## Exploring Analyic Geometry with Mathematica®

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Area

Intuitively, area is the measure of the number of unit squares that can be contained inside a boundary. For a square with sides of length s, the area, A, is given by . For a rectangle with sides a and b, A=ab. As the boundary becomes more complex or contains curved elements, the computation of the area requires more complex considerations. In this chapter we will derive formulas for the areas of Descarta2D objects.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, \$Path.

<<Descarta2D`

Areas of Geometric Figures [Top]

Areas of right, acute and obtuse triangles.

Before exploring formulas for computing areas using analytic geometry, we will look at some formulas from planar geometry. Consider the right triangle ABC shown in Figure [area:fig11] with height h and base b. Clearly, the area of △ABC is one-half of the area of rectangle ABCD, so the area, A, of a right triangle is

Now consider the acute △ABC in the center of Figure [area:fig11]. The area of ABC is given by

Area ABC= Area BCDE- Area ABE- Area ACD

or

Simplifying and using yields

The same formula results for the area of an obtuse triangle (using ), as shown on the right in Figure [area:fig11].

Area of a trapezoid.

Now consider the trapezoid ABCD shown in Figure [area:fig12]. The area of ABCD is given by

Area ABCD= Area ABEF- Area ADF- Area BCE

or

Simplifying and using yields

These formulas from planar geometry will be useful in upcoming sections for deriving the formulas using analytic geometry.

Triangular Area

There are several formulas for the area, A, of a triangle that involve lengths associated with the triangle. The simplest is the familiar A=bh/2, where b is the length of one of the sides of the triangle (the base) and h is the height of the triangle (the distance from the base to the opposite vertex).

The formula of Heron gives the area of the triangle in terms of the lengths of its sides, , alone:

where is the semi-perimeter. This formula is derived in the exploration heron.html.

Area of a triangle by coordinates.

Since a triangle is represented in Descarta2D by the coordinates of the vertices, we wish to derive a formula based on the coordinates. Consider the triangle ABC as shown in Figure [area:fig14], where the coordinates are , and . Projecting A, B and C onto the x-axis produces three points , and . The area of triangle ABC is given by

Area ABC= Area AA'C'C- Area AA'B'B- Area BB'C'C.

The height and base lengths of these trapezoids can be determined as the difference of the coordinates of the points, yielding

Expanding and rearranging will show that the area of a triangle, A, is given by the determinant

where , and are the coordinates of the vertices of the triangle. The sign is selected to yield a positive area.

Alternately, if we multiply the length of the line segment joining two of the points, by the length of the perpendicular line segment on that line from the third point, we have double the area of the triangle determined by the three points.

Example: Find the area of a triangle whose vertices are (1,2), (4,4) and (5,6).

Solution: The function Area2D[triangle] returns the area of a triangle.

Area2D[Triangle2D[{1,2},{4,4},{5,6}]]

Curved Areas [Top]

If we inscribe a polygon inside any closed curve, it is evident that as the number of sides of the polygon is increased, the area of the polygon approaches the area bounded by the curve. Likewise, the perimeter of the polygon approaches the perimeter, or arc length, of the curve. If the number of sides of the polygon is increased ad infinitum, the polygon will coincide with the curve. In like manner, we can see that as the number of sides of a circumscribing polygon is increased, the more nearly its area and perimeter will approach the area and perimeter of the curve. Therefore, when investigating the area or perimeter of a closed curve, we may substitute for the curve an inscribed or circumscribed polygon with an indefinitely increasing number of sides. These notions are formalized in the study of calculus, but they can be applied intuitively in the study of areas of simple curves as will be shown in the following sections.

Circular Areas [Top]

Circle approximated by an inscribed polygon.

To determine the area of a circle, we examine a polygon inscribed in the circle as shown in Figure [area:fig15]. The area of any triangle in the figure is given by , and the area of the entire polygon is given by , because there are n such triangles. As n increases without limit, we find that ns approaches S=2πr and d approaches r. Therefore, the area of the polygon approaches or . Accordingly, the area, A, of a circle is given by , where r is the radius of the circle.

Example: Find the area of a circle centered at (0,0) with a radius of 2.

Solution: The Descarta2D function Area2D[circle] returns the area of a circle.

Area2D[Circle2D[{0,0},2]]

The area of an arc sector of radius r as shown in Figure [area:fig19] may be determined as the ratio of the angular span of the arc to the span of a complete circle ( radians) times the area of the circle. Since the area of a complete circle is , the area of an arc sector is given by

Areas of an arc sector and segment.

Example: Find the area of the sector defined by the arc centered at (0,0) with radius 2 and start and end angles of π/4 and 3π/4.

Solution: The Descarta2D function SectorArea2D[circle,{,}] returns the area of the sector defined by an arc of a circle.

SectorArea2D[Circle2D[{0,0},2],{Pi/4,3Pi/4}]

The area of an arc segment, which is the area bounded by the arc and the chord connecting the end points of the arc as shown in Figure [area:fig19], is calculated as the difference of the areas of the sector and the triangle whose vertices are the end points and the center. Since the area of this triangle is , the formula for the area of the arc segment is

 A = . =

where is the span of the arc, and r is the radius of the arc.

