Answer :
Answer:
You can draw three rectangles:
1- 28 * 1.
2- 14*2
3- 7*4
The permeters will be:
P1 = 2*1 + 2*28 = 58
P2 = 2*2 + 2*14 = 32
P3 = 2*4+2*7 = 22
Step-by-step explanation:
We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.
28 = 1*2*2*7
We can organize them as follows:
R1: 28 = 1*28
R2: 28 = 2*14
R3: 28 = 4*7
Those are the three posible rectangles. The permeters will be:
R1 P1 = 2*1 + 2*28 = 58
R2 P2 = 2*2 + 2*14 = 32
R3: P3 = 2*4+2*7 = 22.
Finally, we can conclude that You can draw three rectangles:
1- 28 * 1.
2- 14*2
3- 7*4
The permeters will be:
P1 = 2*1 + 2*28 = 58
P2 = 2*2 + 2*14 = 32
P3 = 2*4+2*7 = 22
Three different rectangles can be drawn with an area of 28
First rectangle : perimeter = 58
Second rectangle : perimeter = 32
Third rectangle : perimeter = 22
Given :
Area of the rectangle is 28 square units
To identify how many rectangles can be formed , we need to write the factors of 28
Factors of 28 are 1,2,4,7,14,28
Now we combine the factors that gives us 28
[tex]1 \cdot 28=28\\2 \cdot 14= 28\\4 \cdot 7 =28[/tex]
So we can combine and form three different rectangles using area 28
Three different rectangles can be drawn with an area of 28
Now we find the perimeter of each
First rectangle : length =28 and width =1
Perimeter formula = 2(length + width )
Perimeter =[tex]2(28+1)=58[/tex]
Second rectangle : length =14 and width =2
[tex]Perimeter =2(14+2)=32[/tex]
Third rectangle : length = 7 and with =4
Perimeter =[tex]2(7+4)=22[/tex]
Learn more : brainly.com/question/1723776