ArtsAutosBooksBusinessEducationEntertainmentFamilyFashionFoodGamesGenderHealthHolidaysHomeHubPagesPersonal FinancePetsPoliticsReligionSportsTechnologyTravel

How to Solve "Area = Perimeter" Geometry Problems

Updated on March 21, 2017
calculus-geometry profile image

TR Smith is a product designer and former teacher who uses math in her work every day.

Two-dimensional geometric shapes that have equal area and perimeter are called "equable" shapes. For example, in the diagram above, the L-shape has an area of 26 square units and a perimeter of 26 linear units. In fact, if you choose the scale of measurement correctly, any shape's area and perimeter can be made equal. In other words, being equable is dependent on the unit of measurement. A shape that is equable in centimeters and square centimeters will not be equable if you measure the perimeter and area in terms of meters and square meters.

A classic problem in geometry and algebra is to find the dimensions of a particular shape so that the area and perimeter are equal. These problems may have infinitely many solutions, a finite number of correct solutions, or only one unique solution depending on how many free variables there are, and whether the solutions are restricted to the integers. Here are several worked out examples to help you solve these kinds of problems.

(1) Rectangle with Integer Sides

How many rectangles are there with integer side lengths such that the perimeter and area are equal?

Solution: Let x be the width of the rectangle and let y be the length. We want to find two integers x and y such that the area xy equals the perimeter 2x + 2y. Algebraically, this is

xy = 2x + 2y

Since this is one equation in two variables, there may be more than one solution. Solving this equation for y in terms of x gives us

y = 2x/(x - 2) = 2 + 4/(x - 2)

While there are infinitely many non-integer solutions, for example x = 2.5 and y = 10, there are only a finite number of integer solutions. To find them, we need to find values of x so that x - 2 divides evenly into 4. There are three such values of x: 6, 4, and 3. These correspond to the solutions

x = 6, y = 3
x = 4, y = 4
x = 3, y = 6

The first and third are the same rectangle, just oriented differently. Thus, there are only two solution sets. A 3-by-6 rectangle with an area and perimeter of 18, and a 4-by-4 square with an area and perimeter of 16.

(2) Circle

What is the radius of a circle whose area equals its circumference?

Solution: The two geometric equations we need to solve this equable circle problem are

  • area = pi*r^2
  • circumference = 2*pi*r

This gives us

pi*r^2 = 2*pi*r
r^2 = 2r
r = 2

Therefore, a circle with a radius of 2 units has an area equal to its circumference, 4*pi.

(3) Integer Right Triangles

What are the side lengths of an integer right triangle whose perimeter equals its area?

Solution: To find an integer right triangle with equal area and perimeter, we need to find a Pythagorean triple (A, B, C) such that

A + B + C = (1/2)AB

where C = sqrt(A^2 + B^2). Replacing C with its equivalent expression gives us the equation

A + B + sqrt(A^2 + B^2) = (1/2)AB
2*sqrt(A^2 + B^2) = AB - 2A - 2B

Squaring both sides of this equation gives us

4A^2 + 4B^2 = (AB)^2 - 4BA^2 - 4AB^2 + 8AB + 4A^2 + 4B^2
0 = (AB)^2 - 4BA^2 - 4AB^2 + 8AB
0 = AB - 4A - 4B + 8
8 = AB - 4A - 4B + 16
8 = (A-4)(B-4)

Since we are looking for integer solutions, the only possibilities are

  • A - 4 = 8 and B - 4 = 1
  • A - 4 = 4 and B - 4 = 2

This gives the solutions (A, B) = (12, 5) and (8, 6). The Pythagorean triples are (12, 5, 13) and (8, 6, 10). These correspond to a right triangle with an area and perimeter of 30, and a right triangle with an area and perimeter of 24. As in the integer rectangle problem, this one has exactly two distinct solutions.

(4) Equilateral Triangle

Find the side length of an equilateral triangle whose area equals its perimeter.

Solution: Let the side of an equilateral triangle be x. The area of the triangle is [sqrt(3)/4]x^2 and the perimeter is 3x. If the area equals the perimeter, then we have

[sqrt(3)/4]x^2 = 3x

Solving for x gives us

[sqrt(3)/4]x = 3
x = 3/[sqrt(3)/4]
x = 12/sqrt(3)
x = 4*sqrt(3)
x ≈ 6.9282

(5) Heart Shape Made with a Square and Two Semicircles

Imagine a heart constructed by affixing two semicircles to adjacent sides of a square, such that the diameter of each semicircle is equal to the side length of the square. If the perimeter of the heart shape equals its area, find the dimensions of the shape.

Solution: Call the diameter of the circle and the side length of the square D. The area of the heart is the area of a full circle with diameter D plus the area of a square with side length D. In math terms,

heart area = (pi/4)D^2 + D^2

The perimeter of the heart is the circumference of the full circle plus two sides of the square. In math terms,

heart perimeter = pi*D + 2D

Setting these two expressions equal to each other allows us to solve for D.

(pi/4)D^2 + D^2 = pi*D + 2D
[pi/4 + 1]D^2 = [pi + 2]D
[pi/4 + 1]D = pi + 2
D = [pi + 2]/[pi/4 + 1]
D = [4*pi + 8]/[pi + 4]
D ≈ 2.8798

(6) Equable Circular Sector

Suppose a circular sector has a radius of 7 cm. What must the angle of the sector be (in degrees) so that its area in square inches equals its perimeter in linear inches?

Solution: A circular sector with an angle of θ degrees and a radius of r has an area of (θ/360)*pi*r^2. The perimeter of the circular sector is (θ/180)*pi*r + 2r. Since r = 7 cm in this problem, we must solve the equation

(θ/360)*pi*49 = (θ/180)*pi*7 + 14

This gives us

θ[49pi/360 - 7pi/180] = 14
θ = 14/[49pi/360 - 7pi/180] degrees
θ = 144/pi degrees
θ ≈ 45.84 degrees


    0 of 8192 characters used
    Post Comment

    • profile image

      Elana 10 days ago

      Hello, I am trying to solve a problem similar to your Pythagorean triangle example:

      Find all integer right triangles whose area is 3 times its perimeter.

      I'm getting stuck because I found two solutions, but I don't know if there are more or how I would know that I've found them all even if I could find more.

    • calculus-geometry profile image

      TR Smith 9 days ago from Eastern Europe

      Call the legs of the triangle x and y. The equations that relates perimeter and area for this problem is

      3(x + y + sqrt(x^2 + y^2)) = (1/2)xy

      Simplifying it gives you

      xy - 6x - 6y = 6*sqrt(x^2 + y^2)

      (xy - 6x - 6y)^2 = 36x^2 + 36y^2

      (xy)^2 -12xy^2 - 12yx^2 +72xy = 0

      xy(xy - 12y - 12x + 72) = 0

      You can divide both sides by the factor xy because neither x nor y can be equal to 0 in this problem. This gives you

      xy - 12y - 12x + 72 = 0

      (x - 12)(y - 12) - 72 = 0

      (x - 12)(y - 12) = 72

      This problem reduces to finding integers pairs that multiply to 72. You only need to find all the factor pairs of 72 and set the members of each pair equal to x - 12 and y - 12, then solve for x and y. The solutions that come out are

      x = 13, y = 84

      x = 14, y = 48

      x = 15, y = 36

      x = 16, y = 30

      x = 18, y = 24

      x = 20, y = 21

      There are exactly six solutions to this problem. Similar problems are solved with the same method. For example, if the problem is to find all Pythagorean triangles where the area is 10 times the perimeter, when you work out the algebra it reduces to finding factor pairs of 800.

    Click to Rate This Article