increasing the perimeter of a shape increases the area?

Question

I'm confused on a maths question I have, is it true? Surely if you extend a perimeter you increase the area? HELP!

Answer 1

not necessarily. consider u extend the perimeter inwards the shape.. u actually get a smaller area(:

Answer 2

Not necessarily!
Take a shape with an area of 24cm^2.
The sides could be:
4cm x 6cm => gives perimeter of 4+4+6+6 =20cm
3cm x 8cm => gives perimeter of 3+3+8+8 =22cm
2cm x 12cm => gives perimeter of 2+2+12+12=28cm
1cm x 24cm => gives perimeter of 1+1+24+24=50cm
Completely different perimeters but the same area.
OR
Take a shape with a perimeter of 20cm
1+1+9+9=20cm => gives area of 1cm x 9cm = 9cm^2
2+2+8+8=20cm => gives area of 2cm x 8cm = 16cm^2
3+3+7+7=20cm => gives area of 3cm x 7cm = 21cm^2
4+4+6+6=20cm => gives area of 4cm x 6cm = 24cm^2
5+5+5+5=20cm => gives area of 5cm x 5cm = 25cm^2

The shape which gives the largest area for the same perimeter is a square. The smallest area for the same perimeter is a very thin rectangle.

Answer 3

No, not necessarily. Consider the following 'drawing'

/\/\/\
\/\/\/

and imagine the 'ends' closed with straight, vertical lines. Now, if you make the 'notches' in the sides deeper, the perimeter certainly increases, but the surface area enclosed by the figure actually decreases.

Or say you have a rectangle that's 2 by 4 feet. It has a surface area of 8 square feet and a perimeter of 12 feet. Now, look at a rectangle that is 6 feet by 1 foot. It has a perimeter of 14 feet, but encloses an area of only 6 square feet.

There really isn't much direct relationship between perimeter and area contained unless there are other constraints imposed on the figures involved.

HTH ☺


Doug

Answer 4

Not necessarily. For example, if you stretch a square in all 4 directions so that it grows bigger, you've extended the perimeter and also increased the area. But if you had a circle and then took a piece out of it like a pizza slice, you'd have a longer perimeter but a smaller area.

Answer 5

to find if perimeter would increase the area..just substitute it in the equations..
eg:

in case of rectangle if length = 3 and
breadth = 5 then
area = 3*5 =15

so perimeter = 2(3+5)
= 16.

so if u had changed the value of lenght and breadh then normally the perimeter wld also change.. thats if u increase the perimeter it wls increase the area..

like wise for circle if u increase the radius then it shud increase its perimeter too. but in ceratin structures it may not increase

this is my knowledge..if there is any change then kindly do inform me. :-)

Answer 6

Basic fact: for a rectangle of given perimeter, a square will produce the greatest area; the least area would be in a rectangle of extreme "skinniness." Thus the first one would have to get "less square," ie, more rectangular to get more perimeter for a given area. The second one is the opposite -- that one must be getting more square.

Answer 7

it's not necessary .suppose u hav a rectangle 3x2.the perimeter becomes 10 n d area 6 in dere resp. units.now u cnvrt it into a rectngle of 6x1 .d perimeter increases to 14 n area remains 6 sq units only.
In case d 2 figures r similar d given statement becomes true.

Answer 8

No,not necessarily...take two rectangles

1) 6cm x 4 cm.....Perimeter = 20 cm...Area = 24 sq cm.

2) 10cm x 1 cm..perimeter = 22cm (bigger)..but area = 10 sq cm
Does this help??

Answer 9

you will only increase the area if you extend the perimeter outwards. if you invert the perimeter you will decrease the area.

Answer 10

yes,provided that shape is enclosed
and remains proportionately the
same

take the most efficient plane shape
-the circle
C=2pi*r >>> r=C/2pi
A1=pi*r^2=pi(C/2pi)^2=C^2/4*pi
increase the circumference by
dC{a very small proportion of C}
hence,
C+dC=2pi*R{Ris the new radius}
>>>>R=(C+dC)/2pi
A2=pi*R^2=pi((C+dC)/2pi)^2
={C^2+2C*dC+(dC)^2}/4pi
showing that,
{C^2+2C*dC+(dC)^2}/4pi
>C^2/4pi
>>>> A2>A1,therefore,
the area of a circle increases
when the perimeter increases

this can be proved for all 'regular'
shapes

the square is the most efficient
rectangle,but if you 'elongate'
the square,it becomes less
efficient as the ratio of the
length to the breadth increases

i hope that this helps