Example: Find the area of the segment defined by the arc centered at (0,0) with radius 2 and start and end angles of π/4 and 3π/4.

Solution: The Descarta2D function Area2D[arc] returns the area of the segment defined by an arc and its chord.

Area2D[Arc2D[Point2D[{0,0}],2,{Pi/4,3Pi/4}]] //Simplify

Notice that if the angle θ is greater than π radians, the formula is still valid because sinθ will be negative and the area of the central triangle will be added to the sector area producing the correct result.

Elliptic Areas [Top]

The area of an ellipse depends only on the lengths of its semi-major and semi-minor axes, and is independent of its position and orientation. It is shown in calculus that integrating the equation y=f(x) of the curve between limits on the x-axis produces the area between the curve and the x-axis. The equation of an ellipse in standard position is given by

Solving for y yields for the upper portion of the ellipse. The following steps outline the integration process that is used to compute the area:

 A = . = . =

Therefore, the area of the complete ellipse (both the upper and lower portions) is given by

A=πab

where a and b are the lengths of the semi-major and semi-minor axes, respectively, of the ellipse.

Example: Calculate the area of the ellipse .

Solution: The Descarta2D function Area2D[ellipse] returns the area of an ellipse.

Area2D[Ellipse2D[{0,0},3,2,Pi/2]]

Consider an ellipse in standard position with the equation

as shown in Figure [area:fig17]. The area, A, of a segment of the ellipse bounded by the chord defined from (a,0) and a general point on the ellipse, (acosθ,bsinθ), can be determined by subtracting the area under the chord from the area under the ellipse between limits on the x-axis. The equation of the line containing the chord is given by

as can be determined from the two-point form of a line. The area of the segment is determined using integration as follows:

 A = . = . =

The formula is valid for angles θ in the range 0≤θ≤π. Since all segments can be computed as sums or differences of such segments and simple triangles, the area of all ellipse segments can be determined using this basic formula.

Areas of an elliptic segment and sector.

Example: Find the area of the ellipse segment from π/6 to π/3 radians for an ellipse with semi-major axis length of 3 and semi-minor axis length of 1.

Solution: The function SegmentArea2D[ellipse,{,}] returns the area of an ellipse segment defined by two parameter values.

SegmentArea2D[Ellipse2D[{0,0},3,1,0],{Pi/6,Pi/3}]

The area of an ellipse sector, as illustrated in Figure [area:fig17], can be found by adding the area of the triangle formed by the sector sides and the chord of the sector. The area of the triangle is given by

and the resulting formula for the area of the sector is given by

Example: Find the area of the ellipse sector from π/6 to π/3 radians for an ellipse with semi-major axis length of 3 and semi-minor axis length of 1.

Solution: The Descarta2D function SectorArea2D[ellipse,{,}] returns the area of an ellipse sector defined by two parameter values.

SectorArea2D[Ellipse2D[{0,0},3,1,0],{Pi/6,Pi/3}]

Hyperbolic Areas [Top]

Areas of a hyperbolic segment and sector.

Using the integration techniques demonstrated previously for ellipses the areas associated with hyperbolas can also be computed. Consider the hyperbolic segment as shown in Figure [area:fig18]. In Descarta2D the parametric equations for the hyperbola are

x=acosh(st)   and   y=bsinh(st)

where a and b are the lengths of the semi-transverse and semi-conjugate axes, respectively, , and e is the eccentricity of the hyperbola. The exploration areahyp.html uses calculus to derive the formula for the area of the segment, which is given by

Interestingly, the area does not depend on the values of and directly, but only upon the difference between the two parameters.

Since we know the coordinates of the vertex points of the triangle we can compute its area directly as

The area of a hyperbolic sector is the difference between the area of the triangle and the hyperbolic segment as illustrated in Figure [area:fig18]. The area of the hyperbolic sector is given by

 = . =

Example: Find the area of the hyperbolic segment between parameters and for a hyperbola centered at the origin with a=1 and b=1/2 in standard position. Also, find the area of the associated hyperbolic sector.

Solution: The Descarta2D function SegmentArea2D[hyperbola,{,}] returns the area of a segment of a hyperbola defined by two parameters. The function SectorArea2D[hyperbola,{,}] returns the area of the associated hyperbolic sector.

h1=Hyperbola2D[{0,0},1,1/2,0];
{SegmentArea2D[h1,{-2,1}],
SectorArea2D[h1,{-2,1}]} //N

Parabolic Areas [Top]

Area of a parabolic segment.

Consider a parabola in standard position with vertex at (0,0), axis parallel to the x-axis, focal length of f, and opening to the left as shown in Figure [area:fig09]. Such a parabola can be represented with the set of parametric equations

The area of the chordal area defined by the parameters and can be determined by subtracting the area between the parabola and the y-axis from the area between the chord and the y-axis. The end points of the chord are and , and the line through these two points is given by

The appropriate integral is then given by

 A = . =

Performing the integration and making the substitutions and yields the formula for the area, A, of a parabolic segment to be

where and are the parameters of the end points of the chord defining the segment, and for positive areas. A parabola has no sector area definition because a parabola does not have a center point.

