Essential Area Theorems Every Maths Student Should Know
The area of the theorem has basically few properties which are as follows:
A parallelogram is divided into two triangles of equal areas by the diagonal.
The ratio of the areas of 2 triangles with a similar height is equivalent to the ratio of their bases.
The ratio of the areas of 2 triangles on the same base is equivalent to the ratio of their heights.
The area of triangles that are congruent is equal.
Area Theorems
The theorems state some link between the areas of these geometric objects under the condition when they lie between the same parallel lines and on the same base (or equal bases). Following are the area theorem axioms:
Theorem 1
Diagonal of a parallelogram cut it half into 2 triangles of the same area. Parallelograms between the same parallels and on the same base are equal in area. A diagonal of a parallelogram divides it into two triangles of the same area In this case area of (△ABC) = area of (△ADC). Also area of (△ABD) = area of (△BCD)
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Theorem 2
The area of a parallelogram is equivalent to the area of the rectangle of the same altitude and on the same base, i.e., between the same parallels. That is to say, the area of (||gm ABCD) = Area of (rectangle ABFE) since they lie between the same parallels AB and DE and on the same base.
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Theorem 3
The area of a triangle is half of the area of a parallelogram lying between the same parallels and on the same base. From the figure below, the area of (∆ APB) = ½ × Area of (||gm ABCD) seeing that they lie between the same parallels AB and PC and on the same base. Area of a parallelogram is the product of its base and the corresponding height.
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Theorem 4
Triangles between the same parallels as well on the same base are equivalent in area. Area of (∆ ABD) = Area of (∆ ABC) since they remain between the same parallels AB and DC and on the base AB. The area of a triangle is half the product of its corresponding height and any of its sides. This theorem is also called Heron's Theorem.
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Theorem 5
If a parallelogram and a parallelogram lies between the same parallels and on the same base, thus the area of the triangle will be equivalent to the half of the parallelogram.
Area of (△ABCD) = area of (△BCD)
Area of (||gmABCD) = AB × h
Area of (△ABE) = ½ AB × h
Area of (△ABE) = Area of (||gmABCD)
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Theorem 6
This is a Trapezium Area Axiom. According to this theorem, the area of a trapezium is half the product of the sum of its parallel sides and the altitude. A trapezium is a type of a quadrilateral that has two of its sides parallel to each other. There is also a type of trapezium which we call an isosceles trapezium whose non-parallel sides are equal. Having said that, suppose that we have ‘a’ and ‘b’ the parallel sides and ‘h’, the distance between the parallel sides of a parallelogram ABCD. Then Area = (½ A+B) × h
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Theorem 7
Triangles between the same parallels and on the same base share the same area.
Area of (△ABD) = ½ × AB × h
Area of (△ABC) = ½ × AB × h
Hence, Area of (△ABD) = Area of (△ABC)
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Theorem 8
Triangles with equal areas and whose one side of one of the triangles equivalent to one side of the other triangle, with their corresponding heights the same.
Theorem 9
Two triangles whose bases are the same (or equal bases) and equal area remain between the same parallels.
FAQs on Theorems on Area: Definitions, Proofs & Examples
1. What is the fundamental concept of 'area' in geometry as per the CBSE Class 9 syllabus?
In geometry, the area of a plane figure is the measure of the surface or region enclosed within its boundary. It is expressed in square units, such as cm² or m². For the chapter on 'Theorems on Area', the key idea is not just calculating area, but comparing the areas of different figures, like parallelograms and triangles, based on their geometric properties without necessarily knowing the exact numerical value.
2. What are the main theorems on area for parallelograms covered in this chapter?
The primary theorem states that parallelograms on the same base and between the same parallels are equal in area. This is a foundational principle for comparing areas. It implies that if you have multiple parallelograms sharing a common base line and their opposite sides lie on the same parallel line, their areas will be identical, regardless of how 'slanted' they appear.
3. How is the area of a triangle related to a parallelogram based on the theorems in this chapter?
There are two key relationships taught in this chapter:
- The area of a triangle is half the area of a parallelogram if they are on the same base and between the same parallels.
- Consequently, two triangles on the same base (or equal bases) and between the same parallels are equal in area.
4. Why is the condition 'between the same parallels' so important in these area theorems?
The condition 'between the same parallels' is crucial because it ensures that the heights of the figures being compared are equal. The area of a parallelogram (base × height) and a triangle (½ × base × height) directly depends on the perpendicular height. By placing figures between the same parallels, we fix their height, allowing for a direct comparison of their areas based solely on their bases.
5. What does it mean for two geometric figures to be on the 'same base'?
Two figures are considered to be on the 'same base' when they share a common side. For example, two parallelograms ABCD and ABEF share the common base AB. Similarly, a triangle and a parallelogram can be on the same base. This shared side acts as the reference dimension for applying the theorems on area.
6. If two triangles have equal areas and share a common base, what can we conclude about them?
This is the converse of the main triangle theorem. If two triangles have equal areas and are on the same base, we can conclude that they must lie between the same parallels. This means their vertices opposite the common base lie on a line that is parallel to the base. This principle is often used in proofs to establish that two lines are parallel.
7. How can you prove that a median of a triangle divides it into two triangles of equal areas?
You can prove this using the area formula. Let AD be the median to side BC of a triangle ABC. By definition, a median bisects the opposite side, so BD = DC. Now, draw a perpendicular (height) AM from vertex A to the base BC. The area of triangle ABD is ½ × base BD × height AM. The area of triangle ADC is ½ × base DC × height AM. Since BD = DC and the height AM is common to both, their areas are equal. This is a direct application of area concepts.
