Question

Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ what is the area of $ABC?$

$12\sqrt{3}$

$8\sqrt{6}$

$14\sqrt{2}$

$20\sqrt{2}$

$15\sqrt{3}$

Answer: E

Concepts:arithmetic sequence law of cosines triangle area

Difficulty rating: 1630

Solution:

Let the sides be $6,\ 6+d,\ 6+2d.$ The $120^\circ$ angle faces the longest side, so $(6+2d)^2=6^2+(6+d)^2$ $-2\cdot 6\cdot(6+d)\cos 120^\circ.$ Using $\cos 120^\circ=-\tfrac12$ gives $3d^2+6d-72=0,$ so $d=4$ and the sides are $6,10,14.$ The area is $\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ$ $=30\cdot\tfrac{\sqrt3}{2}$ $=15\sqrt3.$

Thus, the correct answer is E.

2023 AMC 12B Problem 17

Answer: E

Difficulty rating: 1630

Solution:

Problem 17 in Other Years