Question

Find the perimeter of a rectangle whose 1 side measures 10m and the diagnoal is 26m.

Answer 1

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

$\sf{The\:Perimeter\:of \:the\:Rectangle\:is\:68 m}$

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

$\textsf{\underline{\underline{Given:}}}$

1 side measures = 10 m

Diagonal = 26 m

$\textsf{\underline{\underline{To find :}}}$

The Perimeter of Rectangle

$\textsf{\underline{\underline{Solution :}}}$

Refer the Attachment for better understanding.

We can find the measure of the unknown side by - Pythagoras' theorem.

Take ∆ ACD

Hypotenuse = AD = 26 m

Base = CD = 10 m

Height = AC = x

Pythagoras' theorem :

$\tt{\star\: Hypotenuse ^{2} = {Height}^{2} + {Base}^{2} }$

$\tt{\implies {26}^{2} = {x}^{2} + {10}^{2} }$

$\tt{\implies676 = {x}^{2} + 100 }$

$\tt{\implies676 - 100 = {x}^{2} }$

$\tt{\implies576 = {x}^{2} }$

$\tt{\implies \: x = \sqrt{576} }$

Square root of 576

$\begin{array}{r|l} 2 & 576 \\\cline{1-2} 2 & 288 \\\cline{1-2} 2 & 144 \\ \cline{1-2} 2 & 72 \\\cline{1-2} 2 & 36 \\\cline{1-2} 2 & 18 \\\cline{1-2} 3 & 9 \\\cline{1-2} 3 & 3 \\\cline{1-2} & 1\end{array}$

$\tt{\Rightarrow{\underline{2 \times 2}} \times {\underline{2 \times 2 }}\times {\underline{2 \times 2 }}\times{\underline {3 \times 3}}}$

Make pairs of common numbers and take only one from them.

$\tt{\Rightarrow3 \times 2 \times {\underline{2 \times 2 }}\times 2}$

$\tt{\Rightarrow6 \times 2 \times 2}$

$\tt{\Rightarrow6 \times 4}$

$\tt{\Rightarrow24}$

$\tt{\implies \: x = 24}$

$\large{\boxed{\bigstar{\sf \: {Another \: side \: = 24 \: m }}}}$

As we know the Length and Breadth of Rectangle, we can find the Perimeter.

$\tt{\implies2(l + b)}$

$\tt{\implies2(10 + 24)}$

$\tt{\implies20 + 48}$

$\tt{\implies68}$

$\large{\boxed{\bigstar{\sf \: { Perimeter = 68 \: m}}}}$

$\therefore$$\textsf{The\:Perimeter\:of\: the\:Rectangle\:is\:68 m}$

Answer 2

Answer:

68m

Step-by-step explanation:

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