Question

The perimeter and the breadth of a rectangle are 82 cm and 20 cm respectively. Calculate the length of its diagonal (in cm

Answer 1

Answer :-

Here we have The perimeter and the breadth of a rectangle are 82 cm and 20 cm respectively.

Perimeter = l + b

Let l as x,

$\\ 82 = 20 + x \\ \\ x = 82 - 20 = 62 \: cm \\ \\ l = 62 \: cm$

We know that rectangle have all angles right angle.Then diagonal is became hypotenuse of rectangle.

From Pythagoras therome length of hypotenus's square is equal to sum of two other sides square.

$= \sqrt{ {(62)}^{2} + {(20)}^{2} } \\ \\ = \sqrt{3844 + 400} = \sqrt{4244} \\ \\ = 65.14 \: cm$

Answer 2

$\textbf{\underline{\underline{Answer :-}}}$

The Measure of the diagonal is 29 cm.

$\textbf{\underline{\underline{Explanation :-}}}$

Given :

Perimeter of Rectangle = 82 cm

Breadth of the Rectangle = 20 cm

To find :

The Length of the diagonal

Solution :

To get the diagonal's length, we need to find the Length first.

$\star$ As we know :

$\boxed{\sf{Perimeter = 2(Length + Breadth)}}$

Consider Length as x

$\tt{\implies}2(x + 20) = 82$

$\tt{\implies}2x + 40 = 82$

$\tt{\implies}2x = 82 - 40$

$\tt{\implies}2x = 42$

$\tt{\implies}x =\dfrac{42}{2}$

$\tt{\implies} x = 21$

Length = 21 cm

$\star$ Diagonal's Length =

By Pythagoras Theorum :-

Consider the diagonal as x

Refer the Attachment for better understanding.

∆ ABC

Hypotenuse = AC = x

Base = BC = 21 cm

Height = AB = 20 cm

$\star$ (Hypotenuse) ² = (Base) ² + (Height) ²

$\tt{\implies}{x}^{2}={21}^{2}+{20}^{2}$

$\tt{\implies}{x}^{2}=441+400$

$\tt{\implies}{x}^{2} =841$

$\tt{\implies}x =\sqrt{841}$

$\begin{array}{r|l}29 & 841 \\\cline{1-2} 29 & 29 \\\cline{1-2} & 1\end{array}$

$\tt{\implies} x = 29$

$\therefore$ The Measure of the diagonal is 29 cm.