Question

A rectangular sheet of paper is 5 2/3cm long and 3 1/5 cups wide find its perimeter

Answer 1

Answer:

• $\sf{perimter\:of\:paper=17\dfrac{11}{15}\;\;cm}$

Explanation:

Given

• Length of rectangular paper = $\sf{5\dfrac{2}{3}\;\;cm}$

• Breadth of rectangular paper = $\sf{3\dfrac{1}{5}\;\;cm}$

To find

• Perimeter of rectangular paper =?

Formula required

• Formula to calculate perimeter of rectangle

$\boxed{\sf{Perimeter\:of\:rectangle=2\times ( length + breadth)}}$

Calculation

Converting length and breadth given in mixed fraction into improper fraction

$\implies\sf{length=5\dfrac{2}{3}\;\;cm}$

$\implies\sf{length=\dfrac{17}{3}\;\;cm}$

and

$\implies\sf{breadth=3\dfrac{1}{5}\;\;cm}$

$\implies\sf{breadth=\dfrac{16}{5}\;\;cm}$

Calculating perimeter of rectangular paper

Using formula for perimeter of rectangle

$\longrightarrow\sf{perimeter\:of\:paper=2\times (length + breadth)}$

$\longrightarrow\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{17}{3} + \dfrac{16}{5}\bigg)}$

$\longrightarrow\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{17(5)+16(3)}{15}\bigg)}$

$\longrightarrow\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{85+48}{15}\bigg)}$

$\longrightarrow\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{133}{15}\bigg)}$

$\longrightarrow\boxed{\boxed{\sf{perimeter\:of\:paper=\dfrac{266}{15}\;\;cm=17\dfrac{11}{15}\;\;cm}}}$

Answer 2

Given

• Length of rectangular paper = $\sf{5\dfrac{2}{3}\;\;cm}$

• Breadth of rectangular paper =$\sf{3\dfrac{1}{5}\;\;cm}$

To find

• Perimeter of rectangular paper =?

Formula required

Formula to calculate perimeter of rectangale

$\boxed{\sf{Perimeter\:of\:rectangle=2\times ( l + b)}}$

where l = length and b = breath

solution

Converting length and breadth given in mixed fraction into improper fraction

$\implies\sf{length=5\dfrac{2}{3}\;\;cm}$

$\implies\sf{length=\dfrac{17}{3}\;\;cm}$

$\implies\sf{breadth=3\dfrac{1}{5}\;\;cm}$

$\implies\sf{breadth=\dfrac{16}{5}\;\;cm}$

Calculating perimeter of rectangular paper

Using formula for perimeter of rectangle

$\implies\sf{perimeter\:of\:paper=2\times (length + breadth)}$

$\implies\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{17}{3} + \dfrac{16}{5}\bigg)}$

$\implies\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{17(5)+16(3)}{15}\bigg)}$

$\implies\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{85+48}{15}\bigg)}$

$\implies\sf{perimeter\:of\:paper=2\times \bigg(\dfrac{133}{15}\bigg)}$

$\implies\boxed{\boxed{\sf{perimeter\:=\dfrac{266}{15}\;\;cm=17\dfrac{11}{15}\;\;cm}}}$