Centres of two circles having radius 2 and 1 units are 5 units apart. The area of the quadrilateral formed by joining the points of contact of external tangents drawn to circles is equal to (in sq. units)

The problem involves two circles with radii 2 and 1 units, and their centers are 5 units apart. We are asked to find the area of the quadrilateral formed by joining the points of contact of the external tangents drawn to the circles.

Step 1: Draw the two circles with their centers 5 units apart. Label the centers as O1 and O2, and the points of tangency as A, B, C, and D.

Step 2: Draw the radii to the points of tangency. Since the radii are perpendicular to the tangents at the points of tangency, quadrilateral ABCD is a cyclic quadrilateral (a quadrilateral inscribed in a circle).

Step 3: The diagonals of a cyclic quadrilateral are perpendicular and bisect each other. Therefore, the line joining the centers O1 and O2 of the circles is the perpendicular bisector of the quadrilateral's diagonal.

Step 4: The length of the line segment O1O2 is the difference of the radii (2 - 1 = 1 unit) plus the distance between the centers of the circles (5 units), which equals 6 units.

Step 5: The length of the diagonal of the quadrilateral is twice the length of O1O2, which is 12 units.

Step 6: The area of a cyclic quadrilateral with diagonals d1 and d2 and the angle θ between them is given by the formula 0.5 * d1 * d2 * sin(θ). In this case, d1 = d2 = 12 units and θ = 90°.

Step 7: Substituting these values into the formula gives the area of the quadrilateral as 0.5 * 12 * 12 * sin(90°) = 72 square units.