a parallelogram divides it into two triangles of equal area. If you draw a diagonal ACAC in parallelogram ABCDABCD, it forms two triangles △ABC\triangle ABC and △ADC\triangle ADC. These triangles are congruent by the SSS criterion and therefore their areas are equal, each being exactly half of the parallelogram's area.

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Triangles with the same vertex and their bases on the same straight line have areas proportional to the lengths of their bases. If △ABC\triangle ABC and △ACD\triangle ACD share vertex AA and bases BCBC and CDCD lie on line BDBD, then the ratio of their areas is Area(△ABC)Area(△ACD)=BCCD\frac{Area(\triangle ABC)}{Area(\triangle ACD)} = \frac{BC}{CD}. This is because they share a common altitude from point AA to the line BDBD.

📐Formulae

Area of Parallelogram=base×heightArea \text{ of Parallelogram} = \text{base} \times \text{height}

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

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

Area of Rhombus=12×d1×d2Area \text{ of Rhombus} = \frac{1}{2} \times d_1 \times d_2

Area of Equilateral Triangle=34×side2Area \text{ of Equilateral Triangle} = \frac{\sqrt{3}}{4} \times \text{side}^2

💡Examples

Problem 1:

In △ABC\triangle ABC, ADAD is the median. If EE is any point on the median ADAD, prove that Area(△ABE)=Area(△ACE)Area(\triangle ABE) = Area(\triangle ACE).

Solution:

  1. In △ABC\triangle ABC, ADAD is the median. Since a median divides a triangle into two triangles of equal area, Area(△ABD)=Area(△ACD)Area(\triangle ABD) = Area(\triangle ACD).
  2. Now, consider △EBC\triangle EBC. EDED is the median of this triangle because DD is the midpoint of BCBC. Therefore, Area(△EBD)=Area(△ECD)Area(\triangle EBD) = Area(\triangle ECD).
  3. Subtracting the area of the smaller triangles from the larger ones: Area(△ABD)−Area(△EBD)=Area(△ACD)−Area(△ECD)Area(\triangle ABD) - Area(\triangle EBD) = Area(\triangle ACD) - Area(\triangle ECD).
  4. This leaves us with Area(△ABE)=Area(△ACE)Area(\triangle ABE) = Area(\triangle ACE).

Explanation:

This solution applies the median property twice—first for the large triangle and then for the smaller triangle formed within it—and uses the subtraction method to isolate the desired areas.

Problem 2:

A triangle PABPAB and a parallelogram ABCDABCD are on the same base ABAB and between the same parallels ABAB and CDCD. If the area of the triangle is 24.5cm224.5 cm^2, find the area of the parallelogram.

Solution:

  1. Let the area of the parallelogram ABCDABCD be XX.
  2. According to the theorem, if a triangle and a parallelogram are on the same base and between the same parallels, Area(△)=12×Area(Parallelogram)Area(\triangle) = \frac{1}{2} \times Area(\text{Parallelogram}).
  3. Given Area(△PAB)=24.5cm2Area(\triangle PAB) = 24.5 cm^2.
  4. Substituting into the formula: 24.5=12×X24.5 = \frac{1}{2} \times X.
  5. X=24.5×2=49cm2X = 24.5 \times 2 = 49 cm^2.
  6. Therefore, the area of the parallelogram ABCDABCD is 49cm249 cm^2.

Explanation:

The problem uses the direct relationship between triangles and parallelograms sharing the same base and parallels, where the parallelogram's area is exactly double that of the triangle.