If the distance between the parallel sides of a trapezium is 4 cm, how can the lengths of the 4 sides be computed?
A trapezium has a perimeter of 22 cm. The 2 parallel sides are such that the length of one is three times the length of the other. The non-parallel sides are equal. If the distance between the parallel sides is 4 cm, find the lengths of the 4 sides.
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answers:
sepia say: A trapezium has a perimeter of 22 cm.
The 2 parallel sides are such that the length of one is three times the length of the other.
The non-parallel sides are equal.
If the distance between the parallel sides is 4 cm, find the lengths of the 4 sides.
A + B + C + D = 22 cm
A = 3D
B = C = 5 cm
Solution:
A = 9 cm, B = 5 cm, C = 5 cm, D = 3 cm
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electron1 say: Let x be the width of the top of the trapezoid. The width of the bottom is 3 x. Let “a” be the length of each non-parallel sides.

x + 3 x + a + a = 22
4 x + 2 a = 22
Eq#1: 2 x + a = 11

The non-parallel side is the hypotenuse of a right triangle. Each right triangle has a height of 4 cm. The following equation shows the relationship between the three sides of a right triangle.

Eq#2: a^2 = 4^2 + x^2

Now we have two equations and two variables. Let’s solve the first equation for a.

a = 11 – 2 x
Let’s square this equation,

a^2 = 121 – 44 * x + 4 x^2

Let’s substitute this for a^2 in the second equation.

121 – 44 * x + 4 x^2 = 16 + x^2
3 x^2 – 44 x + 105 = 0

Let’s solve this quadratic equation for x.

x = [44 ± √(-44^2 – 4 * 3 * 105)]
x = [44 ± 26] ÷ 6

x = [44 + 26] ÷ 6 = 11⅔ cm

OR

x = [44 – 26] ÷ 6 = 3 cm

Let’s substitute these two numbers into the first equation.

2 * 11⅔ + a = 11
a = -12⅓

This is not the correct answer. Let’s use 3 cm for x.

2 * 3 + a = 11
a = 5 cm

To check these two answers, let’s determine the perimeter of the trapezoid.

Perimeter = 3 + 5 + 9 + 5 = 22 cm

I hope this is helpful for you.
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lenpol7 say: The longer base has an overlap of 'l'
Hence 3l = longer base and l - shorter base.
Then by Pythagoras
Sloping line is
'sl'^2 = 4^2 + l^2
sl = sqrt(4^2 + l^2)
Since we have two 'sl' then
2sl + 3l + l = 22
Substituting
2sqrt(4^2 + l^2) + 4l = 22
sqrt(4^2 + l^2) + 2l = 11
sqrt(4^2 + l^2) = 11 - 2l
4^2 + l^2 = (11 - 2l)^2
16 + l^2 = 121 - 44l + 4l^2
3l^2 - 44l + 105 = 0
Apply the Quadratic equation
l = { - - 44 +/- sqrt[(-44)^2 - 4(3)(105)]} / 2(3)
l = { 44 +/- Sqer(1936 - 1260)} / 6
l = { 44 +/- sqrt[676]} / 6
l = { 44 +/- 26} / 6
l = 70/6 = 11 2/3
or
l = 18/6 = 3
Hence
By Pythagoras
2sl + 3 + 3(3) = 22
2sl + 12 = 22
2sl = 10
sl = 5
Hence the sides are
l = 3
3l = 9
sl = 5
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Captain Matticus, LandPiratesInc say: B = 3b

B + b + s + s = 22
3b + b + 2s = 22
4b + 2s = 22
2b + s = 11
s = 11 - 2b

If you draw vertical lines down from the upper base to the lower base, you'll divide the lower base into 3 segments of b length

b^2 + 4^2 = s^2
b^2 + 16 = (11 - 2b)^2
b^2 + 16 = 121 - 44b + 4b^2
0 = 3b^2 - 44b + 105
b = (44 +/- sqrt(44^2 - 4 * 3 * 105)) / 6
b = (44 +/- sqrt(44 * 44 - 4 * 315)) / 6
b = (44 +/- 2 * sqrt(11 * 44 - 315)) / 6
b = (22 +/- sqrt(484 - 315)) / 3
b = (22 +/- sqrt(169)) / 3
b = (22 +/- 13) / 3
b = 9/3 , 35/3
b = 3 , 10.66666.....

b = 3
B = 9
s = 5

b = 10.6666.... is too big. It's extraneous
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