Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.

2. Originally Posted by alexmahone
Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
Unfortunately I can't provide you with an algebraic solution.

In the attachment you see the only possible dimensions of the quadrilateral with approximative values.

Maybe you can use the drawing to control your exact result.

(The radius is of course $r = \dfrac h2 \approx 0.87$

3. Hello, alexmahone!

I have a start on this problem, but haven't finished it.

Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD.
$AB = 3CD$, and the area of the quadrilateral is 4.
If a circle can be inscribed in the quadrilateral, find its radius.
Code:
D - - - x - - - C
-  o-------*-*-*---o
:  |   *     :     *\
:  | *       :       *
:  |*        :r       *
:  |         :         \
:  *         :         *\
2r  *         *         * \
:  *         :         *  \
:  |         :             \
:  |*        :r       *     \
:  | *       :       *       \
:  |   *     :     *          \
-  o-------*-*-*---------------o
A - - - - - - 3x  - - - - - B
$ABCD$ is a trapezoid with two right angles.

Its height is $2r$, its bases are $AB = 3x,\:CD = x.$

The area of a trapezoid is: . $A \;=\;\frac{h}{2}(b_1 + b_2)$

So we have: . $A \;=\;\frac{2r}{2}(x + 3x) \:=\:4 \quad\Rightarrow\quad rx \:=\:1$

Now where can I find another equation?

4. Originally Posted by alexmahone
Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD; AB=3CD; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.
I hope that I now found the solution:

For convenience I set CD = t.

Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:

$A = \dfrac{t+3t}2 \cdot h = 4$ which yield $h = \dfrac2t$ and therefore $r = \dfrac1t$

The length of $x = t- \dfrac1t$ and

$y = 3t- \dfrac1t$

Since $FB = 2t$ you can calculate the sides of the right triangle CFB:

$\left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2$

Expand the brackets and collect like terms:

$\dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{ t=\sqrt{\dfrac43}}$

Re-substitute this value and you'll get $r = \dfrac12 \cdot \sqrt{3}$

EDIT: Too late

5. Originally Posted by earboth
I hope that I now found the solution:

For convenience I set CD = t.

Your quadrilateral is a trapezium with t and 3 t as the lengths of the parallel sides. The area is calculated by:

$A = \dfrac{t+3t}2 \cdot h = 4$ which yield $h = \dfrac2t$ and therefore $r = \dfrac1t$

The length of $x = t- \dfrac1t$ and

$y = 3t- \dfrac1t$

Since $FB = 2t$ you can calculate the sides of the right triangle CFB:

$\left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2$

Expand the brackets and collect like terms:

$\dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{ t=\sqrt{\dfrac43}}$

Re-substitute this value and you'll get $r = \dfrac12 \cdot \sqrt{3}$

EDIT: Too late
Thanks so much!