Answer:
Let X represent the shorter base, and drop the heights from shorter to longer base.
Two congruent right triangles are formed, because the trapezoid is isosceles and
the heights are the same.
The base angles are then 70 degree each, as they total 140.
So the height is 7* sin 70 = 6.5778483455....
The base of each right triangle is 11 - x/2
Now the algebra gets ugly.
By pythagorean theorem:
(11 - x/2)^2 + (7sin70)^2 = 7^2
(11-x/2)^2 + 49sin70^2 = 49
(11 - x/2)^2 = 49 - 49sin70^2
(11 - x/2)^2 = 49(1 - sin70^2)
(11 - x/2)^2 = 49 cos70^2 <-- trig identity : sin^2 + cos^2 = 1 ---> cos^2 = 1 - sin^2
121 - 11x + x^2/4 = 49 cos70^2
484 - 44x + x^2 = 196 cos 70^2
x^2 - 44x + 484 - 196cos70^2 = 0
x^2 - 44x + C = 0 where C = 484 - 196cos70^2 = 461.07235
By quadratic formula:
X = 44 +OR- square-root( (-44)^2 - 4*C) / 2
= 44 +OR- square-root( 91.71)/2
The two roots are:
26.78282 AND 17.2117331
The first one is disqualified and we must reject it
because if used will make the base of the
right triangles negative. Remember this is
supposed to be the smaller base. So it
cannot be greater than 22.
SO the smaller base is 17.2117331
So 11 - x/2 = overhang = 2.39413345
Drawing the diagonal, it's length can
be found by pythagorean theorem.
square-root(17.2117331)^2 + (7sin70)^2)=18.4258472....<-- diagonal
