In this paper, we show that an interesting relationship for a triangle, which is Area/Perimeter = r/2, extends to other shapes with an inradius r (i.e., shapes with an inscribed circle). These shapes include squares, circles, rhombuses, regular polygons, and even irregular polygons and more

Kaufman, Richard

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Area and perimeter relationshipfor different shapes with aninradiusRichard KaufmanFounder, Office Expander (www.of ficeexpander.com)Abstract: In this paper, the author shows that a relationship for a triangle, which is Area/Perimeter=r/2, extends toother shapes with an inradius r (i.e., shapes with an inscribed circle). These shapes include squares, circles, rhombuses,regular polygons, and even irregular polygons and more. Students may find it interesting, or perhaps a challenge, toshow that this is true.Keywords: geometry, measurementIntroductionWe can imagine that a teacher draws a triangle, square, circle, rhombus, and eight-sided polygon (“stopsign” shape) on a whiteboard and says, “Would you believe these shapes all share a relationship?” Ofcourse, a student may correctly state that all the shapes lie on the plane of the board, or that all theshapes have an area. But there is another relationship that they all share. As we show, shapes with aninscribed circle, which therefore have an inradius r, follow the relationship: Area/Perimeter = r/2.This includes triangles, squares, circles, rhombuses, regular polygons, and even irregular polygonsand more. Students might try to show this relationship for shapes that they already know, or theymight try to show the relationship for new shapes that they learn about - reinforcing lessons. We beginby showing the relationship for the triangle, about which much is known and can be learned.Background: The TriangleEvery triangle, whether acute, right, or obtuse, has an inscribed circle. It can be observed, as illustratedin Figure 1, that line segments which are tangent to a circle and intersect each other are the same length.Figure 1: Triangle with inscribed circle (inradius r).Ohio Journal of School Mathematics 91 Page 27Therefore, for any triangle, we have that a = x + y, b = y + z, and c = x + z. By adding any two ofthese sides and subtracting the remaining side, we can obtain each of the following equations (notethat each equation shows the addition and subtraction in the numerator – where substitutions fromthe prior equations can be made):x =a+ c− b2(1)y =a+ b− c2(2)z =c+ b− a2(3)Although students typically learn that the area of a triangle is half the base times the height, the heightof a triangle may not be known (such as in Figure 1). A less familiar equation for a triangle’s area isHeron’s Formula, which only requires the sides of a triangle:Area =√s(s− a)(s− b)(s− c) (4)where s is the semi-perimeter, s = (a+ b+ c)/2. Simple algebra with s, a, b, and c shows that:Area =√(a+ b+ c2)(c+ b− a2)(a+ c− b2)(a+ b− c2)(5)Using Eq. (1), (2), and (3), the area of a triangle is:Area =√(x+ y + z)xyz (6)The last equation was previously known (Incircle and excircles of a triangle, 2022). Also, using theareas shown in Figure 1, we can show the following result, which was previously known (Area oftriangle in terms of inradius, 2022; Chu, 2022).AreaPerimeter=r2(7)Students can try to develop this last equation on their own. A teacher can suggest that students usethe three triangles shown in Figure 1, each with a height of r, and with bases a = x + y, b = y + z,and c = x+ z, respectively. The area of each triangle, which is half the base times the height, can beadded to give the area of the overall triangle, Area = r(x+y)2 +r(y+z)2 +r(x+z)2 =r(x+y+y+z+x+z)2 =r(x+ y + z) = r(Perimeter)/2, which establishes Eq. (7).Using the last two equations, it is easy to show the following, which was previously known (Area oftriangle in terms of inradius, 2022; Chu, 2022; Ogilvy & Anderson, 1988):r =√xyz(x+ y + z)(8)From the last two equations, the area A multiplied by the inscribed radius r is: Ar = xyzArea and Perimeter Relationship for Other Shapes with an InradiusIn the previous section, we saw that the relationship Area/Perimeter = r/2 holds for any triangle. Inthis section, we show that this relationship extends to other shapes with an inradius r.The area of a circle is related to its radius by the formula Area = πr2. Since the circumferenceof a circle is 2πr, then students can see that:Ohio Journal of School Mathematics 91 Page 28A) Circle: AreaPerimeter (i.e., circumference) =πr22πr =r2A square has an incircle with radius r that is easily shown to be half the length of a side of the square.So, the square’s