Respuesta :
Let's denote:
- L as the length of the rectangle box,
- W as the width of the rectangle box.
Given that:
- The perimeter of a rectangle box is 14 feet.
The perimeter P of a rectangle box is given by:
P = 2(L + W)
So, we have:
14 = 2(L + W)
We also know that:
- The width is 3 feet longer than the length, which means W = L + 3
Substitute W = L + 3 into the perimeter equation:
14 = 2(L + (L + 3))
14 = 2(2L + 3)
14 = 4L + 6
Now, solve for L:
4L = 14 - 6
4L = 8
L = 2
So, the length of the rectangle box is 2 feet.
Now, substitute L = 2 into W = L + 3:
W = 2 + 3
W = 5
So, the width of the rectangle box is 5 feet.
To find area A of the rectangle box:
A = L × W
A = 2 × 5
A = 10
So, the area of the rectangle box is 10 square feet.
Answer:
the length is 2 feet and, width is 5feet and area is 10 feet²
Step-by-step explanation:
The perimeter of a rectangle box is 14 feet and width is 3.
so The perimeter of a rectangle is 2(L+w), here width id 3 feet longer than L. so w=L+3.
so the perimeter of rectangle is
14 = 2(L+L+3))
= 2*(2L+3)
14 = 2*2L+2*3
14 = 4L + 6
14-6 = 4L
8 = 4L
8/4 = L
2 = L
so length is 2 feet.
width is L+3= 2+3=5, so W=5 feet and L is 2 feet
and area of rectangle is L*W = 2*5=10 feet²
area of the rectangle box is 10 feet ²
