Question

If the diagonals of a rhombus are 10cm and 24cm, find its perimeter.

Answer 1

Hint: Let us draw the figure as follows.

We solve this problem by using the definition of rhombus, that is, that all the sides of a rhombus are equal. We also use the condition that the diagonals intersect at right angles at mid-points. For finding the side length we use the Pythagoras Theorem, which states that the square of the hypotenuse is equal to the sum of squares of the other two sides, that is for the triangle as shown below.

${{b}^{2}}={{a}^{2}}+{{c}^{2}}$
Complete step by step answer:
We are given the length of the diagonals as 10cm and 24cm. Let us assume that from the figure the length of the diagonals as,
$\begin{align}
  & \Rightarrow AC=24cm \\
 & \Rightarrow BD=10cm \\
\end{align}$
We know that the diagonals of rhombus intersect at right angles at mid-points, that means, the point E is the midpoint of both AC and BD. By using the condition that the length of ED can be calculated as,
$\Rightarrow ED=\dfrac{BD}{2}$
By substituting the required values, we get,
$\Rightarrow ED=\dfrac{10}{2}=5cm$
Similarly by using the condition that point E is the midpoint of AC, we get,
$\Rightarrow AE=\dfrac{AC}{2}=\dfrac{24}{2}=12cm$
Now let us consider the triangle $\Delta AED$, if we apply Pythagoras theorem on it, then,
$\begin{align}
  & \Rightarrow A{{D}^{2}}=A{{E}^{2}}+A{{D}^{2}} \\
 & \Rightarrow A{{D}^{2}}={{\left( 5 \right)}^{2}}+{{\left( 12 \right)}^{2}} \\
 & \Rightarrow A{{D}^{2}}=25+144 \\
 & \Rightarrow A{{D}^{2}}=169 \\
 & \Rightarrow AD=\sqrt{169}=13cm \\
\end{align}$
So, the perimeter of this rhombus is, $4\times 13=52cm$.

Note: We can solve this problem in another method also. If p and q are diagonals of the rhombus, then the area of the rhombus is given as: $A=\dfrac{1}{2}\times p\times q$. If p is the length and a is the side length of the rhombus, then area is:
$\begin{align}
  & A=\dfrac{1}{2}\times p\times \sqrt{4{{a}^{2}}-{{p}^{2}}} \\
 & \Rightarrow \dfrac{1}{2}\times p\times q=\dfrac{1}{2}\times p\times \sqrt{4{{a}^{2}}-{{p}^{2}}} \\
 & \Rightarrow q=\sqrt{4{{a}^{2}}-{{p}^{2}}} \\
\end{align}$
Now by squaring both sides, you get, ${{q}^{2}}=4{{a}^{2}}-{{p}^{2}}$. By substituting the required values here, we get the answer.