CHAPTER 4
Presentation and Discussion
This chapter presents the analysis conducted by the researcher from the different theorems of triangle as a realm of the study. The object is to determine and formulate different theorems about the relationship of perimeter and area of a triangle. Particularly, the study aims to prove and attempted to formulate theorems on;
1. when the triangles have areas and perimeters numerically equal
2. when the triangles have areas and perimeters equal in terms of the length of sides
3. when the triangles have areas and perimeters equal in terms of the length of tangent segments
4. when the triangles have areas and perimeters equal in terms of the deviation of x from the midpoint of the tangent segments
5. when the triangles have areas and perimeters equal in terms of the radius xy/2 where x and y are distance from the midpoint connected to two tangent lines having a common point
The Inscribed Circle
Consider DABC (See figure 1) with sides a,b, and c be any triangle of equal area and perimeter circumscribed about a circle O of radius r. Let P, Q, and R be the points of tangency. By theorem C; OR = OP = OQ = r and by Theorem A, OQ ^ BC, OP ^ AB, and OR ^ AC. Introduce OA, OC, and OB.
Considering the area and perimeter of DABC respectively such that
Area of DABC = area of DCOB + area of DAOC + DAOB
= ½ (ra) + ½ (rb) +½ (rc)
= ½ r(a + b + c)
Perimeter of DABC = a + b + c
Since the area and perimeter of DABC are equal, then it implies that
½ r(a + b + c) = a + b + c , simplifying further
a+b+c
r = 2 (————-)
a+b+c
r = 2
This statement results to the theorem below.
Theorem 1. A triangle has equal area and perimeter if and only if it can be circumscribed about a circle of radius 2.
Consider DABC (see fig. 2). WE know by Theorem B, AP = AR, BP = BQ and CQ = CR. Let AP = AR = x, CQ = CR = y and BP = BQ z such that x,y and z are length of each tangent segments. From figure 2, z + y = a, x + y = b and x + z = c such that the perimeter of DABC = a + b + c in which by substitution it implies that
Perimeter of DABC = (z + y) + (x + y ) + (x + z)
= 2x + 2y +2z
= 2(x + y + z)
Using Heron’s Formula such that
Area of DABC = s(s-a)(s-b)(s-c) , where s = ½ (a + b + c).
By substitution, s= ½ [(z + y) +(x + y)+ (x + z)] which implies that s= x + y + z Also by substitution a = z + y , b = x + y , c = x + z and s = x + y + z to Heron’s Formula, then
Area of DABC = (x+y +x )[(x+y+z)-(z+y)] [(x+y+z)-( x + y)] [(x+y+z)-(x+z)]
= xyz (x+y+z)
Since the area of DABC is equal to its perimeter, then
2(x+y+z) = xyz (x+y+z), squaring both sides
4(x+y+z)2 = xyz (x+y+z)
xyz = 4 (x+y+z)
xyz = 4x + 4y + 4z, solving for z in terms of x and y
xyz – 4z = 4(x + y)
z(xy – 4) = 4 (x + y)
z = 4(x + y)
(xy – 4)
These statements results to the theorem below
Theorem 2. If DABC of side a, b and c has equal area and perimeter, then
Z = 4(x + y)
(xy – 4)
Conditions for choosing x and y:
In figure 1, let DABC with sides a, b and c have equal area and perimeter. We know that the length of sides are nonnegative, then z > 0 which implies
4(x + y) > 0
(xy – 4)
This further implies 4(x + y) > 0 and xy – 4 > 0, therefore xy > 4.
This statement results to the theorem below.
Theorem 3. DABC with sides a, b and c can have equal area and perimeter only if xy > 4, where x and y are tangent segments as it is defined in figure 2.
In figure 2, let DABC of sides a, b and c have equal area and perimeter, and let x, y and c be defined according to Theorem B such that AP = AR = x, CQ = CR = y and BP = BQ = z. From Theorem 3, xy > 4 which implies y >.
By addition Property of Equality, x is added to both sides which gives
x + y > + x
Assuming that the difference between the tangent segment x and the radius r is greater than zero such that x – 2 > 0 because from Theorem 1 r = 2. Taking the square of sides of the equality implies that x2 – 4x + 4 > 0, multiplying both sides by 1/x further implies x – 4 + > 0. Then it follows that x + > 4, but y >, thus by substitution x + y > 4. Since x + y = a, then a > 4. In similar way, we can also show that b > 4 and c > 4.
This statement results to the theorem below.
Theorem 4. A triangle has equal area and perimeter if each side is greater than 4.
Supposing DABC has equal area and perimeter, from fig. 2 it is given that x + y = b. Multiplying both sides of the equality by x implies that
x2 + xy = bx
xy = bx – x2, from Theorem 3, xy > 4 which implies
4< bx –x2
x2 - bx + 4 < 0, using quadratic formula in
solving for x,
then x = -b+ b2 – 4ac , by the property of inequalities
2a
-b- b2 – 4ac < x < -b+ b2 – 4ac , which implies
2a 2a
b - b2 – 16 < x < b + b2 – 16
2 2
By the property of absolute value and inequalities, then
/x – b/ < b2 – 16
2 2
If x- is the deviation of x from , then the deviation of x from is less than b2 – 16 . We know that is the midpoint of side b in DABC
2
This statement results to the theorem below.
Theorem 5. Given DABC with sides a, b and c and with x, y and z be tangent segments ( as shown in figure 2), then DABC has equal area and perimeter if and only if the deviation of x from the midpoint of b is less than
b2 – 16 .
2
Right Triangles
In figure 3, let DABC be circumscribed about circle O of radius 2. By Theorem B, BP = BR. Let BP = 2. Since PB = BR, then BR = 2. Also from fig. 2 if OP = 2, then by Theorem C OR = 2. By Theorem A OP^ BP and OR ^ BR. Since OP=OR=BP=BR, ÐORB and ÐOPB is right, then by Theorem D ÿOPBR is a square. It follows that this implies that ÐC is a right angle or DABC is a right Triangle. Since DABC is circumscribed about a circle O of radius 2, then its area and perimeter are equal, by Theorem 1.
This statement results to the theorem below.
Theorem 6. Given DABC circumscribed about a circle O of radius 2, DABC is a right triangle of equal area and perimeter if a tangent segment from a vertex of a triangle to circle O has length
To be extended: Given that DABC has equal area and perimeter, show that the radius of circle P is xy/2.
Kindly help me solve this extension problem.
Fig. 5
Credit:ivythesis.typepad.com
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