Question

A wire of length 20m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in meters

Answer 1

Answer:

The rectangles formed by folding were 5.

Step-by-step explanation:

Given : A wire of length 20 m is to be folded in the form of a rectangle.

To find : How many rectangles can be formed by folding the wire if the sides are positive integers in meters?

Solution :

A wire of length 20 m is to be folded in the form of a rectangle.

i.e, The perimeter of the rectangle is 20 m.

So, $P=2(\text{Length}+\text{Breadth})$

$20=2(\text{Length}+\text{Breadth})$

$\text{Length}+\text{Breadth}=10$

$\text{Length}=10-\text{Breadth}$

So, The possible integers are

B=1, L=9

B=2, L=8

B=3, L=7

B=4, L=6

B=5, L=5

We cannot take many more as (4,6) and (6,4) were same thing.

So, The rectangles formed by folding were 5.

Answer 2

Answer:

The rectangles formed by folding were 5.

Step-by-step explanation:

Given : A wire of length 20 m is to be folded in the form of a rectangle.

To find : How many rectangles can be formed by folding the wire if the sides are positive integers in meters?

Solution :

A wire of length 20 m is to be folded in the form of a rectangle.

i.e, The perimeter of the rectangle is 20 m.

So, P=2(\text{Length}+\text{Breadth})P=2(Length+Breadth)

20=2(\text{Length}+\text{Breadth})20=2(Length+Breadth)

\text{Length}+\text{Breadth}=10Length+Breadth=10

\text{Length}=10-\text{Breadth}Length=10−Breadth

So, The possible integers are

B=1, L=9

B=2, L=8

B=3, L=7

B=4, L=6

B=5, L=5

We cannot take many more as (4,6) and (6,4) were same thing.

So, The rectangles formed by folding were 5.