Asked by justin
draw an rectangle that has an area of 12sq.in and a perimeter that is greater than 25in.?
Answers
Answered by
MathMate
Think of a long rectangle.
The longer you make the rectangle, the greater is the perimeter.
The longer you make the rectangle, the greater is the perimeter.
Answered by
Bosnian
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
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
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