Quirky Quadrilateral

Geometry Level 2

Quadrilateral $ABCD$ has $AB=CD,$ and the acute angles $B$ and $C$ satisfy $\sin B = \frac{4}{9}$ and $\sin C= \frac{5}{6}.$ Find $\frac{\text{area of }\color{#D61F06}{\triangle ABC}}{\text{area of }\color{#3D99F6}{\triangle BDC}}.$

Give your answer to 3 decimal places.

The image is not drawn to scale.

The answer is 0.5333333.

2 solutions

Paola Ramírez
Oct 6, 2015

Let us use this formula $\boxed{\text{Area of a triangle}=\frac{1}{2} A\cdot B\cdot \sin C}$ to calculate the area of $\triangle ABC$ and $\triangle BDC$

$|\triangle ABC|=\frac{1}{2} AB \cdot BC\cdot \frac{4}{9}$

$|\triangle BDC|=\frac{1}{2} CD \cdot BC\cdot \frac{5}{6}$

$\therefore$ The ratio of areas is $\frac{\sin B}{\sin C}=\frac{\frac{4}{9}}{\frac{5}{6}}=\frac{8}{15}\approx \boxed{0.533}$

Great job!

By the way, if you combine your expressions for the areas of the triangles with the ones I used in the other solution, it actually forms a quick "proof" of why the area of a triangle can be written as $\frac{1}{2}AB\sin C.$

Eli Ross Staff - 5 years, 8 months ago

Thank you!

My solution is more technical, yours is more creative :)

Paola Ramírez - 5 years, 8 months ago

Eli Ross Staff
Oct 5, 2015

Note that the two triangles share a base, so the ratio of their areas is just the ratio of their heights (from AD to BC). Using Right Triangle Trigonometry , we find the following:

Quirky Quadrilateral

The answer is 0.5333333.

2 solutions

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