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Calculate the Area of a Rectangle Given the Diagonal

Updated on January 25, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

How to Find the Area of a Rectangle Using the Length of the Diagonal

The diagonal D of a rectangle whose width and length are W and L is given by the formula D = sqrt(W^2 + L^2), which is just another application of the Pythagorean Theorem. If you are given only the diagonal length of a rectangle, there is no unique solution for either the area or the dimensions of the rectangle. In order to calculate the area of a rectangle given the diagonal, you need one other measurement, which can be

  • one side length
  • perimeter
  • aspect ratio (ratio of length to width)

In order to find the dimensions of a rectangle given the diagonal, you also need one other measurement, which can be one of the three above or the area. Here are several examples of how to find the size of a rectangle given the diagonal length and another quantity.

Rectangular Area from Diagonal and Width or Length

This is the easiest case to solve. If you know the diagonal length of the rectangle D and one of the two side lengths L, then you can solve for the other length W with the formula W = sqrt(D^2 - L^2). This is because D^2 = L^2 + W^2, and solving for W gives you W = sqrt(D^2 - L^2).

For example, suppose a rectangle has a diagonal length of 78.13 cm and one side is 29.55 cm long. Then the length of the other side is computed as

sqrt(78.13^2 - 29.55^2)
= sqrt(6104.2969 - 873.2025)
= sqrt(5231.0944)
≈ 72.33 cm.

The area of the rectangle is 29.55 cm times 72.33 cm, which is approximately 2137.24 cm^2, or equivalently 0.213724 square meters.


Rectangular Area from Diagonal and Perimeter

The perimeter of a rectangle is twice the sum of the width and length. Using the equations D = sqrt(W^2 + L^2) and P = 2(W + L), you could solve for W and L in terms of D and P, but you could also bypass finding the dimensions and solve directly for the area WL instead. The first step is the square the expressions for D and P/2. This gives you

D^2 = W^2 + L^2
(P/2)^2 = W^2 + 2WL + L^2

If you subtract D^2 from (P/2)^2, you get

(P/2)^2 - D^2
= W^2 + L^2 - (W^2 + 2WL + L^2)
= 2WL

In other words, (P/2)^2 - D^2 is equal to twice the area. Therefore,

Area = [(P/2)^2 - D^2]/2

Example: If a rectangle has a diagonal of 20 inches and a perimeter of 56 inches, what is its area? Using D = 20 and P = 56, we have

Area = [(P/2)^2 - D^2]/2
= [(56/2)^2 - 20^2]/2
= (28^2 - 20^2)/2
= (784 - 400)/2
= 384/2
= 192 square inches.


Rectangular Area from Diagonal and Side Ratio

The aspect ratio R of a rectangle is given by the equation R = L/W. Solving this equation for L gives you L = RW. If you plug this into the diagonal equation D = sqrt(L^2 + W^2), you get

D = sqrt((RW)^2 + W^2)
D^2 = (R^2)(W^2) + W^2
D^2 = (R^2 + 1)(W^2)
(D^2)/(R^2 + 1) = W^2
W = sqrt[(D^2)/(R^2 + 1)]

and since L = RW, you also get

L = R*sqrt[(D^2)/(R^2 + 1)]

Area = WL = R(D^2)/(R^2 + 1)

In other words, knowing the aspect ratio and the diagonal of a rectangle allows you to find the width, which in turn gives you the length, which in turn gives you the area.

Example: A rectangle has a diagonal of 20 inches and an aspect ratio of 4/3. To find its area we use D = 20 and R = 4/3. This gives us

Area = R(D^2)/(R^2 + 1)
= (4/3)(20^2)/[(4/3)^2 + 1]
= (4/3)(400)/[25/9]
= (4*400*9)/(3*25)
= 192 square inches.


Rectangular Dimensions from Diagonal and Area

Related to the rectangular area problem is finding the dimensions of a rectangle given its diagonal and area. If we call the area A, we have the equations

A = WL
D = sqrt(W^2 + L^2)

Let's look at the quantities D^2 + 2A and D^2 - 2A. These are

D^2 + 2A
= W^2 + L^2 + 2WL
= (L + W)^2

D^2 - 2A
= W^2 + L^2 - 2WL
= (L - W)^2

Therefore,

sqrt(D^2 + 2A) = L + W
sqrt(D^2 - 2A) = L - W

From these two equations it is easy to recover L and W. Adding and dividing by 2 gives us L, while subtracting and dividing by 2 gives W.

