Question

21. Two wires of length 44cm are bent to
form a square and a rectangle. The
breadth of the rectangle formed is 5cm.
Which of the two figures, square or
rectangle will have a greater area and by
how much? *
(3 Points)
Rec tan gle by 32 cm
Square by 32 cm
Rec tan gle by 36 cm
Square by 36 cm​

Answer 1

→ Area of square= a²

→ The sides of square are equal. So,

$→ \frac{44}{4} = 11$

→ Area of square= side × side

$→ 11cm×11cm =121 cm²$

→ We have to find length of the rectangle.

→ Given, Perimeter of rectangle= 44cm

$→ 44cm= 2(l+b)$

$→ 44cm= 2l+2b$

$→ 44cm= 2l+2(5)cm$

$→ 44cm= 2l+10cm$

$→ 44cm-10cm= 2l$

$→ 34cm= 2l$

$→ \frac{34}{2} = l$

$→ 17cm= l$

→ Area of rectangle= length×breadth

$→ 17cm×5cm= 85cm²$

$→ 121cm² -85cm²= 36cm²$

Hence,

• The square has more area by 36cm².

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

Case 1

When wire of length 44 cm is bent in the form of a square.

• Let assume that side of the square be 'a' cm.

Now,

• As wire is bent in the form of square.

So,

• Perimeter of square = Length of wire

$\rm :\implies\:4 \times a \: = 44$

$\rm :\implies\: \boxed{ \pink{ \bf \: a \:  =  \tt \:11 \: cm }}$

So,

• Area of square is given by

$\rm :\implies\:Area_{(square)} \: = \: {a}^{2}$

$\rm :\implies\:Area_{(square)} = {(11)}^{2}$

$\rm :\implies\: \boxed{ \pink{ \bf \:  Area_{(square)}\:  =  \tt \:121 \: {cm}^{2} }}$

Now,

Case - 2.

When wire of length 44 cm is bent in the form of rectangle.

It is given that

• Breadth of rectangle (b) = 5 cm

• Let length of rectangle be 'l' cm

Since,

• wire is bent in the form of rectangle,

So,

• Perimeter of rectangle = Length of wire

$\rm :\implies\:2(length \: + \: breadth) = 44$

$\rm :\implies\:l \: + \: 5 \: = \: 22$

$\rm :\implies\: \boxed{ \pink{ \bf \: l \:  =  \tt \: 17 \: cm}}$

Now,

• Area of rectangle is given by

$\rm :\implies\:Area_{(rectangle)} = length \times breadth$

$\rm :\implies\:Area_{(rectangle)} = 17 \times 5$

$\rm :\implies\: \boxed{ \pink{ \bf \: Area_{(rectangle)} \:  =  \tt \: 85 \: {cm}^{2} }}$

So,

• From above calculations, we conclude that

$\large \boxed{ \boxed{ \green{ \bf \: Area_{(square)} > Area_{(rectangle)}}}}$

And

Difference in area is

$\rm :\implies\:Area_{(square)} - Area_{(rectangle)}$

$\rm :\implies\:121 - 85$

$\rm :\implies\:36 \: {cm}^{2}$