Example: Find the area between the parabola rotated π/2 radians about its vertex and its focal chord.

Solution: The function SegmentArea2D[parabola,{,}] returns the area of a parabolic segment defined by parameters and . In Descarta2D the end points of the focal chord occur at parameter values and .

p1=Parabola2D[{0,0},1/2,Pi/2];
SegmentArea2D[p1,{-1,1}]

Descarta2D Hint: Notice that the parabola's position and orientation have no bearing on the area of a parabolic segment. The area depends solely on the focal length and the chord position.

Conic Arc Area [Top]

The area between a conic arc and its chord can also be computed by summing infinitesimal rectangles through the use of calculus. Consider, for example, a conic arc whose start point is (0,0), end point is (d,0), apex point is (a,b) and projective discriminant is ρ. Intuitively, since the chord of this conic arc is coincident with the x-axis we can imagine subdividing the area of the conic arc into a large number of horizontal rectangles of very small height. By summing the areas of these small rectangles we can provide an approximation to the area of the conic arc. The methods of integral calculus accomplish this summing process, and in the limit as the height of the rectangles approaches zero, the area approaches the true area of the conic arc segment. The details of this process are captured in the exploration caarea1.html. The area of the conic arc segment is found to be

where (assuming b>0 and d>0). Notice that the abscissa, a, of the apex point has no bearing on the area of the segment bounded by the conic arc and its chord.

If the conic arc is a parabola, then and the formula for the area given above is invalid. The formula for a parabola is much simpler and is given by

as shown in the exploration caarea2.html.

This process can be generalized to find the segment area of any conic arc. Notice that the position of the conic arc in the plane has no bearing on its chordal area. Therefore, we can translate the start point to (0,0) and rotate the conic so that the end point is on the x-axis.

Example: Find the area between the conic arc with start point (1,2), end point (5,0), apex point (3,4) and ρ=0.75 and its chord.

Solution: The Descarta2D function Area2D[cnarc] computes the area between a conic arc and its chord.

Area2D[ConicArc2D[{1,2},{3,4},{5,0},0.75]] //N

Summary of Area Functions [Top]

Table [area:tbl01] summarizes the area functions provided by Descarta2D.

Area2D[object] returns the area enclosed by a closed object (circle, ellipse or triangle).

SectorArea2D[object,{,}] returns the area of a sector defined by two parameters.

SegmentArea2D[object,{,}] returns the area between a chord and the curve.

Descarta2D area functions.

Explorations [Top]

Heron's Formula. [heron.html]

Show that the area, K, of a triangle △ABC is given by

where the semi-perimeter s=(a+b+c)/2, and a, b and c are the lengths of the sides.

Area of Triangle Configurations. [triarea.html]

For the triangle illustrated in the figure, show that the area, , associated with the AAS (angle-angle-side) configuration whose parameters are , and is given by

Show that the area, , associated with the ASA (angle-side-angle) configuration whose parameters are , and is given by

Show that the area, , associated with the SAS (side-angle-side) configuration whose parameters are , and is given by

Area of Triangle Bounded by Lines. [triarlns.html]

Show that the area of the triangle bounded by the lines

is given by

Areas Related to Hyperbolas. [hyparea.html]

Referring to Figure [area:fig18], use calculus to verify that the areas between two parameters and of a segment and a sector of a hyperbola are given by

 = =

where a and b are the lengths of the semi-transverse and semi-conjugate axes, respectively, , and e is the eccentricity of the hyperbola (assuming the parameterization Descarta2D uses for a hyperbola).

Area of a Conic Arc (General) [caarea1.html]

For the conic arc whose control points are (0,0), (a,b) and (d,0), show that the area between the conic arc and its chord is given by

where (assuming b>0 and d>0).

Area of a Conic Arc (Parabola) [caarea2.html]

Show that the area between a conic arc whose projective discriminant is and its chord is given by

when the control points are (0,0), (a,b) and (d,0).

One-Third of a Circle's Area [circarea.html]

Show that the angle, θ, subtended by a segment of a circle whose area is one-third the area of the full circle is the root of the equation

Also, show that θ is within 1/2 percent of 5π/6 radians.

Equal Areas Point [eqarea.html]

Given △ABC with vertices , and show that there are four positions of a point such that △APB, △APC and △BPC have equal areas. The coordinates of are given by

is the centroid of △ABC and . △ABC connects the midpoints of the sides of .

Area of a Tetrahedron's Base [tetra.html]

A tetrahedron is a three-dimensional geometric object bounded by four triangular faces. Given a tetrahedron with vertices O(0,0,0), A(a,0,0), B(0,b,0) and C(0,0,c) show that the areas of the triangular faces are related by the equation

where is the area of the triangle whose vertices are x, y and z. Note the similarity to the Pythagorean Theorem for right triangles.