A geometry problem by Julian Yu

Applying pythagorean theorem in Triangle 1 of my figure, we have

$8^2=a^2+s^2$

$64=a^2+s^2$ $\color{#D61F06}(1)$

Applying pythagorean theorem in Triangle 2 of my figure, we have

$12^2=a^2+(s+8)^2$

$144=a^2+s^2+16s+64$

$80=a^2+s^2+16s$ $\color{#D61F06}(2)$

From $\color{#D61F06}(1)$ , we get

$64=a^2+s^2$

$a^2=64-s^2$

Substituting the above in $\color{#D61F06}(2)$ , we get

$80=a^2+s^2+16s$

$80=64-s^2+s^2+16s$

$16=16s$

or

$s=1$

Therefore the lower base of the trapezoid is $1+8+1=10$ .

and the height of the trapezoid is

$a^2=64-s^2$

$a^2=64-1$

$a^2=63$

$a=\sqrt{63}=3\sqrt{7}$

The area of the trapezoid is

$A=\dfrac{1}{2}(base~1+base~2)(height)$

$A=\dfrac{1}{2}(8+10)(3\sqrt{7})$

$A=27\sqrt{7}$

Finally,

$A^2=(27\sqrt{7})^2=$ $\boxed{5103}$ $\large\boxed{\color{#20A900}\text{answer}}$

The altitude to the side of length 12 in the 8-12-8 isosceles triangle divides the triangle into two congruent right triangles. By the Pythagorean theorem, this altitude has length sqrt (8^2 - 6^2) = 2 sqrt 7, To calculate the height, h, of the trapezoid, compute the area of the isosceles triangle two ways...this gives (1/2)(12)(2 sqrt 7) = (1/2)(8)(h), so h = 3 sqrt 7. From the upper two vertices of the trapezoid, drop perpendiculars to the longer base...these form a rectangle with both bases of the trapezoid and two congruent right triangles on each end, with the shorter leg of each right triangle equal to sqrt (8^2 - (3sqrt 7)^2) = 1. Since opposite sides of a rectangle are congruent the longer base = 1 + 8 + 1 =10, so A = (1/2)(3 sqrt 7) (8 + 10) = 27 sqrt 7 and A^2 = (27 sqrt 7)^2 = 5103.

Since 8,8,12 is isosceles, the perpendicular from vertex bisects the base. So the cosine of acute angle is 6/8=3/4 and sine is root7/4. Applying these values to the equal acute angle in the lower triangle gives height as 3root7 and base as 1+8+1=10. Area of trapezium
=(1/2)(8+10)(3root7) =27root7.