Is it possible to find the are and perimeter of a rectangle with only its length? And same with Square?

Question

Determine the are and perimeter of each figure described:

Rectangle with length of 3.6cm and square with sides of lenght 9mm width 4.2cm

Show your work

So by your answers I can safely judge that my school's summer packet was misprinted?

Answer 1

Hi,

With a square, there is no problem, since by the definition of a square, the lengths of all the sides are equal. This is not true of a rectangle. You need to know the length of the base and the height.

For a square, the perimeter is equal to 4 X length, and the area is equal to (length)².

For a rectangle, the perimeter is equal to 2 X base + 2 X height, and the area is equal to (length X height).

I hope this helps.

James :-)

Answer 2

The question was very well expressed but not the related data. How can a square have a length and a width when all the while its sides are equal in length such that the length and the width are synonymous to each other.

It is impossible to find the area and perimeter of a non-square rectangle with only its length. It is not only possible but we could certainly find the area and the perimeter of a square with only the length of one of its sides.

Answer 3

Square is a special rectangle. All its sides are equal and at right angles, so the data given to you about the square is wrong.

Rectangle will have sides of two different length at right angles. Both of these are two dimensional shapes.

For a rectangle, perimeter = 2(L+B), L= length, B= breadth
For a square, perimeter = 2L, L= length of a side.

I think the data given to you should be length of a side of a square = 3.6cm and for the rectangle it should be 9mm and 4.2cm.
Check it out with you teacher.

Answer 4

Let the sides of a RECTANGLE be of lengths a units,b units,a units,b units.
area of the rectangle=a*b sq.units
perimeter=2(a+b) units

Let the sides of a SQUARE be each of length a units.
are of the square=a square (i.e. a*a) sq.units
perimeter=4a units

As long as it is TOLD that it is a square or a rectangle, you know dt the anles r 90 degree. so there's no problem. but if only the side lenghts r given, the problem crops up, because, a rhombus too has all sides equal like a square, and a parallelogram has opposite sides equal like a rectangle, but there diagonals are of different lengths.

Answer 5

You got them a little mixed up. A rectangle is having two sides and a square only one side (all sides are equal). Now let us do the problem.

A square of side 3.6 cm will have a perimeter of 4 times the side. That means 3.6 x 4 = 14.4 cm

Area = side x side = 3.6 x 3.6 = 12.96 sq.cm

For the rectangle:

Perimeter = 2 x length + 2 x width = 2 x 9 + 2 x 4.2 = 18 + 8.4 = 26.4

Area = Length x width = 9 x 4.2 = 37.8 sq.cm

Answer 6

which you would be able to actual clean up the equ. Xy=one hundred (sideXside=self-discipline) & 2x+2y=40 (x+x+y+y or 2side+ 2side=perimeter) and resolve best equ that x= a hundred/y. Plug that for the time of to get 2(one hundred/y)+2y=40. start up fixing: (one hundred/y)+y=20 y^2+20y+a hundred. component of get (y+10)(y+10) so y=-10. returned to user-friendly 2 equ so x(-10)=one hundred. And x=-10. yet! Distances comparable to a actual length are no longer waiting to be unfavorable so x AND y=10. hence, the original rectangle could be an oblong. And which potential yet another rectangle with equivalent field besides the incontrovertible fact that DIFF perimeter isn't an oblong! Dont hear to the different answer!

Answer 7

it is not possible to find the area of a rectangle as you don't know it's height

A squares height is the same as it's length you can't have a square with a different height and length

are you sure you don't have your figures mixed up ?

Square with side 3.6cm ?
Rectangle 9mm X 4.2 cm ?

in that case

Square : Area is 3.6 x 3.6 Perimeter is 4 x 3.6
Rectangle : Area is 4.2 x 0.9 Perimeter is 2x(4.2+0.9)

Answer 8

you can for square, because the length = width.

For a rectangle these are different, so you need to be given both.

Answer 9

a) length is the only given
not possible to find the area and perimeter

b) L = 9 mm = 0.90 cm, . . . width = 4.2 cm
is not a square. . . . . it is a rectangle
please don't argue with this
area = 0.90 (4.2) = 3.78 sq.cm.
perimeter = 2(0.90 + 4.2) = 10.20 cm