A kosarad üres
The concept of area, area of polygons
It happens often that one has to find the area of some shape, some plane figure. For example, the size of a site can determine its value or the surface of a wall can determine the amount of paint needed to colour it. In many cases, these shapes are polygons, but later we will see problems concerning plane figures with arbitrary shape.
In case of polygons, area is determined by comparison to the unit area. It worth choosing a simple plane figure which can be easily visualized. This role was given to the square with sides of length \latex{ 1 }, the so-called unit square. (Both the length of a side and the area is given using the appropriate unit. If, for example, the size of the square is \latex{ 1\, m }, then the area is \latex{ 1\, m^2 }. If these are not important in some problems, then we omit them.)
To be more precise, we can think of area as a function which maps plane figures to positive numbers such that:
- The area of the unit square is \latex{ 1 }.
- The area of congruent polygons are equal.
- If a polygon is cut into smaller polygons, then the sum of the areas of the parts equals the area of the original polygon.
It can be shown that this function maps a uniquely determined positive number to every polygon, that is, every polygon has area.
The definition can be generalized in the sense that the plane figure is not necessarily bounded by lines, but by arbitrary curves. Out of these plane figures, we will study the disc and its parts.
Using the concept of area, we will be able to study the surface of solids since if the surface can be laid into the plane, then the surface equals the area of this plane figure.
The area of the rectangle
The area of the rectangle is a good starting point in determining the area of other polygons. Thus the following theorem states an important result.
THEOREM: The area of the rectangle with sides \latex{ a } and \latex{ b } is: \latex{t=a\times b.}
Example 1
Let us solve the inequality \latex{x^{2} – 9 \lt 0}.
Proof
The proposition of the theorem is obviously true if the rectangle can be divided into unit squares, that is, if the lengths of the sides are integers. (Figure 28)
If this is not the case, then the proof requires more advanced tools, but it can be verified as well.
It worth noting that for any rectangle, any altitude corresponding to one side equals to the length of the other side. Thus it is also correct to say that the area of the rectangle equals the product of one of its sides and the corresponding altitude.
\latex{ b }
\latex{ a }
\latex{ 1 }
\latex{ 1 }
Figure 28
The area of parallelograms
We shall use the result concerning rectangles to determine the area of parallelograms. Any parallelogram can cut into pieces which can be rearranged as a rectangle with the same area as seen in Figure 29.
The two triangles are congruent, thus their area is the same. Because of this, the area of the parallelogram \latex{ ABCD } equals to the area of the rectangle \latex{ ABC’D’ } that is, the product of the two sides of the rectangle. The length of the segment \latex{ AD’ } is equal to the altitude of the side \latex{ AB } of the parallelogram, therefore the following theorem is true:
\latex{ D }
\latex{ C' }
\latex{ C }
\latex{ D' }
\latex{ A }
\latex{ a }
\latex{ B }
\latex{ b }
\latex{ m_a }
Figure 29
THEOREM: The area of the parallelogram equals the product of the length of any side and the altitude belonging to that side:
\latex{t=a\times m_{a\times }}
The area of triangles
The area of triangles can be traced back to the area of parallelograms: if we reflect the triangle\latex{ ABC } through the midpoint of the side \latex{ AC } as seen in the figure, then we end up with a parallelogram \latex{ ABCB’ }. (Figure 30)
Its area is obviously is double of that of the triangle. Since the altitude belonging to the side a of the parallelogram is equal to the altitude \latex{m_{a}} of the triangle, the following holds for the area of the triangle \latex{ ABC }:
\latex{ m_a }
\latex{ A }
\latex{ B }
\latex{ a }
\latex{ C }
\latex{ B' }
Figure 30
THEOREM: The area of the triangle equals half of the product of the length of one side and the altitude belonging to said side:
\latex{t=\frac{a\times m_{a} }{2}.}
It is worth noting that we have met several equalities during the previous years which gives us connection between the area of the triangle and some other quantities describing the triangle. In the following, we list a few of these equalities, without their proofs.
\latex{t=p\times s}, where \latex{\rho} is the radius of the incircle and s is the half of the circumference;
\latex{t=\sqrt{s\times (s-a)\times (s-b)\times (s-c)},} Heron's formula;
\latex{t=\frac{b\times c\times \sin \alpha }{2},} where \latex{\alpha} is the angle between sides b and c;
\latex{t=\frac{a\times b\times c}{4R},} where R is the radius of the circumscribed circle.
The area of trapezoids
To determine the area of a trapezoid, we can use the formula for parallelogram as well: if we reflext the trapezoid \latex{ ABCD } through the midpoint of its side \latex{ BC }, for example, then we end up with a parallelogram. (Figure 31)
The area of the trapezoid is the half of that of the parallelogram. That is,
The area of the trapezoid is the half of that of the parallelogram. That is,
\latex{ D }
\latex{ c }
\latex{ C }
\latex{ a }
\latex{ m }
\latex{ d }
\latex{ b }
\latex{ c }
\latex{ B }
\latex{ a }
\latex{ A }
\latex{ D' }
\latex{ A' }
Figure 31
THEOREM: The area of the trapezoid equals the product of the arithmetic mean of its two bases and its altitude:
\latex{t=\frac{a+c}{2}\times m}
The area of polygons
Since every polygon can be divided into triangles, their area can be determined by adding the areas of the triangles. However there are some special cases in which we can determine the area in a faster way.
