Ex.6.5 Q8 The Triangle and Its Properties - NCERT Maths Class 7

Go back to  'Ex.6.5'


The diagonals of a rhombus measure \(\rm{}16\,cm\) and \(\rm{}30\, cm.\) Find its perimeter.

 Video Solution
Triangle & Its Properties
Ex 6.5 | Question 8

Text Solution

What is  known?

Diagonals of rhombus are given as \(\rm{}16\,cm\) and \(\rm{}30\,cm.\)

What is unknown?

Perimeter of rhombus.


This question is also based on the two concepts i.e. the concept of rhombus  and Pythagoras theorem. For better understanding of this question understand it with the help of a figure.

As it is mentioned in the question suppose there is a rhombus \(ABCD\) and length of whose diagonals are given as \(16 \rm \,cm\) and \(\rm{}30\,cm.\). As, we know the diagonals of a rhombus bisect each other at \(90^\circ\) i.e. at \(O\). Now, \(O\) divides \(DB\) and \(AC\) into two equal parts i.e. \(OD\) is half of \(DB\) and \(OC\) is half \(AC.\) Now, you can apply Pythagoras theorem in triangle \(DOC\) and get the measure of side \(DC.\) Now, the side of rhombus is known, you can easily find the perimeter of rhombus.


Given, \(AC\) and \(BD\) are the two diagonals of the rhombus.

\(AC = \rm{}30\,cm,\) \(BD =\rm{}16\,cm\)

Since, the diagonal of a rhombus bisect each other at \(90^\circ.\)


\[\begin{align}OD &= \frac{{DB}}{2} = \frac{{16}}{2} = 8\;\rm{cm}\\OC &= \frac{{AC}}{2} = \frac{{30}}{2} = 15\;\rm{cm} \end{align}\]

Now, in right angled triangle\(\Delta \,DOC,\)

\(\begin{align}&\left( \text{Hypotenuse} \right)^2 \\&= \rm{{ }}{{\left(\text {Perpendicular} \right)}^2} + {\rm{ }}{{\left( \text{Base} \right)}^2} \end{align} \)

\(\begin{align} {{\left( {DC} \right)}^2} &= {{\left( {OD} \right)}^2} + {{\left( {OC} \right)}^2}\\{{\left( {DC} \right)}^2} &= \rm{{ }}{{\left( 8 \right)}^2} + {\rm{ }}{{\left( {15} \right)}^2}\\{{\left( {DC} \right)}^2} &= \rm{{ }}64{\rm{ }} + {\rm{ }}225\\{{\left( {DC} \right)}^2} &= {\rm{ }}289\\DC{\rm{ }} &= \rm{{ }}17{\rm{ }}\,cm\end{align}\)

\[\begin{align}\text{Perimeter of rhombus}&=\rm{}4\,\times \,{side}\\&=\rm{} 4 \,\times \,17 \\&= \rm{}68\, \rm{cm}\end{align}\]

Thus, the perimeter of rhombus is \(\rm{}68\,cm.\)

Useful Tip:

Whenever you encounter problem of this kind, it is best to think of the properties related to rhombus.