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

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## 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?

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 $$\rm{}AB = \rm{}40$$$$\rm\,cm$$ 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.

Steps:

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.

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