Question

In a quadrilateral ABCD, BC = 10, CD = 14, AD = 12 and \(\angle CBA = \angle BAD = 60^{\circ}\). If \(AB = a + \sqrt{b}\), where a and b are positive integers, then \(a + b =\)

• A. 193
• B. 201
• C. 204
• D. 207

Hint:

Set up coordinates or drop perpendiculars from C and D onto AB, then use CD = 14 to form an equation in AB.

Answer 1

The Correct Option is C

Step 1: Set up coordinates.
Place A at the origin and B on the positive x-axis, so \(A = (0,0)\) and \(B = (c, 0)\), where c = AB is the length we want to find.

Step 2: Locate D using angle BAD.
Since \(\angle BAD = 60^{\circ}\) and AD = 12, point D lies at \(D = (12\cos 60^{\circ}, 12\sin 60^{\circ}) = (6, 6\sqrt{3})\).

Step 3: Locate C using angle CBA.
At vertex B, the angle between BC and BA is \(60^{\circ}\). BA points in the negative x-direction from B, and since the quadrilateral is convex with C on the same side as D, ray BC makes an angle of \(180^{\circ} - 60^{\circ} = 120^{\circ}\) with the positive x-axis. With BC = 10, \(C = (c + 10\cos 120^{\circ}, 10\sin 120^{\circ}) = (c - 5, 5\sqrt{3})\).

Step 4: Use CD = 14 to form an equation in c.
\[ CD^2 = (c - 5 - 6)^2 + (5\sqrt{3} - 6\sqrt{3})^2 \] \[ 196 = (c - 11)^2 + 3 \] \[ (c - 11)^2 = 193 \]
Step 5: Solve for AB.
\[ c = 11 + \sqrt{193} \] we take the positive root since AB is a length greater than 11. So \(AB = 11 + \sqrt{193}\), which means a = 11 and b = 193.

Final Answer:
\[ a + b = 11 + 193 = \boxed{204} \]