Find Area of Trapezoid ABCD | $(\sqrt 3+1):(3-\sqrt 3)$

[Albert1]

A trapezoid ABCD ,AD // BC ,points E and F are midpoints of AB and CD respectively
(1)area AEFD :area EBCF =($\sqrt 3+1) : (3-\sqrt 3)$
(2) area of $\triangle ABD=\sqrt 3$
please find the area of ABCD
View attachment 1123

 

[Wilmer]

Re: another trapezoid

(1)area AEFD :area EBCF =$\sqrt 3+1:3-\sqrt 3$

What does that mean? ~2.732 / ~1.268 ? Can't be...

 

[Albert1]

Re: another trapezoid

What does that mean? ~2.732 / ~1.268 ? Can't be...

that means the ratio of two areas

 

[Wilmer]

Re: another trapezoid

that means the ratio of two areas

But area AEFD is clearly lesser than area EBCF ;
but your ratio makes it greater...?

 

[Albert1]

Re: another trapezoid

But area AEFD is clearly lesser than area EBCF ;
but your ratio makes it greater...?

the diagram is not scaled

 

[Wilmer]

Re: another trapezoid

I give up!

Hope someone else understands...

 

[Albert1]

Re: another trapezoid

the diagram has been changed now

 

[Opalg]

Re: another trapezoid

A trapezoid ABCD ,AD // BC ,points E and F are midpoints of AB and CD respectively
(1)area AEFD :area EBCF =($\sqrt 3+1) : (3-\sqrt 3)$
(2) area of $\triangle ABD=\sqrt 3$
please find the area of ABCD
View attachment 1123

If $AD = x$, $BC = y$ and the perpendicular distance between $AD$ and $BC$ is $h$, then

area of the yellow region $AEFD$ is $\frac12h\bigl(\frac34x + \frac14y\bigr)$,
area of the cyan region $EBCF$ is $\frac12h\bigl(\frac14x + \frac34y\bigr)$,
area of the triangle $ABD$ is $\frac12xh$.​

Then (2) tells us that $\frac12xh =\sqrt 3$, and so $xh = 2\sqrt3$. From (1) we get $$\frac{\frac12h\bigl(\frac34x + \frac14y\bigr)}{\frac12h\bigl(\frac14x + \frac34y\bigr)} = \frac{\sqrt 3+1}{3-\sqrt 3},$$ from which $(3-\sqrt3)(3x+y) = (1+\sqrt3)(x+3y)$. Thus $(8-4\sqrt3)x = 4\sqrt3y$, from which $y = (2-\sqrt3)x/\sqrt3$, and $x+y = 2x/\sqrt3.$

The area of $ABCD$ is $\frac12(x+y)h = \frac12\,\frac2{\sqrt3}xh = \frac1{\sqrt3}(2\sqrt3) = 2.$

 

[Wilmer]

Re: another trapezoid

Just woke up to your new diagram, Albert; quite a difference;
about same as first showing a circle, then replacing it by an ellipse

Thus $(8-4\sqrt3)x = 4\sqrt3y$, from which $y = (2-\sqrt3)x/\sqrt3$, and $x+y = 2x/\sqrt3.$

VERY clever, Opal; what a "nice" way to get "x + y" (Clapping)

Noticed that the "work" can be reduced quite a bit by letting y = 1.

Quickly leads to:
(3x + 1) / (x + 3) = (1 + SQRT(3)) / (3 - SQRT(3)),
then x = SQRT(3) / (2 - SQRT(3))

 

[Albert1]

Re: another trapezoid

my solution :

View attachment 1133