draw an rectangle that has an area of 12sq.in and a perimeter that is greater than 25in.?

Think of a long rectangle.

The longer you make the rectangle, the greater is the perimeter.

Area=a*b

A=a*b=12 Area=12
12=a*b Divide with b
12/b=a
a=12/b

Perimeter=2*a+2*b
P=25
25=2*(12/b)+2*b
25=(24/b)+2*b Mulitiple with b
25b=24+2*b^2

Solutions of this equation is:
b=[sqroot(433)]/4+(25/4)
=5.202163 +6.25=11.452163
a=12/b=(12/11.452163)=1.047837

a=1.047837 b=11.452163

and :
b=[-sqroot(433)]/4+(25/4)
=-5.202163 +6.25=1.047837

a=11.452163 b=1.047837

Area:
A=a*b=11.452163*1.047837=12

Perimeter:
P=2*a+2*b=2*11.452163+2*1.047837=25

If you want to know solutions of equation: 2b^2-25b+24=0
Go to google and type:
"square equation online"
and you can find solutions step-by-step

To draw a rectangle with an area of 12 sq.in and a perimeter greater than 25 in., we need to first understand the formula for the area and perimeter of a rectangle.

The formula for the area of a rectangle is:

Area = Length * Width

And the formula for the perimeter of a rectangle is:

Perimeter = 2 * (Length + Width)

Now, let's find a suitable combination of length and width that satisfies the given conditions.

Since the area is given as 12 sq.in, we have:

12 = Length * Width

There are several combinations of length and width that could give us an area of 12 sq.in, such as:

1) Length = 3 in, Width = 4 in
2) Length = 2 in, Width = 6 in
3) Length = 1 in, Width = 12 in
4) Length = 6 in, Width = 2 in

Next, we need to calculate the perimeter for each of these combinations. Using the perimeter formula:

Perimeter = 2 * (Length + Width)

1) For Length = 3 in and Width = 4 in:
Perimeter = 2 * (3 + 4) = 2 * 7 = 14 in (Less than 25 in)

2) For Length = 2 in and Width = 6 in:
Perimeter = 2 * (2 + 6) = 2 * 8 = 16 in (Less than 25 in)

3) For Length = 1 in and Width = 12 in:
Perimeter = 2 * (1 + 12) = 2 * 13 = 26 in (Greater than 25 in)

4) For Length = 6 in and Width = 2 in:
Perimeter = 2 * (6 + 2) = 2 * 8 = 16 in (Less than 25 in)

Therefore, the combination that meets the given conditions is a rectangle with a length of 1 inch and a width of 12 inches, which has an area of 12 sq.in and a perimeter of 26 inches.