Question

the perimeter of a rectangle is 296 m. Find the area of the rectangle if it's length os 120 m. Explain also ​

Answer 1

Answer:

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Step-by-step explanation:

A=wl

P=2(l+w)

Solving forA

A=Pl

2﹣l2=296·120

2﹣1202=3360m²

Answer 2

Given :

• The length of Rectangle and the Perimeter of Rectangle

To find :

• We have to find the Breadth and the Area of Rectangle

Solution :

In the given question first we will find the Breadth of the Rectangle because it has given only Perimeter and Length and after finding it we will find the Area also of the given question which Length and Breadth will be same only So, let's solve!

${ \underline{ \underline{ \green { \mathfrak{As \: we \: know \: that}}}}}$

$\boxed{\sf{\purple{Area_{\:(rectangle)} = Length \times Breadth}}}$

$\boxed{\sf{\pink{Perimeter_{\:(rectangle)} = 2 \times Length + Breadth}}}$

First we will find the Breadth

$: \implies \sf {Perimeter} \leadsto 296m \: \\\\ : \implies \: \sf{Length} \: = 2 \: \times (296m \: + L)\\\\ : \implies \: \dfrac{{\cancel 296}}{{ \cancel 2}} = \dfrac{148}{1} \\\\ : \implies \: \sf{148m} \: - \sf{120m} \leadsto \: 28m \\\\ : \implies{ \underline{ \boxed{ \pink{ \mathfrak{b \: \leadsto \: 28m}}}}}$

Now, we have find the Breadth so we can find the Area now

$: \implies \sf{120m} \: \times \sf{28m} \\\\ : \implies \: { \underline{ \boxed{ \pink{ \mathfrak{area \: \leadsto \: 3360 \: {m}^{2} }}}}}$

•°• Hence, verified! that Breadth of Rectangle is 28 m and the Area is $\sf{3,360 {m}^{2} }$

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Diagram :⠀⠀

[Note :- Kindly see the diagram in the attachment]⠀⠀⠀