SOLUTION: Find the height in cm, of trapezoid ABCD A) 56 5/13 B) 60 10/13 C) 59 8/13 D) 55 6/13 E) 57 1/13 https://imagizer.imageshack.com/v2/320xq90/r/923/c58oz8.jpg

Algebra ->  Points-lines-and-rays -> SOLUTION: Find the height in cm, of trapezoid ABCD A) 56 5/13 B) 60 10/13 C) 59 8/13 D) 55 6/13 E) 57 1/13 https://imagizer.imageshack.com/v2/320xq90/r/923/c58oz8.jpg      Log On


   

Question 1166670: Find the height in cm, of trapezoid ABCD
A) 56 5/13
B) 60 10/13
C) 59 8/13
D) 55 6/13
E) 57 1/13
https://imagizer.imageshack.com/v2/320xq90/r/923/c58oz8.jpg
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.


            It was not simple task for me to find the way --- but I found it (!)

            Unfortunately,  the solution goes through  UGLY  calculations --- but I do not see other way
            and therefore,  I was  FORCED  to go through it.


I will show the solution step by step.
Watch my steps attentively. 


1)  We have right angled triangle ABE.  The hypotenuse AB is 61 units long; the leg AE is 60 units long.

    Hence, the leg BE is  sqrt%2861%5E2-60%5E2%29 = 11.



2)  We have right angled triangle ADE.  The hypotenuse AD is 156 units long; the leg AE is 60 units long.

    Hence, the leg ED is  sqrt%28156%5E2-60%5E2%29 = 144.



3)  The area of the triangle ABD is  S%5BABD%5D = %281%2F2%29%2ABD%2AAE = %281%2F2%29%2A%2811%2B144%29%2A60 = 4650 square units.



4)  Triangles ADE and BCD are similar (they are right-angled and have congruent angles DBC and BDA).

    From similarity, we have this proportion 

        abs%28DC%29%2Fabs%28BD%29 = abs%28AE%29%2Fabs%28DE%29,  or  abs%28DC%29%2F%2811%2B144%29 = 60%2F156 = 5%2F13.

        hence  |DC| = %28155%2A5%29%2F13.



5)  Then the area of the triangle BCD  is  S%5BBCD%5D = %281%2F2%29%2ADC%2ABD = %281%2F2%29%2A%28155%2A5%2F13%29%2A155 = 120125%2F26.




6)  The area of the trapezoid ABCD is the sum of areas of triangles  

        S%5BABCD%5D = S%5BABD%5D + S%5BBCD%5D = 4650 + 120125%2F26 = %284650%2A26%2B120125%29%2F26 = 241025%2F26.




7)  From the similarity of triangles ADE and BCD we have this proportion

        abs%28AD%29%2Fabs%28AE%29 = abs%28BC%29%2Fabs%28DC%29,  or   156%2F60 = abs%28BC%29%2F%28%28155%2A5%29%2F13%29%29

    
    From the proportion,  |BC| = %28156%2A155%2A5%29%2F%2813%2A60%29 = %2812%2A155%2A5%29%2F60 = 155.




8)  Now the area of the trapezoid ABCD is half sum of its bases AD and BC multiplied by the height of the trapezoid, or

        %28%28156%2B155%29%2F2%29%2AH = 241025%2F26.


    It gives  H = %282%2A241025%29%2F%28311%2A26%29 = 59 4976%2F%28311%2A26%29 = 59 16%2F26 = 59 8%2F13.    ANSWER

SOLVED.


I hope I deserved your  "THANKS"  for my efforts --- so I am open to accept them (!)




Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
.
AD ∥ CF  is given

AF ∥ CD  because they are both ⊥ DE

AFCD is a parallelogram because both pairs of opposite sides are ∥.

Use the Pythagorean theorem on right triangles △ADE and △ABE, and
we get DE = 144 an BE = 11.

△BEF ∽ △BDC  because  EF ∥ CD

BE%2FBF=BD%2FBC corresponding parts of similar triangles are in the
                  same ratio.

BE%2FBF=%28BE%2BDE%29%2F%28BF%2BCF%29

11%2FBF=%2811%2B144%29%2F%28BF%2B156%29

Solve that for BF and get

BF=143%2F12

Then using the Pythagorean theorem in rt. △BEF, we get

EF=55%2F12

Draw the altitude DG of of △BCD, DG ⊥ BC, where DG is 
the height of the trapezoid ABCD.

rt. △BDG ∽ rt. △BEF

DG%2FBD=EF%2FBF

DG%2F%2811%2B144%29=%2855%2F12%29%2F%28143%2F12%29

Solve that for DG and get

DG=775%2F13

DG=59%268%2F13

Edwin