Question

hi hi help me
4th question is Length of a rectangular field is 5 1/4 m is 1 1/7 m. Find the area of the rectangular field​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar \:{\pmb{\sf{Solving \: Question \: 1^{st}}}}}$

Question: The reciprocal of ${\sf{2\dfrac{1}{3}}}$ is what?

Explaination: To solve this question firstly we have to make the given whole fraction in the form of a simple fraction. Then we have to reciprocate it. Let's see how to solve this question!

Requied solution:

$:\implies \sf 2\dfrac{1}{3} \\ \\ :\implies \sf \dfrac{7}{3} \\ \\ \sf Now \: let's \: reciprocate \: \dfrac{7}{3} \\ \\ :\implies \sf We \: get \: \dfrac{3}{7} \\ \\ {\pmb{\sf{Henceforth, \: \dfrac{3}{7} \: is \: requied \: solution!}}}$

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${\bigstar \:{\pmb{\sf{Solving \: Question \: 2^{nd}}}}}$

Question: We have to divide ${\sf{6 \div \dfrac{2}{3}}}$

Explaination: To solve this question firstly we have to write 6 as that it then have to put the sign of multiply at the place of divide. Then have to reciprocate ${\sf{\dfrac{2}{3}}}$ Afterthat have to solve. Now let's see how to solve this question!

Required Solution:

$:\implies \sf 6 \div \dfrac{2}{3} \\ \\ :\implies \sf 6 \times \dfrac{3}{2} \\ \\ :\implies \sf \cancel{6}^{3} \times \dfrac{3}{\cancel{{2}}} \\ \\ :\implies \sf 3 \times 3 \\ \\ :\implies \sf 9 \\ \\ {\pmb{\sf{Henceforth, \: 9 \: is \: the \: requied \: solution!}}}$

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${\bigstar \:{\pmb{\sf{Solving \: Question \: 3^{rd}}}}}$

Question: We have to find ${\sf{\dfrac{5}{9} \: of \: \dfrac{21}{25}}}$

Explaination: To solve this question we just have to put the sign of multiply at the place of 'of' Then have to carry on. Let us see how to solve this question!

$:\implies \sf \dfrac{5}{9} \: of \: \dfrac{21}{25} \\ \\ :\implies \sf \dfrac{5}{9} \times \: \dfrac{21}{25} \\ \\ :\implies \sf \dfrac{\cancel{{5}}}{9} \times \: \dfrac{21}{\cancel{{25}}} \\ \\ :\implies \sf \dfrac{1}{9} \times \: \dfrac{21}{5} \\ \\ :\implies \sf \dfrac{1}{\cancel{{9}}} \times \: \dfrac{\cancel{{21}}}{5} \\ \\ :\implies \sf \dfrac{1}{3} \times \dfrac{7}{5} \\ \\ :\implies \sf \dfrac{7}{15} \\ \\ {\pmb{\sf{Henceforth, \: \dfrac{7}{15} \: is \: requied \: solution!}}}$

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${\bigstar \:{\pmb{\sf{Solving \: Question \: 4^{th}}}}}$

Correct question(meaning): The length of the rectangle is ${\sf{5\dfrac{1}{4}}}$ m and the breadth is ${\sf{1\dfrac{1}{7}}}$ m We have to find out the area of rectangle.

Figure regard this question:

$\begin{gathered} \sf 5\dfrac{1}{4} \: m\: \: \: \: \: \: \: \: \: \: \: \\ \begin{gathered}\begin{gathered}\boxed{\begin{array}{}\bf { \red{}}\\{\qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{}\\ { \sf{ }}\\ { \sf{ }} \\ \\ { \sf{ }}\end{array}}\end{gathered}\end{gathered} \sf 1\dfrac{1}{7} \: m \end{gathered}$

Given that:

• Length of rectangle = ${\sf{5\dfrac{1}{4}}}$ m

• Breadth of rectangle = ${\sf{1\dfrac{1}{7}}}$ m

To find:

• Area of rectangle

Solution:

• Area of rectangle = 6 m²

Using concept:

• Formula to find out the area of the rectangle

Using formula:

• Area of rectangle = Length×Breadth

Full Solution:

$:\implies \sf Area \: of \: rectangle = Length \times Breadth \\ \\ :\implies \sf Area \: of \: rectangle = 5\dfrac{1}{4} \times 1\dfrac{1}{7} \\ \\ :\implies \sf Area \: of \: rectangle = \dfrac{21}{4} \times \dfrac{8}{7} \\ \\ :\implies \sf Area \: of \: rectangle = \dfrac{\cancel{{21}}}{4} \times \dfrac{8}{\cancel{{7}}} \\ \\ :\implies \sf Area \: of \: rectangle = \dfrac{3}{4} \times \dfrac{8}{1} \\ \\ :\implies \sf Area \: of \: rectangle = \dfrac{3}{\cancel{{4}}} \times \dfrac{\cancel{{8}}}{1} \\ \\ :\implies \sf Area \: of \: rectangle = \dfrac{3}{1} \times \dfrac{2}{1} \\ \\ :\implies \sf Area \: of \: rectangle = 6 \: m \: sq. \\ \\ {\pmb{\sf{6 \: m \: sq. \: is \: requied \: solution!}}}$

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• Dear web users, you can see the step of cancelling from the attachment. Thanks for understanding!