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

## Question

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.

**Reasoning:**

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 \(AB = \rm{}40 \)\(\rm\,cm\) 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.

**Steps:**

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

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

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

Let breadth of rectangle be \(x\).

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

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

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

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

Now, we know that

Perimeter of rectangle

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

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.