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# maths (pls help)

Sat 26 May, 2012 05:31 am
A trapezium is drawn inside the semicircle of radius 1m such that all the corners of trapezium are at the circumference of semicircle and one side of trapezium is diameter of semicircle. what is the maximum sum of other three sides of trapezium. plssss help....
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Type: Discussion • Score: 0 • Views: 544 • Replies: 9
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fresco

1
Sat 26 May, 2012 06:29 am
@sandy111,
I suggest the trapezium forms one half of a regular hexagon inscribed in the circle. Hence the maximum length of the three sides should be 3R (R=Radius of circle)
sandy111

1
Sat 26 May, 2012 06:34 am
@fresco,
how can say its regular hexagon with equal sides/
fresco

1
Sat 26 May, 2012 06:41 am
@sandy111,
Because the greater the regular "curvature" of the shape (in this case hexagon) the greater the perimeter.
Note the closer the side opposite the diameter gets to it, the closer the total reduces to 2R.
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fresco

1
Sat 26 May, 2012 07:49 am
@sandy111,
Note also as you move the side parallel to the diameter away, i.e. reduce its size, the maximum total for the three sides approaches 2Rxsqrt[2]=2.88R
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markr

1
Sat 26 May, 2012 12:20 pm
@sandy111,
According to Wolfram's Mathworld,
"There are two common definitions of the trapezium. The American definition is a quadrilateral with no parallel sides; the British definition is a quadrilateral with two sides parallel (e.g., Bronshtein and Semendyayev 1977, p. 174)--which Americans call a trapezoid."

For the British definition, it would be rather easy to prove that fresco's answer is correct as there is only one variable (distance between the parallel sides).

I don't have a proof, but for the American definition, it is essentially the same polygon. You can construct an American trapezium with a perimeter length that is arbitrarily close to the length of the perimeter of the British trapezium.
raprap

1
Sat 26 May, 2012 01:06 pm
@markr,
beat me to it.
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engineer

1
Sat 26 May, 2012 03:24 pm
@sandy111,
What level is this course? I can easily prove the angles are equal therefore the sides are equal, but I'd use calculus.

There are three unknown sides of the trapezoid, s1,s2 and s3 with angles a1, a2 and a3.

From law of cosines
s1^2 = r^2 + r^2 - 2r^2 cos a1 = 2r^2 (1 - cos a1)
s2^2 = 2r^2 (1 - cos a2)

Using a3 = pi - a1 - a2 and cos(pi - x) = - cos x
s3^3 = 2r^2 (1 + cos a1+a2 )

The sum of all three is 2r^2 ( 3 - cos a1 - cos a2 + cos (a1+a2))

The angle where a maximum occurs is where the partial derivative wrt a1 and a2 is zero.

The partial derivative wrt a1 = 2r^2 (sin a1 - sin(a1 + a2)) = 0
sin a1 = sin a1+a2 = 2(sin a1)(cos a2)
1 = 2 cos a2
1/2 = cos a2
a2 = pi/3

Doing the same for a2 will yield a1= pi/3 and a3 will be the same so the sides are equal and the sum is 3R. It seems like there should be a more direct way.
markr

1
Sat 26 May, 2012 03:40 pm
@engineer,
Nice.
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raprap

1
Sat 26 May, 2012 10:59 pm
@engineer,
From constructive geometry you can show that a1=a3 & s1=s3.

If you draw the trapazoid in a hemicircle then side s2 is parallel to the hemicircle diameter. Then the height of triangles created by angles a1 and a3 are the same. As the hypotaneus of these triangles is the hemicirclle radius and the height chord is a right angle with the base, the triangles are identical and the angles chords are identical. So a1=a3 and s1=s3

Now the following relations hold;
2a1+a2=180 (pi)
and length of the three sides is
l=2s1+s2

then you can use the derivitive to maximize l

The result is the same, a1=a2=60deg (pi/3)

Rap
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