Question

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Q1.A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 13 cm, find its breadth. Which shape encloses more area?​​

Answer 1

The breadth of the rectangle measures 7 cm and the square encloses more area.

Step-by-step explanation

In the question, it's been stated that a wire which is in the shape of a square with side 10 cm is rebent into a rectangle with length 13 cm. We've been asked to calculate the breadth of the rectangle and determine which shape encloses more area.

In order to ascertain the shape that encloses more area, we need the measure of breadth of the rectangle. So, as per the question, we can say that the perimeters of square and rectangle are identical.

Breadth of the rectangle:

$\twoheadrightarrow \quad{ \sf{4s = 2(l +b)}}$

Plugging in the values.

$\twoheadrightarrow \quad{ \sf{4 \times 10 = 2(13 + b)}}$

Multiplying the numericals in the LHS.

$\twoheadrightarrow \quad{ \sf{40 = 2(13 + b)}}$

Simplifying the RHS by multiplying 2 with the term inside the bracket.

$\twoheadrightarrow \quad{ \sf{40 = 26 + 2b}}$

Transposing the like terms from RHS to LHS. [While transposing, 26 will be changed to (– 26) in the LHS]

$\twoheadrightarrow \quad{ \sf{40 - 26 = 2b}}$

Subtracting the numericals in the LHS.

$\twoheadrightarrow \quad{ \sf{14 = 2b}}$

Again transposing the like terms from RHS to LHS. This time 2 is transposed to the RHS. [2 will be written with division symbol in the LHS, since it is multiplied in the RHS]

$\twoheadrightarrow \quad{ \sf{14 \div 2 = b}}$

Writing the LHS in fractional form.

$\twoheadrightarrow \quad{ \sf{ \dfrac{14}{2} = b}}$

Reducing the fraction to its lowest term.

$\twoheadrightarrow \quad{ \sf{ \dfrac{7}{1} = b}}$

This can be also written as,

$\twoheadrightarrow \quad{ \sf{7 = b}}$

Hence, the measure of breadth of the rectangle is 7 cm.

Now, in order to find which shape encloses more area, we need the measurements of areas of the square and rectangle.

Area of the square:

$\twoheadrightarrow \quad{ \sf{(s)^{2} \: sq.units }}$

Plugging in the values.

$\twoheadrightarrow \quad{ \sf{(10)^{2} }}$

Squaring the side of the square.

$\twoheadrightarrow \quad{ \sf{100 \: sq.cm}}$

Area of the rectangle:

$\twoheadrightarrow \quad{ \sf{lb \: sq.units}}$

Plugging in the values.

$\twoheadrightarrow \quad{ \sf{13 \times 7}}$

Multiplying the length with the breadth.

$\twoheadrightarrow \quad{ \sf{91 \: sq.cm}}$

Now we have the measurements of both of their areas. On comparing them [[ 100 cm² > 91 cm² ]], it's obvious that the square encloses more area.