Diagonal of Parallelogram Formula

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other.

Let’s consider the following parallelogram whose sides are known, say a and a. To find the length of one of its diagonals, we can use the cosine rule for triangles.

What is a Diagonal?

When students learn about quadrilaterals they get to learn about their various types and names. Quadrilaterals or polygons are geometric figures on a plane with 4 sides, 4 vertices, and 4 angles. Depending upon the length of the sides and the degree of the angle they form at the joining vertices they have been classified and named differently. The different types of quadrilaterals are as follows :

• Square: Having equal sides and angle of 90 degrees it is the most common type and simple quadrilateral that we are all familiar with.

• Rectangle: It is similar to square for having equal all 5 angles of 90 degrees but differs due to unequal sides. The opposite sides are equal but adjacent sides are unequal.

• Irregular Quadrilateral: It is the most irregular form with unequal sides and angles.

• Parallelogram: In a parallelogram the adjacent sides and angles are unequal but opposite sides and angles are equal to each other.

• Rhombus: Two sides adjacent to each other are equal in pairs. And only two angles opposite to each other are equal.

• Trapezoid or Trapezium: This is a quadrilateral with one pair of sides that are opposite to each other and vary in length and run parallel to each other. 

• Isosceles Trapezium: It is a special type of Trapezium with the non-parallel sides having an equal length.

These are some of the simple quadrilaterals we have studied in the previous class. 

Every quadrilateral has two diagonals that bisect each other at the center of a quadrilateral. Likewise, the diagonals of a parallelogram bisect each other but are not equal in length. Students can easily find out the diagonals of a parallelogram or any other quadrilateral by joining the opposite vertices. In square and rhombus, they always intersect perpendicular to each other while for others it is not perpendicular.

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In \[\Delta\] ADC, using the cosine rule, we calculate the length of AC as follows.

\[A{C^2} = A{D^2} + D{C^2} - 2 \times AD \times DC \times \cos \,\angle ADC \Rightarrow {p^2} = {a^2} + {b^2} - 2ab\cos \,\angle ADC\]

Similarly in \[\Delta\] \[ADS \] we have the following equation.

\[D{B^2} = A{D^2} + A{B^2} - 2 \times AD \times DB \times \cos \,\angle DAB \Rightarrow {q^2} = {a^2} + {b^2} - 2ab\cos \,\angle DAB\]

We know that adjacent angles in a parallelogram are supplementary. So we have the following equations.

\[\cos \angle DAB = \cos \pi - \angle ADC = - \cos \angle ADC\]

So we have the following property.

\[{p^2} + {q^2} = {a^2} - {b^2} - 2ab\cos \angle ADC + {a^2} + {b^2} + 2ab\cos \angle ADC\]

\[{p^2} + {q^2} = 2\left( {{a^2} + {b^2}} \right)\]

Therefore, to calculate the diagonal lengths individually, we can use the following formulas.

p = \[\sqrt {{a^2} + {b^2} - 2ab\cos \theta }\]

q = \[\sqrt {2\left( {{a^2} + {b^2}} \right) - {p^2}}\] 

Also note that the longer diagonal is opposite to the larger of the parallelogram’s angles, which is a direct result of the cosine rule.

In the case of a rectangle, which is a type of parallelogram in which the interior angles are 90°, the diagonal formula reduces to the following. The following result can also be obtained by applying the Pythagorean theorem to the rectangle.

\[{p^2} = {q^2} = {l^2} + {b^2}\]

p = q = \[\sqrt {{l^2} + {b^2}}\] 

Let’s look at the following example to understand how to use these formulas.

Question:

If one of the angles of a parallelogram is 60° and its adjacent sides are 4 cm and 6 cm long, then evaluate the lengths of its diagonals.

Solution:

\[a = 4\,cm,\,\,b = 6\,cm,\,\,\theta = {60^ \circ }\] 

p = \[\sqrt {{a^2} + {b^2} - 2ab\cos \angle \theta }\]

= \[\sqrt {{4^2} + {6^2} - 2 \times 4 \times 6 \times \cos {{60}^ \circ }}\]

= \[\sqrt {16 + 52 - 48 \times \frac{1}{2}} = \sqrt {28} = 2\sqrt 7 \],cm  

q = \[\sqrt {2\left( {{a^2} + {b^2}} \right)}\]

= \[\sqrt {2 \times \left( {{4^2} + {6^2}} \right) - {{\left( {2\sqrt 7 } \right)}^2}}\]

= \[\sqrt {104 - 28} = \sqrt {76} = 2\sqrt {19} \] ,cm

Why don’t you try to solve the following problem to see if you have mastered these formulas?

Question:

One of the angles of a parallelogram is 135° and its diagonals are \[\$3\,cm\$ and \$3\sqrt 5 \,cm\$\] respectively. Find the perimeter of the parallelogram.

Options:

(a) 4\[\left( {3 + \sqrt 2 } \right)\]cm 

(b) 3\[\left( {2 + \sqrt 2 } \right)\]cm

(c) 12cm

(d) 6\[\left( {1 + \sqrt 2 } \right)\]cm

Answer: (d)

Solution:

\[p = 3\,cm,\,\,q = 3\sqrt {5\,} \,cm,\,\,\theta = {135^ \circ }\]

\[{p^2} = {a^2} + {b^2} - 2ab\cos \theta\]

\[\Rightarrow {\left( {3\sqrt 5 } \right)^2} = {a^2} + {b^2} - 2ab\cos {135^ \circ }\]

\[\Rightarrow {a^2} + {b^2} - 2ab\left( { - \frac{1}{{\sqrt 2 }}} \right) = 45 \]

\[\Rightarrow {a^2} + {b^2} + \sqrt 2 ab = 45\]

\[{p^2} + {q^2} = 2\left( {{a^2} + {b^2}} \right)\]

\[\Rightarrow {3^2} + {\left( {3\sqrt 5 } \right)^2} = 2\left( {{a^2} + {b^2}} \right)\]

\[\Rightarrow {a^2} + {b^2} = \frac{{9 + 45}}{2} = 27\] 

Now, we can solve these two equations to calculate a and b.

27 + \[\sqrt 2 ab\] = 45 

\[\Rightarrow\] \[\sqrt 2 ab\ = 18 \Rightarrow\] ab = 9\[\sqrt 2\] \[\Rightarrow\] b = \[\frac{{9\sqrt 2 }}{a}\] 

Substituting in one of the equations, we get the following.

\[{a^2} + {\left( {\frac{{9\sqrt 2 }}{a}} \right)^2} = 27\]

\[\Rightarrow {a^2} + \frac{{162}}{{{a^2}}} = 27\]

\[\Rightarrow {a^4} - 27{a^2} + 162 = 0\]

Solving this quadratic and ignoring the negative answers (because lengths cannot be negative), we get the following values. 

a = 3cm,3\[\sqrt 2\]

For the corresponding values of a, we get values of b as follows.

b = 3\[\sqrt 2\] cm,3cm

Therefore, the dimensions of the parallelogram are \[3,cm and 3\sqrt 2 ,cm\]. Hence its perimeter is calculated as follows.

\[{\text{perimeter}} = 2\left( {a + b} \right) = 2\left( {3 + 3\sqrt 2 } \right) = 6\left( {1 + \sqrt 2 } \right)\,cm\]