Question

A wire is in the shape of a square of side 10cm. Find the length of the wire. The same wire is bent to form a rectangle of length 12 cm. Find its breadth.

Answer 1

Step-by-step explanation:

, Side of the square: 10 cm

Hence, Perimeter of the square = 4 X 10 = 40 cm = Length of the wire.

When wire is re-bend to make a rectangle,

Given, Length of the rectangle= 12 cm

Let the breadth of the rectangle be b

Thus, the perimeter of both the rectangle and the square is same since the square is stretched to make a rectangle.

We know that the Perimeter of rectangle

\begin{gathered}\begin{array} { c } { = 2 ( l + b ) = 2 ( 12 + b ) } \\\\ { 40 = 24 + 2 b } \\\\ { 2 b = 40 - 24 \Rightarrow 2 b = 16 } \\\\ { b = 8 \mathrm { cm } } \end{array}\end{gathered}

=2(l+b)=2(12+b)

40=24+2b

2b=40−24⇒2b=16

b=8cm

Area of enclosed by the square

= ( \text {side} ) ^ { 2 } = 10 \times 10 = 100 \mathrm { cm } ^ { 2 }=(side)

2

=10×10=100cm

2

Area enclosed by the rectangle

= ( \text {length} ) \times ( \text {breadth} ) = 12 \times 8 = 96 \mathrm { cm } ^ { 2 }=(length)×(breadth)=12×8=96cm

2

Thus, Difference in their area

= ( 100 - 96 ) c m ^ { 2 } = 4 c m ^ { 2 }=(100−96)cm

2

=4cm

2

Hence, the square encloses more area by 4 cm^24cm

2

.