Question

A wire in shape of square of side 8m is required to form a rectangle of breadth 4 less than half of its length. Find the change in area

Answer 1

Question :-

• A wire in shape of square of side 8m is required to form a rectangle of breadth 4 less than half of its length. Find the change in area.

Answer

Side of square = 8 m

• We know,

$\boxed{ \pink{ \rm \: Area_{(square)} = {(side)}^{2} }}$

$\rm :\implies\:Area_{(square)} \: = {8}^{2} = 64 \: {m}^{2} - - (i)$

Now,

• We know,

$\boxed{ \green{ \rm \: Perimeter_{(square)} \: = 4 \times side}}$

$\rm :\implies\:Perimeter_{(square)} \: = 4 \times 8$

$\rm :\implies\:Perimeter_{(square)} \: = \: 32 \: m$

Since,

• The square is rebent to form a rectangle.

So,

$\red{\rm :\implies\:Perimeter_{(square)} \: = \: Perimeter_{(rectangle)}}$

Now,

According to statement,

• Breadth of rectangle is 4 less than half of its length.

So,

Let

• Length of rectangle = 2x

and

• Breadth of rectangle = x - 4

Now,

$\green{\rm :\implies\:Perimeter_{(square)} \: = \: Perimeter_{(rectangle)}}$

$\rm :\implies\:32 = 2(length \: + \: breadth)$

$\rm :\implies\:16 = 2x + x - 4$

$\rm :\implies\:20 = 3x$

$\rm :\implies\:x = \dfrac{20}{3} \:$

Hence,

• Dimensions of rectangle are

$\rm :\implies\:Length = 2x = \dfrac{40}{3} \: m$

$\rm :\implies\:Breadth \: = x - 4 = \dfrac{20}{3} - 4 = \dfrac{8}{3} m$

Hence,

$\rm :\implies\:Area_{(rectangle)} = Length \times Breadth$

$\rm :\implies\:Area_{(rectangle)} = \dfrac{40}{3} \times \dfrac{8}{3}$

$\rm :\implies\:Area_{(rectangle)} = \dfrac{320}{9} \: {m}^{2}$

Hence,

$\rm :\implies\:Difference \: in \: area = Area_{(square)} - Area_{(rectangle)}$

$\rm :\implies\:Difference \: in \: area =64 - \dfrac{320}{9}$

$\rm :\implies\:Difference \: in \: area =\dfrac{576 - 320}{9}$

$\rm :\implies\:Difference \: in \: area =\dfrac{256}{9} \: {m}^{2}$