Question:

Step 1: Define the geometry and variables. Let the quadrilateral ABCD be a right trapezoid with AB parallel to CD, and AD perpendicular to AB and CD. Let $CD = a$. Given $AB = 2CD$, we have $AB = 2a$. Let $AD = h$. The area of the trapezoid is given as 18. Step 2: Relate the variables using the area formula. The area of a trapezoid is given by $\frac{1}{2}(\text{sum of parallel sides}) \times \text{height}$. $$ \text{Area} = \frac{1}{2}(AB + CD) \times AD $$ $$ 18 = \frac{1}{2}(2a + a) \times h $$ $$ 18 = \frac{3ah}{2} $$ $$ 36 = 3ah $$ $$ ah = 12 $$ Step 3: Apply the condition for an inscribed circle. For a quadrilateral to have an inscribed circle (i.e., be a tangential quadrilateral), the sums of opposite sides must be equal. $$ AB + CD = AD + BC $$ $$ 2a + a = h + BC $$ $$ 3a = h + BC $$ Step 4: Calculate the length of side BC. To find BC, construct a right triangle by drawing a perpendicular from C to AB, let's call the intersection point E. Then $AE = CD = a$. $EB = AB - AE = 2a - a = a$. $CE = AD = h$. Using the Pythagorean theorem in $\triangle CEB$: $$ BC^2 = EB^2 + CE^2 $$ $$ BC^2 = a^2 + h^2 $$ $$ BC = \sqrt{a^2 + h^2} $$ Step 5: Solve for 'h' in terms of 'a'. Substitute the expression for BC into the tangential condition from Step 3: $$ 3a = h + \sqrt{a^2 + h^2} $$ Isolate the square root term: $$ \sqrt{a^2 + h^2} = 3a - h $$ Square both sides: $$ a^2 + h^2 = (3a - h)^2 $$ $$ a^2 + h^2 = 9a^2 - 6ah + h^2 $$ Subtract $h^2$ from both sides: $$ a^2 = 9a^2 - 6ah $$ Rearrange the terms: $$ 6ah = 8a^2 $$ Since $a$ represents a length, $a \neq 0$. Divide by $a$: $$ 6h = 8a $$ $$ h = \frac{8a}{6} = \frac{4a}{3} $$ Step 6: Determine the side lengths of the trapezoid. Substitute $h = \frac{4a}{3}$ into the area equation $ah = 12$ from Step 2: $$ a \left(\frac{4a}{3}\right) = 12 $$ $$ \frac{4a^2}{3} = 12 $$ $$ 4a^2 = 36 $$ $$ a^2 = 9 $$ Since $a$ is a length, $a = 3$. Now, calculate the lengths of all sides: $$ AB = 2a = 2(3) = 6 $$ $$ CD = a = 3 $$ $$ AD = h = \frac{4a}{3} = \frac{4(3)}{3} = 4 $$ $$ BC = \sqrt{a^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$ Step 7: Calculate the radius of the inscribed circle. For a right trapezoid with an inscribed circle, the height of the trapezoid is equal to the diameter of the inscribed circle. Thus, $AD = 2r$. Since $AD = 4$, we have: $$ 2r = 4 $$ $$ r = 2 $$ Alternatively, the area of a tangential quadrilateral is given by $Area = r \times s$, where $r$ is the inradius and $s$ is the semiperimeter. The perimeter $P = AB + CD + AD + BC = 6 + 3 + 4 + 5 = 18$. The semiperimeter $s = \frac{P}{2} = \frac{18}{2} = 9$. Using the area formula: $$ 18 = r \times 9 $$ $$ r = \frac{18}{9} = 2 $$ Both methods confirm that the radius of the inscribed circle is 2.