Calculate the Area of a Rectangle Given the Diagonal
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
- 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)
≈ 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)
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
= 192 square inches.
Related Geometry Help
Rectangle Diagonal, Area & Perimeter Calculator: Input any two measurements to find the third.
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]
= 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
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.