Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The height of a trapezoid is $10$ cm. The lengths of the two diagonals of the trapezoid are $30$ cm and $50$ cm. Calculate the area of the trapezoid.

On the homework I solved this using $${D_1D_2\over 2}$$ and my teacher marked me wrong. So I don't know what I did wrong. Please help. I know I can only use the formula if the diagonals are $90$ degrees. But how do I check that ?

share|cite|improve this question
that is the formula of the area of a rhombus. – lab bhattacharjee Dec 20 '12 at 5:02

To help you think about the solution, draw two parallel lines 10 cm apart. The bottom will contain the B1 and the top will contain the B2. Draw the two diagonals at the proper lengths from the base to the top line making sure they cross. Now imagine that you slide one of the diagonals along the bottom and top lines. Notice that (B2+B2)/2 does not change (i.e one stretches and one shrinks). So no matter where the diagonals cross you have the same area. Now slide them apart until the intersection point reaches the top line (i.e. B2 = 0). Now you have triangle with the same area as the trapezoid. You know the height and two sides of the triangle, a little geometry and you can compute the base and your're there.

share|cite|improve this answer

enter image description here

The area will be $\frac12\cdot 10\cdot (y+x+y+z)=5(x+2y+z)$

Now, $(y+z)^2+10^2=50^2$ and $(x+y)^2+10^2=30^2$

$(y+z)=\sqrt{50^2-10^2} CM=20\sqrt6 CM$

$(x+y)=\sqrt{30^2-10^2} CM=20\sqrt2 CM$

SO, the area will be $5(20\sqrt2(\sqrt3+1)) CM^2$

share|cite|improve this answer

Gluing two copies of the trapezoid together with one rotated by 180 degrees can yield a parallelogram. One diagonal from each trapezoid cuts the parallelogram into two triangles, each of which has an altitude of 10 and sides adjacent to that altitude's vertex of lengths 30 and 50.

So the area is the same as that of a triangle with two sides of 30 and 50, and an altitude between them of 10. Is this enough of a help?

enter image description here

share|cite|improve this answer
Trying to understand what you said... Is the answer 193.1 cm? – NGPP1 Dec 20 '12 at 0:35
This is the right answer, but I don't think your gluing construction works. – Peter Shor Dec 20 '12 at 0:36
I am still confused – NGPP1 Dec 20 '12 at 0:42
@PeterShor, in my method, the area is around $386.37 cm^2$ – lab bhattacharjee Dec 20 '12 at 5:03
@PeterShor I have added imagery that demonstrates the gluing construction works. I leave the labeling of edges and confirmation to you. The trapezoid's area is the same as that of the triangle at the end of the process: a triangle with altitude 10 and two surrounding sides of length 30 and 50. – alex.jordan Dec 20 '12 at 10:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.