Question 253415: 4. If the perimeter of a rectangle is 8√2 cm, what is the smallest possible value of the length of one of its diagonals in cm?
I've tried:
let length be l, width be w, and diagonal be d
let l be x, then w will be:
(8√2-2x)/2 = 4√2-x
d^2=l^2+w^2
=x^2+(4√2-x)^2
=x^2+(32-8√2x-x^2)
=32-8√2x...
Found 4 solutions by stanbon, edjones, scott8148, ralhp09:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If the perimeter of a rectangle is 8√2 cm, what is the smallest possible value of the length of one of its diagonals in cm?
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Perimeter = 2(L + W) = 8sqrt(2) cm
L+W = 4sqrt(2) cm
Note:
W = 4sqrt(2)-L
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Diagonal = sqrt(L^2 + W^2)
Substitute for W:
D(L) = sqrt(L^2 + (4sqrt(2)-L)^2)
D(L) = sqrt(L^2 + [L^2 - 8sqrt(2)L +32]
D(L) = sqrt(2L^2 -8sqrt(2)L + 32)
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You have a quadratic with a = 2 ; b = -8sqrt(2)
Minimum occurs when L = -b/2a = 8sqrt(2)/(4) = 2sqrt(2)
W = 4sqrt(2)-2sqrt(2) = 2sqrt(2)
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Minimum Diagonal:
D^2 = L^2 + W^2
D^2 = (2sqrt(2))^2 + (2sqrt(2))^2
D^2 = 8 + 8
D = 4 cm
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Cheers,
Stan H.
Answer by edjones(8007)  (Show Source):
You can put this solution on YOUR website! A square has the smallest possible diagonals.
s=8sqrt(2)/4
=2sqrt(2) One side of the square
a^2+b^2=c^2
2sqrt(2)^2 + 2sqrt(2)^2 = c^2
8+8=16
c=4 cm
.
Ed
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! let length be l, width be w, and diagonal be d
let l be x, then w will be:
(8√2-2x)/2 = 4√2-x
d^2=l^2+w^2
=x^2+(4√2-x)^2
=x^2+(32-8√2x-x^2) ___ not quite ___ the x^2 in the expansion is positive (negative times negative is positive)
=32-8√2x... ___ 2x^2 - 8√2x + 32
this is a quadratic (parabola) with a minimum on the axis of symmetry
x = -b / 2a = 8√2 / 4 = 2√2
substituting ___ d^2 = 2(2√2)^2 - 8√2(2√2) + 32 = 16 - 32 + 32 = 16
d = 4
Answer by ralhp09(3) (Show Source):
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