Below are images of an it is intended triangle, a square, a regular hexagon, and also a circle each havingthe same perimeter:
This trouble is part of a an extremely rich heritage of troubles looking to maximizethe area attached by a shape with addressed perimeter. Just three shapes are taken into consideration here because the difficulty is daunting for much more irregular shapes. Forexample, of all triangles, the one with solved perimeter $P$ and largest areais the it is provided triangle whose side lengths space all $fracP3$ but thisis complicated to show because it is not basic to discover the area of triangle in termsof the 3 side lengths (though Heron"s formula accomplishes this). Noris it simple to compare the area of 2 triangles v equal perimeter withoutknowing their individual areas. Because that quadrilaterals, a similar problem arises:showing the of all rectangles v perimeter $P$ the one with the largestarea is the square whose side lengths space $fracP4$ is a an excellent problem i m sorry students should think about. However comparing a square come an irregularly shaped quadrilateral of same perimeter will be difficult.
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For this problem, very explicit shapes have been chosen aiming in ~ providingan chance for students to practice using their understanding of differentgeometric formulas for area and also perimeter. The teacher, in order come reinforcepart (d), may wish to attract examples that a hexagon and also octagon (both sharingthe very same perimeter together the various other figures). The main idea is that as the variety of sides that the polygon increase, the polygon looks more and much more like the circle and also in certain its area is acquiring closer come the area that the circle.
Note the the last part of component (e), questioning students come think about the regular pentagon, is draft to assist them watch that as the variety of sides of theregular polygon increases, the polygon looks much more and much more like a circle. Theresults the (a), (b), (c), and also (d) additionally might result in the conjecture that as the number of sides of the polygon (with resolved perimeter) grows, the areaincreases. While students space not in a place to verify this conjecture, that is good for them to think about the problem. The teacher may additionally wish come share an exciting historical context of this problem which is described here:
Finally, effort is made in the equipment to develop numerical feeling in compare the size of the locations found in components (a), (b), (c), and also (d). If the teacher is interested in stressing this aspect of the task, climate it would certainly be ideal toinstruct the students no to usage a calculator when doing part (e) of the task.If this aspect is no important, then the answers to (a), (b), (c), and also (d)can merely be contrasted with the assist of a calculator.
In stimulate to usage the formula, Area = $frac12 imes$ base $ imes$ elevation for atriangle, we need to select a base and also then recognize the elevation of the triangle withthis base. Labelling the vertices that the triangle $A$, $B$, and $C$, allow $M$ bethe suggest where the perpendicular line from $A$ meets $overlineBC$:
Note that $|AB| = |AC| = |BC| = frac13$ because the perimeter the triangle $ABC$is $1$ and all 3 sides space congruent. Since$overleftrightarrowAM$ is perpendicular to $overlineBC$ us can use the Pythagorean theorem to the right triangles $AMC$ and $AMB$:
We recognize that $|AC| = |BC| = frac13$ therefore these 2 equations present that $|CM| = |BM|$. Because $|BC| = frac13$ this means that both $|AM|$and $|BM|$are $frac16$. We nowapply the Pythagorean to organize again to triangle $AMC$ to uncover $|AM|$:
So $|AM|^2 = frac336$ and$$|AM| = fracsqrt36.$$Now us have uncovered the elevation of triangle $ABC$, making use of $overlineBC$ together base, and also so the area is$$frac12 imes frac13 imes fracsqrt36 = fracsqrt336.$$The square through perimeter one unit has 4 congruent sides and also since their lengthsadd approximately one this means each next is $frac14$ unit in length. The areaof the square is length times height so this is $frac14 imes frac14 = frac116.$
A continuous hexagon with perimeter $1$ have the right to be split into $6$ it is intended triangles, each having side size $frac16$ together pictured below.
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We have discovered in component (a) the area that an equilateral triangle whose side length was $frac13$ unit. We have the right to repeat the debate from part (a) to discover the area of each of thesetriangles and will uncover that it will be $frac14$ the the prize from part (a) together the base and also height the the equilateral triangle from part (a) have both been scaled through a variable of $frac12$. So the area of the hexagon will be $frac64$ times the area that the it is provided triangle in component (a):$$ extArea(Hexagon) = frac64 imes fracsqrt336 =fracsqrt324.$$The circumference of the circle is $pi imes$ diameter therefore if this is $1$ unitthen the diameter should be $frac1pi$ units. This means that the radiusof the one is $frac12pi$ units. The area that the one is $pi imes m radius^2$ therefore this is$$pi imes left( frac12pi ight)^2.$$Multiplying this out and simplifying offers $frac14pi$.Since $sqrt3 lt 2$ us have$$fracsqrt336 lt frac236 = frac118.$$We know $frac118 lt frac116$ for this reason the area of the triangle is lessthan the area the the square.To to compare $frac116$ through $fracsqrt324$ keep in mind that if we convert to a denominator the $48$ we have actually $frac116 = frac348$and $fracsqrt324 = frac2sqrt348$. We know that $sqrt3 gt frac32$ together $left(frac32 ight)^2 = frac94$. So$frac116 lt fracsqrt324$.Lastly to to compare $fracsqrt324$ come $frac14pi$we could use the reality that $sqrt3 lt 1.8$ so $fracsqrt324 ltfrac1.824 = 0.075.$ meanwhile $4pi lt 13$ and so$frac14pi gt frac113 gt 0.076$.So$$fracsqrt324 lt 0.075 lt 0.076 lt frac113 lt frac14pi.$$The triangle through perimeter one unit has actually the smallest area, complied with by the square, climate the hexagon andfinally the circle. It is reasonable to conjecture, based on this evidence,that as the variety of sides that the constant polygon through perimeter one unit increases so go the area. This would mean that the constant octagon would certainly have higher area 보다 the regular hexagon but less 보다 the circle.
Below are images of an equilateral triangle, a square, a continuous hexagon, and also a circle every havingthe same perimeter: