Question

A wire is in the form of a sqaure of side 26cm .It is bent in form of rectangle,whose length and breadth are in the ratio 2:1 What is the area of rectangle so formed
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Answer 1

ANSWER:

Area of Rectangle = 598.58 cm²

Step-by-step explanation:

given that,

A wire is in the form of a sqaure of side 26cm

here,

side of the wired square = 26 cm

so,

length of the wire = perimeter of square

perimeter of square = 4 × side

= 4 × 26

= 104 cm

now,

same wire is in the form of rectangle

where,

ratio of sides = 2 : 1

let the common ratio be x

so,

length = 2x

breadth = x

since,

wire is same so,

perimeter of square = perimeter of rectangle

perimeter of rectangle

= 2(length + breadth) .

ACCORDING TO THE QUESTION,

2(2x + x) = 104

3x = 104/2

3x = 52

x = 52/3

x = 17.3

so,

breadth of the rectangle = 17.3 cm

length of the rectangle = 2(17.3) =

34.6 cm

now,

Area of rectangle = length × breadth

= 17.3 × 34.6

= 598.58 cm²

so,

Area of Rectangle = 598.58 cm²

Answer 2

Area = 598.58cm²

Given:-

side of square = 26 cm

Ratio of length and breadth of rectangle = 2 : 1.

To find :-

Area of rectangle so formed.

Solution:-

The wire is in the form of square.

So, we can use the formula to find its perimeter.

$\boxed{\sf</strong><strong>\</strong><strong>g</strong><strong>r</strong><strong>e</strong><strong>e</strong><strong>n</strong><strong>{</strong><strong>P</strong><strong>e</strong><strong>r</strong><strong>i</strong><strong>m</strong><strong>e</strong><strong>t</strong><strong>e</strong><strong>r</strong><strong> </strong><strong>of</strong><strong> </strong><strong>Square</strong><strong> </strong><strong>=</strong><strong> </strong><strong>4</strong><strong> </strong><strong>×</strong><strong> </strong><strong>side</strong><strong>}}$

Now,

p. of square = 4 × 26

= 104 cm

Since, same wire is bent in the form of rectangle.

Let,the length and breadth of rectangle in the ratio be 2x and x

respectively.

$\boxed{\sf</strong><strong>\</strong><strong>p</strong><strong>i</strong><strong>n</strong><strong>k</strong><strong>{ </strong><strong>Perimeter</strong><strong> </strong><strong>of</strong><strong> </strong><strong>rectangle</strong><strong> </strong><strong>=</strong><strong> </strong><strong>2</strong><strong> </strong><strong>(</strong><strong> </strong><strong>l</strong><strong> </strong><strong>+</strong><strong> </strong><strong>b</strong><strong>)</strong><strong> </strong><strong>}}$

$\implies$ 104 = 2 ( 2x + x)

$\implies$ 104 = 2 × 3x

$\implies$ 104 = 6x

$\large{\underline{\boxed{ </strong><strong>x</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>7</strong><strong>.</strong><strong>3</strong><strong> </strong><strong>cm</strong><strong>}}}$

length of rectangle = 2x

= 2 × 17.3

= 34.6 cm

Now, we have

l = 34.6 cm

b = 17.3 cm

area of rectangle so formed = (length × breadth )

= ( 34.6 × 17.3 )

= 598.58 cm² ~ 598.6 cm².

hence, the area of rectangle so formed by benting the wire is 598.58cm².