Question

The area of a Rectangle is (12m² - 17m - 5) cm² and, Length of the Rectangle is given (4m + 1) cm. Find the Breadth and Perimeter of the Rectangle.

Please Answer this Question.​

Answer 1

AnswEr :

• Area = (12m² - 17m - 5) cm²
• Length = (4m + 1) cm

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(7.3,2){\mathsf{\large{? cm}}}\put(7.7,1){\large{B}}\put(8.7,0.7){\matsf{\large{(4m + 1) cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,3){\large{D}}\end{picture}$

$\rule{150}{2}$

• According to the Question Now :

$\longrightarrow\texttt{Area of Rectangle = Length \times Breadth}\\\\\\\longrightarrow\tt (12m^2 -17m - 5) \:cm^2= (4m +1) \:cm \times Breadth\\\\\\\longrightarrow \tt Breadth = \dfrac{(12m^2 -17m - 5) \:cm^2}{(4m +1) \:cm}\\\\\\\longrightarrow\blue{\tt Breadth = (3m-5) \:cm}$

$\boxed{\begin{minipage}{6.4 cm}\quad \begin{array}{m{3.5em}cccc}&&\sf3m& \sf-5&\\\cline{1-6}\multicolumn{2}{l}{\sf4m+1\big)}&\sf12m^2&\sf-17m&\sf-5\\&& \sf-(12m^2&\sf+3m)&&\\\cline{3-4}&&&\sf-20m&\sf-5&\\&&&\sf-(-20m&\sf-5)&\\\cline{4-5}&&&\sf0&\sf0&\\\end{array}\end{minipage}}$

⠀

$\therefore\:\underline{\textsf{Breadth of the Rectangle is \textbf{(3m - 5) cm}}}$

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• P E R I M E T E R :

$:\implies\tt Perimeter = 2(Length + Breadth)\\\\\\:\implies\tt Perimeter = 2[(4m + 1) + (3m - 5)]\\\\\\:\implies\tt Perimeter = 2[4m + 3m + 1 - 5]\\\\\\:\implies\tt Perimeter = 2[7m - 4]\\\\\\:\implies\boxed{\green{\tt Perimeter = (14m - 8) \:cm}}$

⠀

$\therefore\:\underline{\textsf{Perimeter of the Rectangle is \textbf{(14m - 8) cm}}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Area of rectangle is (12m² - 17 m -5)cm²
• Length is 4m + 1

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

As we know that :

$\large{\boxed{\sf{Area \: of \: Rectangle \: = \: Length \: \times \: Breadth}}} \\ \\ \implies {\sf{Breadth \: = \: \dfrac{Area}{Length}}} \\ \\ \implies {\sf{Breadth \: = \: \dfrac{12m^2 \: - \: 17m \: - \: 5}{4m \: + \: 1}}}$

See attached Picture for value of Breadth

$\large{\boxed{\sf{Breadth \: = \: (3m \: - \: 5) \: cm}}}$

Now,

$\rule{200}{1}$

$\large \star {\boxed{\sf{Perimeter \: = \: 2( Length \: + \: Breadth)}}} \\ \\ \implies {\sf{Perimeter \: = \: 2 \big[ \big(4m \: + \: 1 \big) \: + \: \big( 3m \: - \: 5 \big) \big] }} \\ \\ \implies {\sf{Perimeter \: = \: 2 \big( 4m \: + \: 1 \: + \: 3m \: - \: 5 \big)}} \\ \\ \implies {\sf{Perimeter \: = \: 2(7m \: - \: 4)}} \\ \\ \implies {\sf{Perimeter \: = \: 14m \: - \: 8}}$

$\large {\boxed{\sf{Perimeter \: = \: (14m \: - \: 8)cm}}}$⠀