[sqrt(D^2 + 2A) + sqrt(D^2 - 2A)]/2 = L
[sqrt(D^2 + 2A) - sqrt(D^2 - 2A)]/2 = W

Example: A rectangle has a diagonal length of 20 inches and an area of 192 square inches. Using D = 20 and A = 192, we have

L = [sqrt(D^2 + 2A) + sqrt(D^2 - 2A)]/2
= [sqrt(400 + 384) + sqrt(400 - 384)]/2
= [sqrt(728) + sqrt(16)]/2
= [28 + 4]/2
= 16 inches

W = [sqrt(D^2 + 2A) - sqrt(D^2 - 2A)]/2
= [sqrt(400 + 384) - sqrt(400 - 384)]/2
= [sqrt(728) - sqrt(16)]/2
= [28 - 4]/2
= 12 inches

So a rectangle with dimensions 16 by 12 has a diagonal of 20 and an area of 192.


Area and Perimeter Bounds Given Diagonal

If you only know the diagonal length, you can't determine the perimeter and area of the rectangle, but you can determine the bounds on the values. If a rectangle has a diagonal length of D, then the most extreme cases are when the rectangle is so long and skinny that it resembles a line of length D, and when the rectangle is a square with diagonal D.

For a rectangle that is almost a line, the limiting dimensions are 0 by D, which produces a rectangle whose area is 0, and whose perimeter is 2D.

For a rectangle that is a square, the side lengths are D/sqrt(2), which produces a rectangle whose area is (1/2)D^2, and whose perimeter is 2*sqrt(2)*D.

Therefore, if you know the diagonal length D, you can obtain the following bounds on the area and perimeter:

0 < Area ≤ (1/2)D^2

2D < Perimeter ≤ 2*sqrt(2)*D

In other words, the area of a rectangle is no more than half the square of the diagonal length; the perimeter is at least twice the diagonal length, but no more than 2*sqrt(2) ≈ 2.828 times the diagonal.

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      :) 2 years ago

      Cool fact about the ratio of a diagonal to the perimeter, I did not know it was bounded between 2 and 2.8. I was attempting to solve a rectangle problem were the perimeter was 3 times the diagonal, so I set it up like 3sqrt(x^2+y^2) = 2x+2y, which gave me 5(x^2+y^2)=8xy but I couldn't find any values of x and y to make it work with guess and check. Now I know that it can't be done.

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      Author

      TR Smith 2 years ago

      Glad to have helped. If you solve the equation 5(x^2 + y^2) = 8xy for x with the quadratic formula, you find x is a complex multiple of y

      x = [8y +/- sqrt(64y^2 - 100y^2)]/10

      = y(4 +/- 3i)/5

      so this problem has no real-valued solutions.

    • profile image

       24 months ago

      Can you help with this geometry problem, find the dimensions of a rectangular box with a diagonal of 31 and a surface area of 720 if the length width and height are whole numbers.

    • calculus-geometry profile image
      Author

      TR Smith 24 months ago

      Thanks for the interesting question, Anon. Here's one way to approach the problem. You have the equations

      a^2 + b^2 + c^2 = 961

      2ab + 2ac + 2ab = 720

      If you add them together you get

      a^2 + b^2 + c^2 + 2ab + 2ac + 2ab = 1681

      (a + b + c)^2 = 41^2

      a + b + c = 41

      which tells you that the side lengths all add up to 41, a very helpful fact to know when testing possible solution sets.

      Since 961 is of the form 4k+1 and all square numbers are of the form 4k or 4k+1, you can deduce that one of a, b, and c is odd and the other two are even. And since even square numbers end in {0, 4, 6} and odd square numbers end in {1, 5, 9} you can deduce that the ending digits of a^2, b^2 and c^2 are either {0, 6, 5} or {4, 6, 1}. This tells you that one of the squares is either 16, 36, 196, 256, 576, or 676.

      With this information you can whittle down the possibilities enough to do guess-and-check efficiently. With trial and error you will find that the solution is {5, 6, 30} for the edge lengths.

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