Question

Show that the area of the rhombus is half the product of the lengths of its diagonals.

Answer 1

Hint: In this question, we need to prove that the area of the rhombus is half the product of the lengths of its diagonals. For this, we will use the geometrical, mathematical formulae along with the properties of the rhombus.

Complete step-by-step answer:
Let ABCD be a rhombus with AC and BD as the diagonals with the length $d_2$ and $d_1$

As the diagonals of the rhombus bisect each other at 90 degrees. So, the following figure has been generated.

From the above figure, we can write that the area of the rhombus ABCD has been divided into the area of the triangles ABC and ADC.
Now, the triangles ABC and ADC are two right-angled triangles with BO perpendicular to AC and DO perpendicular to AC.
Half of the product of the base and the height of a triangle is the area of the triangle. Mathematically, $A={\displaystyle \frac{1}{2}}\times b\times h$ .
So, the area of the triangle ABC is given as
 $$\begin{gathered} \Rightarrow A_{ABC}={\displaystyle \frac{1}{2}}\times b\times h \\ ={\displaystyle \frac{1}{2}}\times AC\times BO----(i) \end{gathered}$$
Similarly, the area of the triangle ADC is given as
 $$\begin{gathered} \Rightarrow A_{ADC}={\displaystyle \frac{1}{2}}\times b\times h \\ ={\displaystyle \frac{1}{2}}\times AC\times DO----(ii) \end{gathered}$$
As, the sum of the areas of the triangles ABC and ADC is equal to the area of the rhombus ABCD. Mathematically, $A_{rhombus}=A_{ABC}+A_{ADC}$ .
Substituting the values of the areas of the triangles in the equation $A_{rhombus}=A_{ABC}+A_{ADC}$ to determine the area of the rhombus.
 $$\begin{gathered} \Rightarrow A_{rhombus}=A_{ABC}+A_{ADC} \\ ={\displaystyle \frac{1}{2}}\times AC\times BO+{\displaystyle \frac{1}{2}}\times AC\times DO \\ ={\displaystyle \frac{1}{2}}\times AC\left(BO+DO\right)----(iii) \end{gathered}$$
From the figure, we can see that the sum of BO and DO is equalled to the length of the diagonal BD. So, substituting (BO+DO=BD) in the equation (iii), we get
 $$\begin{gathered} \Rightarrow A_{rhombus}={\displaystyle \frac{1}{2}}\times AC\left(BO+DO\right) \\ ={\displaystyle \frac{1}{2}}\times AC\times BD----(iv) \end{gathered}$$
Also, from the figure, we can see that the length of the diagonal AC is $d_1$ and the diagonal BD is $d_2$ . So, substituting the values in the equation (iv), we get
 $$\begin{gathered} \Rightarrow A_{rhombus}={\displaystyle \frac{1}{2}}\times AC\times BD \\ ={\displaystyle \frac{1}{2}}\times d_1\times d_2 \end{gathered}$$
Hence, we can say that the area of the rhombus is half of the product of the length of the diagonals.

Note: Students must note that the length of the diagonals in a rhombus are different and they bisect each other at 90 degrees. Moreover, rhombus differs from the square in one thing only that the length of the diagonals in the square are the same.