how do you answer this problem????

Let length = L and width = W then L = W + 13

Also half perimeter = L +W = 2W + 13 = 51

Thus 2W = 38 , W = 19 and L = 19 + 13 =32

Area of rectangle = LW = 32 x 19 = 608

Answer: 608 square cm

x = length
y = width

x - y = 13 (solve by using substitution)
x + y = 51

x - y = 13
x = 13 + y

x + y = 51
(13 + y) + y = 51
13 + y + y = 51
2y = 51 - 13
2y = 38
y = 38/2
y = 19

x - y = 13
x - 19 = 13
x = 13 + 19
x = 32

Area of the rectangle:
xy
= (32)(19)
= 608 square centimeters

Let 'l' be the length of the rectangle and 'b' be the breadth of the rectangle.

Perimeter of the rectangle is 2(l + b). But it's given that half of the perimeter is 51cm. SO:

½ * 2(l + b) = 51.
=> l + b = 51-----------1.
=> l - b = 13------------2.

Adding 1 and 2:
2'l' = 64.
So l = 32 cm.

Since l + b = 51, so:
b = 51 - 'l'
=> b = 19 cm.

The length of the rectangle is 32 cm and
the breadth of the rectangle is 19 cm.

Area of he rectangle is l × b, which is:
32 × 19 = 608 cm². Hope that helped...

if half the perimeter is 51 then thats two sides. call them x and y
the x + y = 51
but one side is 13 longer (say x)
so
y + (y+13) = 51
2y = 51 -13
2y = 38
y = 38/2
y = 19

Check you answer
there fore x = 32 (y+13)
check x + y === 19 + 32 = 51

Area = 19 * 32 = 608 cm sm

P - perimeter
L - length
W - width
A - area

Given :
1/2P = 51

1/2P=L+W if W+13 = L
then 1/2P = W+W+13
51 = 2W+13
2W = 51-13
2W = 38
W = 19cm
since L = W+13
then L = 32cm

A = (L) (W)
A = 32 x 19
A = 608cm
hope this helps

EASY! Perimeter = 2L + 2W, so L+ W = 51. Also L-W = 13

L+W = 51
L-W = 13............Add these two equations
-------------
2L = 64
L = 32
W = 51-32 = 19
Area = LW = 19 x 32 = 608 cm^2

ok? did I help?

51 is half the perimeter, so it is also the sum of two sides.

32 and 19 are 13cm apart and add to make 51.

32*19 = 608cm squared

Perimeter = 2L + 2W, so L+ W = 51. Also L-W = 13

L+W = 51
L-W = 13............Add these two equations
-------------
2L = 64
L = 32
W = 51-32 = 19
Area = LW = 19 x 32 = 608 cm^2