Bunuel wrote:

The figure shown is a rhombus in which the measure of angle A = 120°. What is the ratio of the length of AC to the length of DB?

A. \(1:2\sqrt{3}\)

B. \(1:2\sqrt{2}\)

C. \(1:2\)

D. \(1:\sqrt{3}\)

E. \(1:\sqrt{2}\)

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Geometry plus clues . . .

THREE clues suggest that a (30-60-90)

triangle is likely to help solve:

• The figure is a rhombus

Equal side lengths,

diagonals bisect vertices,

diagonals are perpendicular

(create 90° angles),

opposite angles are equal

• Vertex A = 120° (= C)

A and

C's bisected angle = 60° + 60°

From above, diagonals create 90° angles

180° in a triangle: 60 . . . 90 . . . 30

• Answer choices (not foolproof, but a hint)

Four options' lengths are square roots

\(\sqrt{2}\) suggests a 45-45-90 triangle

\(\sqrt{3}\) suggests a 30-60-90 triangle

1) Fill in angle measures

A = C = 120 (opposite angles equal)

A + C = 240

360° = sum of interior angles

360° - 240° = 120° remains, to be split equally:

B = D = 60

2) Draw diagonals AC and BD

=> 4 identical right triangles

Angles: 90° (diagonals' intersection)

60° (bisected A and C)

30° (

bisected B and

D)

3) Lengths for

one triangle

30-60-90 triangles:

sides

lengths opposite those angles correspond, respectively,

in ratio

\(x: x\sqrt{3}: 2x\)Diagonal

AC\(= (x + x) =\) \(2x\)Diagonal

BD\(= x\sqrt{3} + x\sqrt{3}=\) \(2x\sqrt{3}\)Ratio of AC: BD?

\(\frac{AC}{BD} = \frac{2x}{2x\sqrt{3}} = \frac{1}{\sqrt{3}}\)Answer D
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