Question 1132048
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If the lengths of the bases of an isosceles trapezoid {{{highlight(cross(is))}}} {{{highlight(cross(a))}}} {{{highlight(cross(circle))}}} are 10 cm and 22 cm, and if one of the legs is 10 cm, 
then what is the length of the diagonal in cm?
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            As posted,  this text is nonsense.


            I edited it in that unique way to make sense.


            Hope you agree with my editing.


            If so,  then the solution is below.



<pre>
Make a sketch.


Let ABCD be the given isosceles trapezoid.
The bases are AB = 22 cm long;  CD = 10 cm;
the lateral sides are AD = BC = 10 cm.


Draw the perpendiculars CE and DF from the vertices C  and D to the base AB.

It is clear that the right-angled triangles ADF  and BCE are congruent.

Then the segments AF and BE have the length  {{{(AB - CD)/2}}} = {{{(22-10)/2}}} = {{{12/2}}} = 6 cm each.


The height of the trapezoid CE is the leg of the right angled triangle BCE and its length is equal to  {{{sqrt(10^2-6^2)}}} = {{{sqrt(100-36)}}} = {{{sqrt(64)}}} = 8 cm.


Now from the right-angled triangle AEC you have

    AC = {{{sqrt((10+6)^2+8^2)}}} = {{{sqrt(16^2+8^2)}}} = {{{sqrt(256+64)}}} = {{{sqrt(320)}}} = {{{sqrt(16*20)}}} = {{{4*sqrt(20)}}} = {{{8*sqrt(5)}}}.


Thus the diagonal of the trapezoid AC is  {{{8*sqrt(5)}}} cm long.   <U>ANSWER</U>
</pre>

Solved.  &nbsp;&nbsp;The answer is option &nbsp;B).