Question

Find the area of the rectangular park of length 12
$\frac{3}{5}$
m and breadth 8
$\frac{2}{5}$
​

Answer 1

Given Information : Length of the rectangular park is $\sf {12 \dfrac{3}{5} \: m }$ and breadth is $\sf {8 \dfrac{2}{5} \: m }$.

Need to calculate : Area of the rectangular park.

F⠀I⠀G⠀U⠀R⠀E :

$\sf{ 8 \dfrac{2}{5} \: m }\huge\boxed{ \begin{array}{cc} \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{array}} \\ \: \: \: \: \: \sf{12 \cfrac{3}{5} \: m}$

$\dag$ As we know that,

$\star \: \:\underline{ \boxed{\sf \purple{Area_{(Rectangle)} = \ell ength \times breadth }}}\\ \\ \\ \bf {\underline{Substituting \: the \: values :} }\\ \\ \\ \dashrightarrow \sf{ Area_{(park)} =12 \cfrac{3}{5} \: m \times 8 \dfrac{2}{5} \: m } \\ \\ \\ \dashrightarrow \sf{ Area_{(park)} = \cfrac{63}{5} \: m \times \dfrac{42}{5} \: m } \\ \\ \\ \dashrightarrow\sf{ Area_{(park)} = \cfrac{63 \times 42}{5 \times 5} \: m^2 } \\ \\ \\ \dashrightarrow \sf{ Area_{(park)} = \cfrac{2646}{25} \: m^2} \\ \\ \\ \dashrightarrow \boxed{ \mathfrak \purple { Area_{(park)} =105 \cfrac{21}{25} \: \: m^2}}$

Therefore, area of the park is $\pmb { \mathfrak \gray { 105 \cfrac{21}{25} \: \: m^2 }}$.

More to know :

• Perimeter of rectangle = 2 (length + breadth)
• Area of rectangle = length × breadth

★ Important fraction rules :

$\small \boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\textsf{Fraction Rules :}}\\\\\bigstar\:\:\sf\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A+B}{C} \\\\\bigstar\:\:\sf{\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A-B}{C}}\\\\\bigstar\:\:\sf\dfrac{A}{B} \times \dfrac{C}{D} = \dfrac{AC}{BD}\\\\\bigstar\:\:\sf\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD+BC}{BD} \\\\\bigstar\:\:\sf\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD-BC}{BD}\\\\\bigstar \:\:\sf \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C} = \dfrac{AD}{BC}\end{array}}$