Missing leg of trapezoid

[eternenyi23]

So I have a trapezoid problem and I forgot how to find the missing leg, I have no angles only the height, the bases, and one side. I know I should be able to figure this out, but I forgot how to do it. I would like some help with this, thank you.

 

[Deleted member 4993]

So I have a trapezoid problem and I forgot how to find the missing leg, I have no angles only the height, the bases, and one side. I know I should be able to figure this out, but I forgot how to do it. I would like some help with this, thank you.

Please post the EXACT problem statement.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment

 

[lev888]

Post a drawing. I think right triangles may be helpful.

 

[eternenyi23]

This is the problem.

 

[eternenyi23]

I tried to see if there were any special triangles in there, but I couldn't solve it for any. I also divided it into a square and 2 triangles, but I didn't know what to do from there without any angles

 

[lev888]

I tried to see if there were any special triangles in there, but I couldn't solve it for any. I also divided it into a square and 2 triangles, but I didn't know what to do from there without any angles

Draw the other height. Look at the triangle on the right. Would it help you if you calculated its base?

 

[eternenyi23]

Draw the other height. Look at the triangle on the right. Would it help you if you calculated its base?

I drew the other height, it would help if I calculated it's base. That would be all I needed. After that, I could do everything else.

 

[HallsofIvy]

You don't need angles. If you drop a perpendicular from the right end of the upper base, you cut off a right triangle that has hypotenuse of length 10 and one leg of length 6. By the Pythagorean theorem, the other leg has length \(\displaystyle \sqrt{100- 36}= \sqrt{64}= 8\) (a "3-4-5 right triangle"). Since the lower leg has length 17 the two perpendicular cut it into length 8, 7, 2. That is, the left perpendicular cuts off a base of length 2 so that you have a right triangle with legs of lengths 2 and 6. Use the Pythagorean theorem again to determine y, the length of the hypotenuse.

 

[Deleted member 4993]

I tried to see if there were any special triangles in there, but I couldn't solve it for any. I also divided it into a square and 2 triangles, but I didn't know what to do from there without any angles

Draw some more lines as follows:



mAA'D = mBB'D = 90^o

And

AD || BD'

Now what do you get?

 

[eternenyi23]

Ok, I got it now. Thank you for your help.