The following theorem is easy to prove:
The following theorem is easy to prove:
THEOREM: The area of an arbitrary quadrilateral is equal to the half of the product of its diagonals and the sine of the angle between them:
\latex{t=\frac{1}{2}\times e\times f\times \sin \varphi.}
Proof
Use the notation seen in Figure 32.
\latex{180°-\varphi}
\latex{ u }
\latex{ v }
\latex{ x }
\latex{ y }
\latex{ B }
\latex{ A }
\latex{ D }
\latex{ C }
Figure 32
It is clear that the area of an arbitrary convex quadrilateral equals the sum of the four triangles determined by the diagonals, therefore
\latex{t=\frac{1}{2}\times x\times y\times \sin \varphi+\frac{1}{2}\times y\times u\times \sin (180°-\varphi)+}
\latex{+\frac{1}{2}\times u\times v\times \sin\varphi+\frac{1}{2}\times v\times x\times \sin (180°-\varphi).}
Using that \latex{\sin (180°-\varphi)=\sin \varphi} the expression can be transformed by factoring out:
\latex{t=\frac{1}{2}\times \left[y\times \sin \varphi\times (x+u)+v\times \sin \varphi\times (x+u)\right]=}
\latex{=\frac{1}{2}\times (x+u)\times (y+v)\times \sin \varphi=\frac{1}{2}\times e\times f\times \sin \varphi.}
It follows from the result, that if the diagonals of a convex quadrilateral are perpendicular to each other (this is the case, for example, in a kite), then its area is the half of the product of the diagonals:
\latex{t=\frac{e\times f}{2}.}
For regular polygons it is a good idea to divide them into isosceles triangles which are obtained by connecting the centre of the polygon with its vertices. The sum of the areas of these congruent triangles is equal to the area of the polygon. (Figure 33)
For a regular \latex{ n }-sided polygon, its area is
\latex{t=n\times \frac{r^{2}\times \sin \frac{2\Pi }{n} }{2},}
where \latex{ r } denotes the radius of the circumscribed circle of the polygon.
We have seen during our previous studies that while congruences do not change the area of a polygon, this is not true for a similarity with ratio \latex{ k }.
It is enough to examine the magnitude of change for triangles. The similarity multiplies by \latex{ k } the parameters of a triangle with side of length \latex{ a } and altitude \latex{m_{a}.} Therefore the area of the image is
We have seen during our previous studies that while congruences do not change the area of a polygon, this is not true for a similarity with ratio \latex{ k }.
It is enough to examine the magnitude of change for triangles. The similarity multiplies by \latex{ k } the parameters of a triangle with side of length \latex{ a } and altitude \latex{m_{a}.} Therefore the area of the image is
\latex{t=\frac{k\times a\times k\times m_{a} }{2}=k^{2}\times \frac{a\times m_{a} }{2}.}
It follows that a similarity with ratio \latex{ k } changes the area of a polygon by a factor of \latex{k^{2}} .
\latex{ r }
\latex{ r }
\latex{ \frac{2p}{n} }
\latex{ n=6 }
Figure 33
Exercises
{{exercise_number}}. Determine the area of an equilateral triangle with sides of length \latex{ a }.
{{exercise_number}}. One side of a parallelogram is \latex{ 7\, cm } long, the altitude belonging to it is \latex{ 6\, cm }, while the other altitude is \latex{ 3\, cm }. Find the length of the other side and the angles of the parallelogram.
{{exercise_number}}. Determine the length of the diagonals and the angle between them in the parallelogram whose sides are of length \latex{ 6 } and \latex{ 10 }, and whose area is \latex{40\,cm^{2}.}
{{exercise_number}}. The sides of triangle \latex{ ABC } were increased by their own lengths, \latex{ AB } over \latex{ B }, \latex{ BC } over \latex{ C } and \latex{ CA } over \latex{ A }, to end up with the triangle \latex{ PQR }. Express the area of the triangle \latex{ PQR } in terms of the area of \latex{ ABC }.
{{exercise_number}}. The trisecting points of a triangle are connected to the vertices as seen in the figure. Express the area of the triangle \latex{ PQR } in terms of the area of \latex{ ABC }.
\latex{ R }
\latex{ P }
\latex{ Q }
\latex{ B }
\latex{ A }
\latex{ C }
{{exercise_number}}. Divide triangle \latex{ ABC } into two parts with equal areas using a line through one of its vertices.
{{exercise_number}}. The length of the side of a regular pentagon is \latex{ 10\, cm }. Find its area.
{{exercise_number}}. Prove that the total area of the darker coloured parts in each figure below equals the area of the lighter coloured parts.
paralelogramm,
\latex{ P } is arbitrary
\latex{ P } is arbitrary
convex, quadrilateral,
\latex{ P },\latex{ Q } are midpoints
\latex{ P },\latex{ Q } are midpoints
regular octagon
\latex{ P }
\latex{ P }
\latex{ Q }
a)
c)
b)
{{exercise_number}}. The diagonals of a rhombus are \latex{ 2 } and \latex{ 4 } unit long. The rhombus is rotated around its centre by \latex{ 90º }. Compute the area of the intersection of the original rhombus and its image.
{{exercise_number}}. Is there any rectangle with integer sides with measure of circumference and area being equal?
{{exercise_number}}. Reflect the circumcentre of a regular polygon across each side of the polygon. Let \latex{ T } denote the area of the original polygon, and \latex{ T’ } denote the area of the polygon determined by the images.
- Prove that \latex{1\leq \frac{T'}{T}\lt 4.}
- For what polygon will \latex{\frac{T'}{T}} be an integer?
