Respuesta :
L=W+5
A=LW, and using L from above this becomes:
A=(W+5)W
A=W^2+5W and we are told A=1800 so:
W^2+5W=1800
W^2+5W-1800=0 factor
W^2+45W-40W-1800=0
W(W+45)-40(W+45)=0
(W-40)(W+45)=0 Since length cannot have a negative value in this problem (because it would have no meaning)
W=40m, and we were told L=W+5, L=45m
A=LW, and using L from above this becomes:
A=(W+5)W
A=W^2+5W and we are told A=1800 so:
W^2+5W=1800
W^2+5W-1800=0 factor
W^2+45W-40W-1800=0
W(W+45)-40(W+45)=0
(W-40)(W+45)=0 Since length cannot have a negative value in this problem (because it would have no meaning)
W=40m, and we were told L=W+5, L=45m
The dimensions of the rectangle field, whose length is 5m greater than its width and area 1800 m² is 45×40 meters.
What is the area of a rectangle?
Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,
[tex]A=a\times b[/tex]
Here, (a)is the length of the rectangle and (b) is the width of the rectangle
Suppose the dimension of the field is x×y, where x is the length and y is the width of the field.
As, the area of the field is 1800 m². Therefore,
[tex]xy=1800\\x=\dfrac{1800}{y}[/tex] ....1
The length of this field is 5 m greater than its width. Therefore,
[tex]x=y+5[/tex]
Put the value of x from the equation 1 in the above expression as,
[tex]\dfrac{1800}{y}=y+5\\y^2+5y-1800=0\\y^2+45y-40y-1800=0\\y(y+45)-40(y+45)=0\\(y+45)(y-40)=0[/tex]
By equation the values, we get the width -45 and 40. As the width can not be negative. Thus, the width of the field is 40 m.
Put this value in equation one as,
[tex]x=40+5\\x=45[/tex]
Hence, the dimensions of the rectangle field, whose length is 5m greater than its width and area 1800 m² is 45×40 meters.
Learn more about the area of rectangle here;
https://brainly.com/question/11202023
