Given=
length of the segment AD is 28 cm
distance between the midpoints of segments AB and CD is 16 cm
find out length of BC
To proof
AD = 28 cm
let the midpoint of the AB is E.
let the midpoint of the CD is F.
E & F are the midpoints i.e these points divide AB & CD in two equal parts.
Let BC = z
Let AE = EB = x ( E is midpoint)
Let CF = FD = y (F is midpoint)
the equation becomes
2x + 2y + z = 28
x + y + z = 16
mulitipy above equation by 2
we get
2x + 2y + 2z = 32
thus solving the equations
2x + 2y + 2z = 32
2x + 2y + z = 28
we get
z = 4 cm
i.e BC = 4 cm
Hence proved
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9514 1404 393
Answer:
a square 175 m on a side
Step-by-step explanation:
Let x and y represent the sides of the rectangle. Then the perimeter is ...
P = 2(x + y) = 700
x + y = 350 . . . . . . . divide by 2
y = 350 -x . . . . . . . . subtract x
The area of the fenced field is ...
A = xy
A = x(350 -x)
This is a quadratic function that has zeros at x = 0 and x = 350. The axis of symmetry is x = (0 +350)/2 = 175. The vertex (maximum area) is on the axis of symmetry, so corresponds to x = 175. The y-value there is ...
y = 350 -x = 350 -175 = 175
That is, the maximum area will be obtained when the fenced area is a square. Each side of the square is 175 m, which is 1/4 of the total length of the fence.
The dimensions of the space are 175 m by 175 m.
