Question

A wire bent in the form of a square of side 30 cm is bent in the form of a rectangle with length 35 cm find width of the rectangle.
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Answer 1

Question:-

A wire bent in the form of a square of side 30 cm is bent in the form of a rectangle with length 35 cm find width of the rectangle.

Given:-

. peri of rectangle=$2(l + h)$

Solution:-

$= > 2(30 + 35cm)$

$= > 2(88)$

$= > 176$

Perimeter remain same so,

$2\pi \: r$

${\sf{\implies\cancel\dfrac{22} {7}×R=\cancel{176}}}$

• 88⁴

$\sf\fbox{20 cm}$

Answer 2

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✰Given :-

A wire bend in form of square of side 30 cm.

Then wire is again bend in form of rectangle of length 35 cm.

✰To find :-

Width of the rectangle.

✰Solution :-

Here, Concept is : If we are bending wire in form of square than again bending it in rectangle. Than, perimeter of square will equal to perimeter of rectangle because we are not increasing length of wire by one measure we are bending it in square and rectangular shape.

So,

Perimeter of square = 4 × side

➠ Perimeter = 4 × 30

➠Perimeter = 120

Thus,

Perimeter of square is 120 cm.

According to concept, Perimeter of square and perimeter of rectangle are equal.

So, Perimeter of rectangle is 120 cm.

Let, Breadth or width or rectangle be x cm.

We know,

Perimeter of rectangle = 2(Length + Breadth)

➺ 120 = 2×(35 + x)

➺ 120 = 70 + 2x

➺ 120 - 70 = 2x

➺50 = 2x

➺50/2 = x

➺ x = 25

We take, Width of rectangle be x.

Therefore,

Width of rectangle is 25 cm.

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