2005 AMC 8 Problem 9

Attempt Problem 9 of the 2005 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

In quadrilateral

ABCD,

sides

\overline{AB}

and

\overline{BC}

both have length

10,

sides

\overline{CD}

and

\overline{DA}

both have length

17,

and the measure of angle

ADC

is

60^\circ.

What is the length of diagonal

\overline{AC}?

13.5

14

15.5

17

18.5

Answer: D

Concepts:isosceles triangle equilateral triangle angle sum

Difficulty rating: 1020

Small hint:

Look only at

\triangle ADC

.

Big hint:

It is isosceles and has a

60^\circ

angle.

Solution:

Note that

\triangle ADC

is isosceles. This means that

\angle DAC = \angle DCA.

We get that

\begin{gather*} \angle DAC + \angle DCA \\ {}+ \angle ADC = 180^{\circ} \\ \angle DAC + \angle DCA = 120^{\circ} \\ \angle DAC = \angle DCA = 60^{\circ}. \end{gather*}

This shows that

\triangle ADC

is equilateral. This gives us that

AC = CD = 17.

Thus, D is the correct answer.