Ex.6.5 Q7 The Triangle and Its Properties - NCERT Maths Class 7

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Find the perimeter of the rectangle whose length is \(\rm{}40\, cm\) and diagonal is \(\rm{}41\,cm\).

Text Solution

What is known?

Length of rectangle is given \(\rm{}40\, cm\) and diagonal is \(\rm{}41\,cm. \)

What is unknown?

Breadth and perimeter of rectangle.


This question is based on the two concepts i.e. the concept of rectangle 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 rectangle \(ABCD\) and whose length is given \(\rm{}40\,cm.\) As, we know the opposite sides of a rectangle are equal if \(\rm{}AB = \rm{}40 \)\(\rm\,cm\) that means side opposite to \(AB\) i.e. \(CD\) will also be \(\rm{}40\, cm.\) Now, one of the diagonals of a rectangle is given \(AC = 41\,\rm\,cm\) it divides the rectangle into two right-angled triangles. Now, you can apply Pythagoras theorem and find the third side i.e. breadth of the rectangle. Now, you have the measure of both the sides of a rectangle i.e. length and breadth, you can easily find out the perimeter of the rectangle.


Given,in rectangle \(ABCD,AB \) and \(CD\) are the lengths of the rectangle.

\(\rm{}AB =\rm{}40\,cm,\) \(\rm{}CD = \rm{}40\,cm\) and \(AC\) is the diagonal

Therefore, \(\rm{}AC = \rm{}41\,cm\)

Let breadth of rectangle be \(x\).

Now, in triangle \(ADC,\) by Pythagoras theorem

\[\begin{align}{{\left(\text {Hypotenuse} \right)}^2} &= \rm{{ }}{{\left( \text{Perpendicular} \right)}^2} + {\rm{ }}{{\left(\text {Base} \right)}^2}\\{{\left( {AC} \right)}^2} &= \rm{{ }}{{\left( {DC} \right)}^2} + {\rm{ }}{{\left( {AD} \right)}^2}\\{{\left( {41} \right)}^2} &= \rm{{ }}{{\left( {40} \right)}^2} + {\rm{ }}{{\left( x \right)}^2}\\1681 &= \rm{{ }}1600{\rm{ }} + {{\left( x \right)}^2}\\1681-1600{\rm{ }} &= \rm{{ }}{{\left( x \right)}^2}\\{x^2} &= \rm{{ }}81\\\rm{or} \,x &= \rm{{ }}9{\rm{ }}cm\end{align}\]

Therefore, breadth of rectangle \(= \rm{}9\,cm\)

Now, we know that

\[\begin{align}\text{Perimeter}{\text{ }}of{\text{ }}\text rectangle{\rm{ }} &= \rm{{ }}2\left( {l{\rm{ }} + {\rm{ }}b} \right)\\&= {\rm{ }}2{\rm{ }}\left( {40{\rm{ }} + {\rm{ }}9} \right)\\& = {\rm{ }}2\left( {49} \right)\\& = {\rm{ }}98{\rm{ }}\rm\,cm\end{align}\]

Hence, the perimeter of rectangle is \(98\rm\,cm\)

Useful Tip:

Whenever you encounter problems of this kind, remember the Pythagoras property of right -angled triangle and the formulas related to rectangle.