side is of length 2r. Therefore, the square has an area of (2r)2 and it has a perimeter of4(2r), so students can easily demonstrate that:B) Square: AreaPerimeter =(2r)24(2r) =r2The area of a rhombus is related to the inradius r by the formula Area = 2ar, where a is a side lengthof the rhombus (Rhombus, 2022). Since the perimeter of the rhombus is 4a, we have:C) Rhombus: AreaPerimeter =2ar4a =r2A regular polygon is one that has congruent sides and congruent angles. The area of a regular polygonis Area = nsa/2, where a is the apothem, or inradius r, n is the number of sides, and s is the sidelength (Apothem, 2022). The perimeter of a regular polygon is the product of the number of sides andside lengths. Therefore, we have:D) Regular Polygon: AreaPerimeter =nsr/2ns =r2It can be observed that line segments which are tangent to a circle and intersect each other are thesame length (for example, see Figure 1). This is indicated by length v in Figure 2, which shows a shapethat we will call Shape E.Figure 2: Shape E with Lines Tangent to a Circle.The line segments v are perpendicular to the line segment r, which is the radius of the circle. The areaof Shape E is:Area = 2(rv2)+(2π − 2θ2π)πr2 = r(v + (π − θ)r)The perimeter of the Shape E is:Perimeter = 2v + (2π − 2θ)r = 2(v + (π − θ)r)Therefore, for Shape E, we again have the relationship:E) Shape E: AreaPerimeter =r(v+(π−θ)r)2(v+(π−θ)r) =r2Shape E is made up of a circle and a shape that is shown in Figure 3, which we will call Shape F.Although the relationship Area/Perimeter = r/2 holds for Shape E and also for the circle, it doesnot hold for Shape F. The area of Shape E is equal to the area of the circle plus the area of Shape F.However, the perimeter of Shape E is not equal to the circumference of the circle plus the perimeter ofShape F since the arc of Shape F cannot be included (twice at that!) in the overall perimeter of Shape E.This is why Shape F does not follow the relationship Area/Perimeter = r/2 itself.Ohio Journal of School Mathematics 91 Page 29Figure 3: Shape F (from the top of Shape E).The choice of the tangent lines used in Figure 2 was arbitrary and extends to any number of non-overlapping tangent lines of any length that intersect. Therefore, the following relationship holds forany irregular polygon with an inradius:F) Irregular Polygon with inradius: AreaPerimeter =r2And it also holds for unusual shapes such as Figure 4.Figure 4: Shape including a circle with different-length tangent lines that intersect.In all cases of a triangle, circle, square, rhombus, regular polygon, irregular polygon (with an inradius),and shapes formed with tangent lines to a circle, the following relationship between the area, perimeter,and inradius r holds:G) AreaPerimeter =r2ConclusionStudents and teachers may find it interesting how one relationship between the area and perimeterexists in shapes with an inradius. Depending on the shapes learned by students, then a teacher coulddevelop the relationship for some or all of the shapes in this paper. Alternatively, the relationship canprovide a challenge for students to demonstrate what they have learned about a shape (or shapes). Itis hoped that this paper provides the inspiration for students to explore mathematics and find newrelationships and patterns.ReferencesIncircle and excircles of a triangle. (July 10, 2022). In Wikipedia. https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#cite_note-9Area of Triangle in Terms of Inradius. (July 10, 2022). In ProofWiki. https://proofwiki.org/wiki/Area_of_Triangle_in_Terms_of_InradiusOhio Journal of School Mathematics 91 Page 30Chu, T. (2005). Problem 584. The Pentagon, 65(1), 50. Available online at https://www.kappamuepsilon.org/Pentagon/Vol_65_Num_1_Fall_2005.pdfOgilvy, C. S., & Anderson, J. T. (1988). Excursions in number theory. Dover Publications, 68.Rhombus. (July 10, 2022). In Wikipedia. http://en.wikipedia.org/w/index.php?title=Rhombus&oldid=650988704Apothem. (July 10, 2022). In Wikipedia. http://en.wikipedia.org/w/index.php?title=Apothem&oldid=650696918Richard Kaufman (rdkaufman01@gmail.com) received his B.S. in InformationTechnology in 2014, his M.S. in Applied Math in 2011, and his M.S. in MechanicalEngineering in 2000, all from UMass Lowell, USA. Richard is the founder of OfficeExpander (www.OfficeExpander.com) and is a registered Professional Engineer.Ohio Journal of School Mathematics 91 Page 31

Area and perimeter relationship for different shapes with an inradius

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Area and perimeter relationship for different shapes with an inradius

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