30 30 - 60 60 - 90 90 Triangle in the Rhombus

Geometry Level 2

The diagram above illustrates Δ A B D \Delta ABD inscribed in the regular rhombus A C E F ACEF of 6 0 60^{\circ} and 12 0 120^{\circ} angles, where D D is the midpoint of E C \overline{EC} , and B D \overline{BD} is perpendicular to A C \overline{AC} .

What is the ratio of red area to green area ?

3 16 \dfrac{3}{16} 13 16 \dfrac{13}{16} 1 3 \dfrac{1}{3} None of the choices. 3 13 \dfrac{3}{13}

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1 solution

Michael Huang
Jul 3, 2017

In the given diagram, set A B = 1 |\overline{AB}| = 1 , so B D = 1 3 |\overline{BD}| = \dfrac{1}{\sqrt{3}} . Since Δ A B D Δ D B C \Delta ABD \sim \Delta DBC , B C = 1 3 |\overline{BC}| = \dfrac{1}{3} , which is clearly 3 3 times shorter than A B |\overline{AB}| .

Let's now look at the next diagram, which involves section of rhombus into small triangles...

Here, the area of Δ A B D \Delta ABD is three times the area of Δ B D C \Delta BDC , which is shown by the sub-rectangle of six small triangles hovering Δ A B D \Delta ABD . Since Δ A B D \Delta ABD covers half the rectangle, this is equivalent to half the area of the rectangle.

Counting the number of small triangles for different-colored region, we see that there are 13 13 of Δ B D C \Delta BDC that are not colored red. Thus, the ratio is 3 13 \boxed{\dfrac{3}{13}} .

One can also view the rhombus as the triangle by moving two small triangles to the dashed regions.

The angles of rhombus A C E F ACEF need to be specified in the problem.

Jon Haussmann - 3 years, 10 months ago

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