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What is the ratio $ A_I:A_{II} $ ?

Ratio of Areas

I know that the given angles and common angle prove the triangles are similar. Using proportionality, I found the length of the middle lengthed side of $\triangle_I$ to be $\frac 34 \times 7$. The length of the long side of the quadrilateral is $ 8 - (\frac 34 \times 7) $.

I am stuck here. I cannot think of how to compute the area only given two sides of a scalene triangle, although I imagine the next step is finding the last side and then using Hero's formula. Thanks for any help.

This problem is from Geometry for Enjoyment and Challenge, Chapter 11: Areas of triangles, parallelograms, trapezoids, kites, etc. It does not yet cover the Law of Sines or Cosines, although I couldn't imagine how one could find the angle either way.

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up vote 3 down vote accepted

Hint: For the ratio of the areas we need very little. If you scale a triangle by the linear factor $s$, then area is scaled by the factor $s^2$.

Remark: The idea of scaling is a very powerful tool, with important uses all over Physics, and elsewhere.

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The answer's $\frac97$.

In similar triangles, $(\text{side}:\text{side})^2 = \text{area}:\text{area}$. Thus $\left(\frac{6}{8}\right)^2= \frac{9}{16}$, and $\text{tri}_1 : \text{tri}_{\text{whole}} = \frac9{16}$. Then $\text{tri}_2$ must equal $\frac7{16}$, thus the ratios are $\frac97$.

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Welcome to math.SE: I have tried to improve the readability of your question by introducing Tex. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. – A.P. May 21 '13 at 5:30

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