Question

14. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the 5.5
same wire is re-bent in the shape of a square, what will be the measure of each side?
Also, find which shape encloses more area?
(NCERT)
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Answer 1

Answer:

Hi

Here is your answer

L=40cm

B=22cm

The perimeter of the wire = Perimeter of the rectangle = 2 (l + b) = 2 x (40 cm + 22 cm) = 2 x 62 cm = 124 cm

Then, the perimeter of the square = 124 cm

The side length of the square = (124/4) cm = 31 cm

Area of the rectangle = l x b = 40 cm x 22 cm = 880 cm^2

Area of the square = s^2 = 31 cm x 31 cm = 961 cm^2

The shape which encloses more are = Area of square - area of reactangle

=>961cm^2-880cm^2

=>81cm^2

This proves that the square encloses more area than the rectangle.

By the way, whenever you have a certain perimeter length, the largest quadrilateral that has this perimeter is always a square....

hope it helps...

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Answer 2

⭐ Question

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. 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?

${\underline{\bf { Given \: that } }}$

• A wire is in the shape of a rectangle has length of 40 cm and breadth of 22 cm.

${\underline{\bf { To \: find: } }}$

• 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 and by how much?

${\underline {\bf { Solution : } }}$

According to question-

$\rm\blue {Perimeter \: of \: rectangle = Perimeter \: of \: square </p><p>}$

$\rm { \therefore 2(Length + Breadth) = 4 \times Side </p><p>}$

$\rm { \longrightarrow 2 (40 + 22) = 4 \times Side </p><p>}$

$\rm { \longrightarrow 2 \times 62 = 4 \times Side </p><p>}$

$\rm { \longrightarrow 124 = 4 \times Side </p><p>}$

$\rm { \longrightarrow Side = \dfrac{124}{4}</p><p>}$

$\rm\red { Side = 31 cm}$

Now,

$\rm {\implies Ar. \: of \:rectangle = length \times breadth }$

$\rm {\implies Ar. \: of \:rectangle = 40 \times 22 = 880 {cm}^{2} }$

and,

$\rm { \implies Area \: of \: square = {(Side)}^{2} }$

$\rm { \implies 31 \times 31 = 961 {cm}^{2} </p><p>}$

$\rm\purple { 880 {cm}^{2} < 961 {cm}^{2} }$

Therefore, the square-shaped wire encloses more area.