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.

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

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 $\Delta ADB$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.

$\begin{gathered}

Â Â p = \sqrt {{a^2} + {b^2} - 2ab\cos \theta } \hfill \\

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

\end{gathered} $

Also note that the longer diagonal is opposite to the larger of the parallelogramâ€™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*90Â°*, the diagonal formula reduces to the following. The following result can also be obtained by applying the Pythagorean theorem to the rectangle.

$\begin{gathered}

Â Â {p^2} = {q^2} = {l^2} + {b^2} \hfill \\

Â Â p = q = \sqrt {{l^2} + {b^2}} \hfill \\

\end{gathered} $

Letâ€™s look at the following example to understand how to use these formulas.

**Question:**

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

**Solution****:**

$a = 4\,cm,\,\,b = 6\,cm,\,\,\theta = {60^ \circ }$** **

$\begin{gathered}

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

Â Â \,\,\,\,\, = \sqrt {{4^2} + {6^2} - 2 \times 4 \times 6 \times \cos {{60}^ \circ }} \hfill \\

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

\end{gathered} $** **

$\begin{gathered}

Â Â q = \sqrt {2\left( {{a^2} + {b^2}} \right)} \hfill \\

Â Â \,\,\,\, = \sqrt {2 \times \left( {{4^2} + {6^2}} \right) - {{\left( {2\sqrt 7 } \right)}^2}} \hfill \\

Â Â \,\,\,\, = \sqrt {104 - 28} = \sqrt {76} = 2\sqrt {19} \,cm \hfill \\

\end{gathered} $** **

Why donâ€™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*135Â°* 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) $12\,cm$

(d) $6\left( {1 + \sqrt 2 } \right)\,cm$

**Ans****wer****:**** **(d)

**Solution:**

$p = 3\,cm,\,\,q = 3\sqrt {5\,} \,cm,\,\,\theta = {135^ \circ }$** **

$\begin{gathered}

Â Â \,\,\,\,\,{p^2} = {a^2} + {b^2} - 2ab\cos \theta \hfill \\

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

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

Â Â Â \Rightarrow {a^2} + {b^2} + \sqrt 2 ab = 45 \hfill \\

\end{gathered} $** **

$\begin{gathered}

Â Â \,\,\,\,\,{p^2} + {q^2} = 2\left( {{a^2} + {b^2}} \right) \hfill \\

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

Â Â Â \Rightarrow {a^2} + {b^2} = \frac{{9 + 45}}{2} = 27 \hfill \\

\end{gathered} $** **

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

$\begin{gathered}

Â Â 27 + \sqrt 2 ab = 45 \hfill \\

Â Â Â \Rightarrow \sqrt 2 ab = 18 \Rightarrow ab = 9\sqrt 2 \Rightarrow b = \frac{{9\sqrt 2 }}{a} \hfill \\

\end{gathered} $

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

$\begin{gathered}

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

Â Â Â \Rightarrow {a^2} + \frac{{162}}{{{a^2}}} = 27 \hfill \\

Â Â Â \Rightarrow {a^4} - 27{a^2} + 162 = 0 \hfill \\

\end{gathered} $

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

$a = 3\,cm,3\sqrt 2 \,cm$

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

$b = 3\,\sqrt 2 \,cm,3\,cm$

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

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

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

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 $\Delta ADB$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.

$\begin{gathered}

Â Â p = \sqrt {{a^2} + {b^2} - 2ab\cos \theta } \hfill \\

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

\end{gathered} $

Also note that the longer diagonal is opposite to the larger of the parallelogramâ€™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

$\begin{gathered}

Â Â {p^2} = {q^2} = {l^2} + {b^2} \hfill \\

Â Â p = q = \sqrt {{l^2} + {b^2}} \hfill \\

\end{gathered} $

Letâ€™s look at the following example to understand how to use these formulas.

If one of the angles of a parallelogram is

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

$\begin{gathered}

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

Â Â \,\,\,\,\, = \sqrt {{4^2} + {6^2} - 2 \times 4 \times 6 \times \cos {{60}^ \circ }} \hfill \\

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

\end{gathered} $

$\begin{gathered}

Â Â q = \sqrt {2\left( {{a^2} + {b^2}} \right)} \hfill \\

Â Â \,\,\,\, = \sqrt {2 \times \left( {{4^2} + {6^2}} \right) - {{\left( {2\sqrt 7 } \right)}^2}} \hfill \\

Â Â \,\,\,\, = \sqrt {104 - 28} = \sqrt {76} = 2\sqrt {19} \,cm \hfill \\

\end{gathered} $

Why donâ€™t you try to solve the following problem to see if you have mastered these formulas?

One of the angles of a parallelogram is

(a) $4\left( {3 + \sqrt 2 } \right)\,cm$

(b) $3\left( {2 + \sqrt 2 } \right)\,cm$

(c) $12\,cm$

(d) $6\left( {1 + \sqrt 2 } \right)\,cm$

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

$\begin{gathered}

Â Â \,\,\,\,\,{p^2} = {a^2} + {b^2} - 2ab\cos \theta \hfill \\

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

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

Â Â Â \Rightarrow {a^2} + {b^2} + \sqrt 2 ab = 45 \hfill \\

\end{gathered} $

$\begin{gathered}

Â Â \,\,\,\,\,{p^2} + {q^2} = 2\left( {{a^2} + {b^2}} \right) \hfill \\

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

Â Â Â \Rightarrow {a^2} + {b^2} = \frac{{9 + 45}}{2} = 27 \hfill \\

\end{gathered} $

Now, we can solve these two equations to calculate

$\begin{gathered}

Â Â 27 + \sqrt 2 ab = 45 \hfill \\

Â Â Â \Rightarrow \sqrt 2 ab = 18 \Rightarrow ab = 9\sqrt 2 \Rightarrow b = \frac{{9\sqrt 2 }}{a} \hfill \\

\end{gathered} $

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

$\begin{gathered}

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

Â Â Â \Rightarrow {a^2} + \frac{{162}}{{{a^2}}} = 27 \hfill \\

Â Â Â \Rightarrow {a^4} - 27{a^2} + 162 = 0 \hfill \\

\end{gathered} $

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

$a = 3\,cm,3\sqrt 2 \,cm$

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

$b = 3\,\sqrt 2 \,cm,3\,cm$

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

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