Circle Inscribed in a Trapezium

Let the foot of perpendicular from $O$ to $AB,BC,CD,DA$ be $E,F,G,H$ respectively.

Observe that $EO=OH=R$ (radius of circle), $AE = AH$ (tangents to circle), thus $\angle OEH = \angle OHE$ ; thus $\angle AEH = \angle AHE$ . Thus $AOH$ and $AOE$ are congruent triangles, which means that they have the same area. Similarly, $BOE$ and $BOF$ , $COF$ and $COG$ , $DOG$ and $DOH$ are congruent.

Thus $\frac {1}{2} (AD*R) + \frac {1}{2} * (BC*R) = \frac {1}{2} (600)$ , $50*R = 600$ , so $R = 12$ .

Let the tangent point of AB is P , tangent point of BC is Q , tangent point of CD is R , tangent point of DA is S. And let: AP = x then by tangent line theory AS = x and we know that SD = 25-x , and by tangent line theory we know that DR = 25-x.

Let PB = y , then according to logic that has been mentioned before, we know that BQ = y QC = 25-y and CR = 25-y

Now we use the area of trapezoid ABCD information to get the height of the trapezoid. $Area = \frac{(AB+CD)height}{2}$

$600 = \frac{(x+y+25-x+25-y)height}{2}$

$1200 = 50height$

$height = 24$

And the radius of the circle is $= 1/2 height = 12$

Since every isosceles trapezoid is symmetric about the common perpendicular to its bases the height of the given trapezoid is $2r$ . The area of a trapezoid is given by $\frac {1}{2} \cdot h \cdot (a + b)$ where $a$ and $b$ is the length of the bases and since the area of the trapezoid is 600 we combine to get:

$600 = \frac {1}{2} \cdot 2r \cdot (a + b) = r(a + b)$

From that we conclude:

$r = \frac{600}{(a + b)}$ (*)

By the Pitot Theorem the sum of the lengths of the legs are equal to the sum of the length of the bases in an isosceles tangential trapezoid. Therefore:

$a + b = 25 + 25 = 50$

From (*) we finally get what we want:

$r = \frac{600}{50} = 12$ .

The given trapezium is an isosceles trapezium, let $CD=2y$ where y is the length of the tangent either from $D$ or $C$ to the circle,also let $d$ be the height of trapezium. let the tangent from $D$ & $C$ meet circle at $T$ & $U$ respectively from above $DT = CU =y$ also $AT =BT = 25-y$ , $AB = 25-y+25-y = 50-2y$ & $CD = 2y$ from above using area of trapezium = (sum of opposite parallel sides)d/2 = 600 => $(50-2y+2y)d =600$ => $d= 24$ . If you draw a diagram you can clearly see that this d is nothing but the diameter of the circle hence the radius will be 12.

Let $E$ , $F$ , $G$ and $H$ be the points that $\Gamma$ is tangent to $AB$ , $BC$ , $DC$ and $AD$ , respectively. Let $AE = a$ . By symmetry we have $AE = BE$ and $DG = GC$ . Thus from circle properties, we have $AH = AE = BE = BF = a$ and $DH = DG = GC = CF =25 - a$ .

Let $r$ be the radius of $\Gamma$ . By symmetry, we have that $2r$ is the height of the trapezium. The area of $ABCD$ is $600 = \frac{1}{2}(AB + DC)(2r) = r(2AE + 2DG) = r(2a + 50 - 2a) = 50r$ . Hence $r = \frac{600}{50} = 12$ .