Question

In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively. Find the radius of the circle that passes through all four of the points A, B, C and D? 

(a) 2√3 

(b) 2√5 

(c) 2√11 

(d) 2√7

Answer 1

(d) 2√7.

Let ABCD be the given trapezoid. Let E and F be the mid-points of the sides BC and AD respectively of trapezoid ABCD. By symmetry, the centre of the circle passing through the points A, B, C and D lies on EF. Let the centre be O. 

Then, OB = OA = r (say)          (radii of the circle) 

Let h be the height of the altitude EF of the trapezoid ABCD. Then, as we can see in the diagram, in ΔCGD,

h² + 4² = (8)² ⇒ h² = 64 – 16 = 48 ⇒ h = 4√3 

Now let EO = x, so OF = h – x = 4√3 − x

In ΔOBE, OB = r = \(\sqrt{OE^2+BE^2}\) = \(\sqrt{x^2+1}\)         ...(i)

In ΔOAF, OA = r = \(\sqrt{OF^2+AF^2}\) = \(\sqrt{(4\sqrt3-x)^2+5^2}\)         ...(ii)

∴ From (i) and (ii)

\(\sqrt{x^2+1}\) = \(\sqrt{(4\sqrt3-x)^2+5^2}\)

⇒ x² + 1 = 48 - 8√3x + x² + 25

⇒ 8√3x  = 72 ⇒ x = \(\frac{72}{8\sqrt3}\) = 3√3 

∴ r = \(\sqrt{x^2+1}\) = \(\sqrt{(3\sqrt3)^2+1}\) = \(\sqrt{28}\)

⇒ r = 2√7.