Class 9: Area – Lecture Notes

$\displaystyle \textbf{Points to Remember}$

$\displaystyle \textbf{1. Equal figures: }\text{Two plane figures having equal area are called equal figures.}$

$\displaystyle \textbf{2. Congruent figures: }\text{Two plane figures having the same shape and size are}$
$\displaystyle \text{called congruent figures. But two plane figures having equal areas need not}$
$\displaystyle \text{be congruent.}$

$\displaystyle \textbf{3. Results on Area of Polygon Regions:}$

$\displaystyle \text{(i) Parallelograms on the same base and between the same parallels are}$
$\displaystyle \text{equal in area.}$

$\displaystyle \text{(ii) The area of a parallelogram is equal to the area of the rectangle on the}$
$\displaystyle \text{same base and of the same altitude, i.e. between the same parallels.}$

$\displaystyle \text{(iii) Triangles on the same base and between the same parallels are equal}$
$\displaystyle \text{in area.}$

$\displaystyle \textbf{4. Some more results:}$

$\displaystyle \text{(i) Area of a parallelogram }=\text{Base}\times\text{Height.}$

$\displaystyle \text{(ii) Area of a triangle }=\frac{1}{2}\times\text{Base}\times\text{Height.}$

$\displaystyle \text{(iii) Area of a trapezium }=\frac{1}{2}\times\text{(Sum of parallel sides)} \times\text{Height.}$

$\displaystyle \text{(iv) Area of a rhombus }=\frac{1}{2}\times\text{Product of diagonals.}$

$\displaystyle \textbf{5.}$

$\displaystyle \text{(i) If a triangle and a parallelogram are on the same base and between the}$
$\displaystyle \text{same parallels, then the area of the triangle is half of the area of the}$
$\displaystyle \text{parallelogram.}$

$\displaystyle \text{(ii) Parallelograms on equal bases and between the same parallels are}$
$\displaystyle \text{equal in area.}$

$\displaystyle \\$

$\displaystyle \textbf{Area Axioms}$
$\displaystyle \textbf{Area Axiom:}\ \text{Every polygonal region has an area and is measured in square units.}$
$\displaystyle \textbf{Congruent Area Axiom:}\ \text{If }\triangle ABC\text{ and }\triangle DEF\text{ are two congruent triangles, then}$
$\displaystyle ar(\triangle ABC)=ar(\triangle DEF)$
$\displaystyle \text{i.e., two congruent regions have equal area.}$
$\displaystyle \textbf{Rectangle Area Axiom:}\ \text{If }ABCD\text{ is a rectangular region such that}$
$\displaystyle AB=a\text{ units and }BC=b\text{ units, then}$
$\displaystyle ar(ABCD)=ab\text{ square units}$
$\\$

$\displaystyle \textbf{Parallelograms on the Same Base and Between the Same Parallels}$
$\displaystyle \textbf{Theorem 1:}\ \text{A diagonal of a parallelogram divides it into two triangles of equal area.}$ 2019-02-09_22-32-31
$\displaystyle ar(\triangle ABC)=ar(\triangle ADC)$
$\displaystyle ar(\triangle ABD)=ar(\triangle BCD)$
$\\$

$\displaystyle \textbf{Theorem 2:}\ \text{Parallelograms on the same base and between} \\ \text{the same parallels are equal in area.}$ 2019-02-10_9-54-47
$\displaystyle ar(ABCD)=AB\times h$
$\displaystyle ar(ABFE)=AB\times h$
$\displaystyle \therefore\ ar(ABCD)=ar(ABFE)$
$\\$

$\displaystyle \textbf{Theorem 3:}\ \text{The area of a parallelogram is the product of its} \\ \text{base and the corresponding altitude.}$ 2019-01-12_12-45-40
$\displaystyle \text{Let the adjacent sides of the parallelogram be }a\text{ and }b.$
$\displaystyle \text{Area}=\text{Base}\times\text{Height}$
$\\$

$\displaystyle \textbf{Triangle Area Axioms}$

$\displaystyle \textbf{Theorem 4:}\ \text{The area of a triangle is half the product of any side} \\ \text{and its corresponding altitude.}$ 2019-01-12_12-49-14
$\displaystyle \text{Let }a,\ b,\ c\text{ denote the sides of the triangle.}$
$\displaystyle \text{Area}=\frac{1}{2}\times\text{Base}\times\text{Height}=\frac{1}{2}bh$
$\displaystyle \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$
$\displaystyle \text{This is known as Heron's Theorem.}$
$\\$

$\displaystyle \textbf{Theorem 5:}\ \text{If a triangle and a parallelogram are on the same base} \\ \text{and between the same parallels, then the area of the triangle is half the} \\ \text{area of the parallelogram.}$ 2019-02-10_10-00-24
$\displaystyle ar(ABCD)=AB\times h$
$\displaystyle ar(\triangle ABE)=\frac{1}{2}\times AB\times h$
$\displaystyle ar(\triangle ABE)=\frac{1}{2}\,ar(ABCD)$
$\\$

$\displaystyle \textbf{Trapezium Area Axiom}$

$\displaystyle \textbf{Theorem 6:}\ \text{The area of a trapezium is half the product of its height} \\ \text{and the sum of the lengths of its parallel sides.}$ 2019-01-12_12-45-06
$\displaystyle \text{A trapezium is a quadrilateral with one pair of parallel sides.}$
$\displaystyle \text{A trapezium whose non-parallel sides are equal is called an isosceles trapezium.}$
$\displaystyle \text{Let }a\text{ and }b\text{ be the parallel sides and }h\text{ be the height.}$
$\displaystyle \text{Area}=\frac{1}{2}(a+b)\times h$
$\\$

$\displaystyle \textbf{Triangles on the Same Base and Between the Same Parallels}$

$\displaystyle \textbf{Theorem 7:}\ \text{Triangles on the same base and between the same} \\ \text{parallels have equal areas.}$
$\displaystyle ar(\triangle ABD)=\frac{1}{2}\times AB\times h$ 2019-02-10_10-05-29
$\displaystyle ar(\triangle ABC)=\frac{1}{2}\times AB\times h$
$\displaystyle \therefore\ ar(\triangle ABD)=ar(\triangle ABC)$
$\\$
$\displaystyle \textbf{Theorem 8:}\ \text{Triangles having equal areas and one side of} \\ \text{one triangle equal to one side of the other have equal corresponding altitudes.}$
$\\$
$\displaystyle \textbf{Theorem 9:}\ \text{Two triangles having the same (or equal) bases and} \\ \text{equal areas lie between the same parallels.}$