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

Instead of full coordinates, drop perpendiculars from D and C onto line AB and use right-triangle trigonometry, applying the distance formula only in the last step.

1. Drop a perpendicular from D to AB, meeting it at M. In right triangle AMD, angle A = 60 degrees and AD = 12, so $AM = 12\cos60^\circ = 6$ and $DM = 12\sin60^\circ = 6\sqrt3$.
2. Drop a perpendicular from C to AB, meeting it at N. In right triangle BNC, angle B = 60 degrees and BC = 10, so $BN = 10\cos60^\circ = 5$ and $CN = 10\sin60^\circ = 5\sqrt3$. Since N is measured from B towards A, N sits at a distance $AB - 5$ from A.
3. Find the horizontal gap between M and N. M is at distance 6 from A, N is at distance $AB - 5$ from A, so the horizontal separation is $(AB - 5) - 6 = AB - 11$.
4. Find the vertical gap between M and N. $DM = 6\sqrt3$ and $CN = 5\sqrt3$, both measured upward from AB, so the vertical separation is $6\sqrt3 - 5\sqrt3 = \sqrt3$.
5. Apply the distance formula for CD. $CD^2 = (AB-11)^2 + (\sqrt3)^2$. With CD = 14: $196 = (AB - 11)^2 + 3$, so $(AB-11)^2 = 193$ and $AB = 11 + \sqrt{193}$.

Comparing with $AB = a + \sqrt b$, we get $a = 11$ and $b = 193$, so $a + b = 11 + 193 = 204$, which is option C.

Let's summarize:

• Dropping perpendiculars from D and C onto AB turns the 60 degree angle conditions into simple right-triangle trig.
• Side CD then becomes the hypotenuse of a right triangle formed by the horizontal and vertical gaps between the two feet of the perpendiculars.

So $a + b = 204$.