Trapezoid height?

[Demonic]

 

I tried splitting up the image into 3 parts, 2triangles, retangle, and tried doing the Pytagorean Theorem on the triangle, no luck because each triangles base can't be 5 since it isn't a isosceles trapezoid.  My teacher had problems with it in the teacher manual said they split it into a parallelogram and a triangle, but I still won't get the height of the triangle because I can't define if its a isosceles triangle.

 

[discomatt]

Unless you're given an angle, i don't think it's possible. Too many unknowns. I would love to be proven wrong, but I don't think it's possible.

 

A  ____________ B
   /            \
  /              \
/                \
/__________________\
D                   C

 

The easiest way (i find) to calculate a non-isosceles trapezoid area with unknown height is to divide it into 2 triangles by drawing a line through either AC or BD. After doing this, you have 2 side lengths and no angles for each triangle... and that doesn't compute sadly.

 

http://www.math.psu.edu/geom/koltsova/section13.html covers the math behind a trapezoid fairly well, and I still think you have too many unknowns

[mb81]

Use some creative geometry and the law of cosines

 

If you draw two lines perpendicular to the bottom line segment that intersect the top vertices and put the triangle together, you have a scalene triangle with the following properies: base of 10 (difference between the top line segment and the bottom), one side of 6, and one side of 8.

 

The law of cosines says:

For a triangle with sides a, b, c opposite angles A, B, and C respectively then, a^2 = b^2 + c^2 - 2bccosA

 

If your side a = 6, your side b=8, and your side c = 10, then

36 = 64+100-2*8*10cosA

-128 = -160cosA

cosA = 128/160

 

If you draw another perpendicular line down through the triangle from the top point to the bottom line, then you have a right triangle with a known angle and a known side. The sin of A will be equal to the opposite (the height of the trapezoid) /  the hypotenuse , so the height of the trapezoid is equal to the the hypotenuse * the sin of A.

 

Hope that helps.

 

 

[mb81]

Here's the long version of the formula.

 

height = side1 * sine ( inverse cosine ( side1^2-side2^2-(top-bottom)^2 / -2 (side2) (top-bottom) ) )

 

side1 being one the length of one of the sides, side 2 being the length of the other side, top being the length of the top, bottom being the length of the bottom.

 

THERE MAY BE ERRORS IN THIS WORK, DEFINITELY DOUBLE CHECK ME

[GingerRobot]

I think there's an error there - the side1s and side2s in the cosine expression need to be swapped around. Also, a couple brackets missing:

 

h= s1*sin(cos^-1((s2^2-s1^2-(t-b)^2)/(-2*s1*abs(t-b))))

 

And if i was being picky, the top-bottom would give a negative answer, hence the abs().

 

Good work though - took me a while to spot it!

 

I make h equal to 4.8

 

Oh, and Demonic: I think your assumption that you get a right angle by splitting into a parallelogram and triangle is wrong. Otherwise this would be even easier with some normal trig.

[Demonic]

and I still think you have too many unknowns

Me and my teacher thought the same and said that problem was bazaar for a "normal" problem.

 

Oh, and Demonic: I think your assumption that you get a right angle by splitting into a parallelogram and triangle is wrong.

Uh the teachers manual said it was a right triangle. (Teacher was trying ti figure this out and she ended up looking, but didn't want to tell us the height)

 

Sorry for my bad English

the line that is || the side of length 6 is right. the triangles bottom sides length is 10 (32-22=10)

 

So you get a triangle with sides 6 8 10 these numbers are specific for a triangle with 90*. to get the height u firstly need to get the triangles square. square = 6*8/2.

And finally the height = 2*square/a . Where a = 10.

If the triangle you got is without the 90* corner then the formula for its square is S=sqrt(p(p-a)(p-b)(p-c)) the formula is called Heron formula. And p = (a+b+c)/2 a,b,c are the sides of the triangle

 

square of triangle(sot) = a*b/2

sot = 6*8/2

sot = 48/2

sot = 24

 

 

//a is the side of triangle to which the height is pointing

// if you imagine a triangle ABC and the height is A to a point on BC then BC is that "a"

height = 2*square/a

height = 2*24/10

height = 48/10

height = 4.8

 

trapizoid area(ta) = (h(b1+b2))/2

ta = 4.8(22+32) / 2

ta = 4.8*54 / 2

ta = 259.2 / 2

ta = 129.6 units squared

[AndyB]

 

Looks pretty straightforward to me.

[Demonic]

 

Looks pretty straightforward to me.

 

How did you get the 10-X

[GingerRobot]

Probably an easier approach andy!

 

The 10-X comes from the fact that the bases of the two triagles must be equal to 10, since the bottom of the trapeziod has length 32 and the top has length 22. The difference (which is what the bases of the triangles are) is therefore 10.

 

If you then say one of the bases has length X, then the other one must have whatever's left from 10: 10-X

[Barand]

 

Looks pretty straightforward to me.

 

Leaving a rectangle 22 x H

 

and a 6 - 8 - 10 rightangled triangle

 

so

 

$H = 8 * sin( asin(6/10) );

[mb81]

 

Looks pretty straightforward to me.

 

Leaving a rectangle 22 x H

 

and a 6 - 8 - 10 rightangled triangle

 

so

 

$H = 8 * sin( asin(6/10) );

 

Are you sure that the resulting triangle from when you remove the center rectangle is a right triangle? I don't think that is going to happen in most cases, so I am not sure you can apply that to every case.

 

 

[Barand]

But for the purposes of solving this particular problem, it is, and I expect the dimensions were specifically chosen so it would be.