Perimeter = 2(width+length) =28
Let x=width of base, then length of base = (28/2)-x = 14-x
Volume of box = length*width*height
=x(14-x)*5 = 5x(14-x)
2b+h = 4.82
b+2h = 3.70
multiply the second equation by 2:
2b + 4h = 7.40
subtract the first equation from it:
3h = 2.58
h = 0.86
We compute for the side lengths using the distance formula √[(x₂-x₁)²+(y₂-y₁)²].
AB = √[(-7--5)²+(4-7)²] = √13
A'B' = √[(-9--7)²+(0-3)²] = √13
BC = √[(-5--3)²+(7-4)²] = √13
B'C' = √[(-7--5)²+(3-0)²] =√13
CD = √[(-3--5)²+(4-1)²] = √13
C'D' = √[(-5--7)²+(0--3)²] = √13
DA = √[(-5--7)²+(1-4)²] = √13
D'A' = √[(-7--9)²+(-3-0)²] = √13
The two polygons are squares with the same side lengths.
But this is not enough information to support the argument that the two figures are congruent. In order for the two to be congruent, they must satisfy all conditions:
1. They have the same number of sides.
2. All the corresponding sides have equal length.
3. All the corresponding interior angles have the same measurements.
The third condition was not proven.
Answer:
2.8 feet
Step-by-step explanation:
tan(29)=x/5
5tan(29)=x
2.8=x
