It’s so unbelievably frustrating that the exercises hinge on concepts that are not explained in the PrepSwift videos. For example, nowhere does the video talk about how when you compare the area of two triangles, their ratios get squared?
Like how am I suppose to know this?
Also, I don’t quite understand why the assumption talks about the base of 5 being 5y or 7 being 7y. Why can’t it just be 5 and x and 7 and x?
@Leaderboard can you please check this and adjust/remove the question if its testing concepts not being taught?
If we denote the side lengths of triangle 1 as (l_1, l_2, l_3) and the side lengths of triangle 2 as (l_4, l_5, l_6), then by similarity we have:
\frac{l_1}{l_4} = \frac{l_2}{l_5} = \frac{l_3}{l_6} = k
for appropriate side labeling.
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In your case, you have:
\frac{b_1}{b_2} = \frac{h_1}{h_2} = k = \frac 57
where A_1 = \frac 12 b_1 h_1 and A_2 = \frac 12 b_2 h_2.
The area ratio is then:
\frac{A_1}{A_2} = \left( \frac{b_1}{b_2} \right) \left( \frac{h_1}{h_2} \right) = \left( \frac{b_1}{b_2} \right)^2 =\left( \frac{h_1}{h_2} \right)^2 = k^2 = \left(\frac 57\right)^2
I personally don’t think you needed anything other than being taught similarity and how to compute the area of a triangle. In fact, by observation you could’ve just surmised that you’re dealing with two corresponding length ratios, so you’d square the singular length ratio given in the problem.
This. @user2135, the question was carefully written to allow students that have covered only this video to get it correct, and the solution does the same too. Hence the fact that P and Q were right-angled triangles.
