Question

a wire bent in the form a square of side 15 cm.if this wire is reshaped to form a rectangle of length 20 cm find the :1)perimeter of rectangle 2)breadth of the rectangle

Answer 1

Step-by-step explanation:

Perimeter of equilateral triangle = perimeter of square ( as the wire is same)

3*side = 4*side

3*20 = 4*side

60 = 4*side

60/4 = side

15 cm is the side of square

Answer 2

Answer:

Given Information:-

• Side of the square shaped wire = 15 cm
• If it is reshaped into a rectangle, it's length is = 20 cm

To Find:-

1. Perimeter of the Rectangle
2. Breadth of the Rectangle

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At First, Perimeter of the Rectangle:-

Side of the Square shaped wire = 15 cm

We know,

Perimeter of the square = 4(Side)

Hence, of side is given, the perimeter is:

4(side)

= 4(15) {Step, as per the formula of perimeter of square}

= 60 cm (As we have multipled 15 with 4, the product is 60 cm, which is the perimeter of the square)

Hence,

Perimeter of Square = Perimeter of Rectangle

{As it is reshaped}

Therefore, Perimeter of the wire (in shape of a Rectangle) is 60 cm.

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Now, Breadth of the Rectangle:-

As we know,

Perimeter of Rectangle = 2(l+b),

Value of Length = 20 cm

Let, the value of Breadth be = b cm

Hence,

2(20+b)=60 {Suitable Equation formation}

= 40+2b=60 {Multiplied the Left Handed expression}

=2b=60-40 {Taken 40 to Right Hand Side, it's value is now opposite}

=2b=20 {Subtrated 60 with 40}

=b=20/2 {Taken 2 to the RHS}

=b=10 {Suitable value of b after Division of 20 and 2}

Therefore, breadth of the Rectangle is 10 cm.

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REQUIRED ANSWER:-

• Perimeter of Rectangle is $\rm\blue{60 \ centimeters}$.
• Breadth of Rectangle is $\rm\blue{10 \ centimeters}$.