The PERIMETER of isosceles trapezoid ABCD is equal to 80cm. The length of the line AB is 4 times bigger than lenght of a CD line which is 2/5 the lenght of the line BC (or the lines which are the same in lenght). What is the area of the trapezoid?

I'm sorry for the grammar mistakes etc.

Thanks!

1 Answer
Dec 18, 2016

Answer:

Area of trapezium is #320# #cm^2#.

Explanation:

Let the trapezium be as shown below:
http://mathworld.wolfram.com/IsoscelesTrapezoid.html
Here, if we assume smaller side #CD=a# and larger side #AB=4a# and #BC=a/(2/5)=(5a)/2#.

As such #BC=AD=(5a)/2#, #CD=a# and #AB=4a#

Hence perimeter is #(5a)/2xx2+a+4a=10a#

But perimeter is #80# #cm.#. Hence #a=8# cm. and two paallel sides shown as #a# and #b# are #8# cm. and #32# cm.

Now, we draw perpendiculars fron #C# and #D# to #AB#, which forms two identical right angled trianges, whose

hypotenuse is #5/2xx8=20# #cm.# and base is #(4xx8-8)/2=12#

and hence its height is #sqrt(20^2-12^2)=sqrt(400-144)=sqrt256=16#

and hence as area of trapezium is #1/2xxhxx(a+b)#, it is

#1/2xx16xx(32+8)=8xx40=320# #cm^2#.