Question

A wire is in the shape of a rectangle. Its length is 40cm and breadth is 22cm. If the same wire is rebent in the shape of a square,what will be the measure of each side. also find which shape encloses more area?​

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Answer 1

$\Large\text{\underline{\underline{Question 1}}}$

What is the length of each side of a new square?

31cm.

$\Large\text{\underline{\underline{Explanation 1}}}$

The perimeter of the rectangle is $2\times(40+22)=124\text{cm}.$

We know that the side lengths are all equal in a square.

The average of lengths of four sides is $31\text{cm}.$

Hence, each side of the new square measures $\text{31cm.}$

$\Large\text{\underline{\underline{Question 2}}}$

Which area between the rectangle or square is greater?

Square.

$\Large\text{\underline{\underline{Explanation 2}}}$

Which area of the rectangle and the square is broader? We can determine the greater area by A.M.-G.M. inequality.

Statement

If the perimeter is equal, the area of a square is broader.

Proof

Let the length and breadth be $\largea, b.$

The average side lengths are equal.

There is a relation between the A.M. and the G.M. The two are equal only when $\large\text{ a=b .}$

$\cdots\longrightarrow\boxed{\dfrac{a+b}{2}\geq\sqrt{ab}}$

This can be taken as -

$\cdots\longrightarrow\boxed{\text{(Average side length)}\geq\sqrt{\text{(Area)}}.}$

As we see here, the average length of the two shapes is equal. Then, the rectangles attain the maximum area, $\largeab$, only when $\large\text{ a=b .}$ So, a square is the most optimal figure.

Hence, the area of a square is always greater than that of a rectangle!

$\Large\text{\underline{\underline{Verification}}}$

→ 40×22=880cm² (The area of the rectangle)

→ 31×31=961cm² (The area of the square)

Answer 2

Step-by-step explanation:

$\Large\text{\underline{\underline{Question 1}}}$

Question 1

What is the length of each side of a new square?

31cm.

$\Large\text{\underline{\underline{Explanation 1}}}$

Explanation 1

The perimeter of the rectangle is 2×(40+22)=124cm.

We know that the side lengths are all equal in a square.

The average of lengths of four sides is

$31\text{cm}.$

31cm.

Hence, each side of the new square measures 31cm.

$\Large\text{\underline{\underline{Question 2}}}$

Question 2

Which area between the rectangle or square is greater?

Square.

$(\Large\text{\underline{\underline{Explanation 2}}})$

Explanation 2

Which area of the rectangle and the square is broader? We can determine the greater area by A.M.-G.M. inequality.

Statement

If the perimeter is equal, the area of a square is broader.

Proof

Let the length and breadth be \large\text{$a, b.$}a,b.

The average side lengths are equal.

There is a relation between the A.M. and the G.M. The two are equal only when

$\large\text{ a=b .}$

a=b.

$\cdots\longrightarrow\boxed{\dfrac{a+b}{2}\geq\sqrt{ab}}⋯⟶$

This can be taken as -

$\cdots\longrightarrow\boxed{\text{(Average side length)}\geq\sqrt{\text{(Area)}}.}$

)

.

As we see here, the average length of the two shapes is equal. Then, the rectangles attain the maximum area,

$\largeab$

ab , only when

$\large\text{ a=b .}$

a=b. So, a square is the most optimal figure.

Hence, the area of a square is always greater than that of a rectangle!

$\Large\text{\underline{\underline{Verification}}}$

→ 40×22=880cm² (The area of the rectangle)

→ 31×31=961cm² (The area of the square)

$\bf\green{Hlo \: Chudail}$