The distances between the dots in the figure are equal to 1, both horizontally and vertically. What is the measure of the common area of the triangle and the square which you can see shaded in the figure?
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$\hspace{4cm}$$(1) \ 9/10 \qquad (2) 15/16 \qquad (3) \ 8/9 \qquad (4) \ 11/12$
It's obvious the area will be area of square minus area of the little white right triangle. Even after writing all the coordinates down, how can I determine the coordinates where bigger triangle is cutting the square. If that can be found the problem is done. I do not see any similar triangles here either.
Any help or idea is appreciated. Thanks.
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$\begingroup$Refer to the figure:
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The area of the grey shaded region: $$S_{ABC}-S_{CDE}-S_{AFG}-1.$$ The areas of the two triangles: $$\frac{S_{CDE}}{S_{ABC}}=\left(\frac{CE}{BC}\right)^2 \Rightarrow S_{CDE}=\left(\frac12\right)^2\cdot \frac12\cdot 2\cdot 3=\frac34,\\ \frac{S_{AFG}}{S_{ABC}}=\left(\frac{AF}{AB}\right)^2 \Rightarrow S_{AFG}=\left(\frac13\right)^2\cdot \frac12\cdot 2\cdot 3=\frac13.\\$$ Hence: $$S_{\text{grey region}}=S_{ABC}-S_{CDE}-S_{AFG}-1=3-\frac34-\frac13-1=\frac{11}{12}.$$
$\endgroup$ $\begingroup$Consider the labeled points $p_{1}$ to $p_{16}$
The segment $\overline{p_{1}p_{12}}$ bisects $\overline{p_{6}p_{7}}$, giving you one of your coordinates. The coordinate on the right side of the square can be found considering the slope of $\overline{p_{1}p_{12}}=-\frac{2}{3}$.
So it is just up to you to decide where you want to place your origin. If you consider $p_{1}$ to be $(0,0)$, then the top point on the square is $(\frac{3}{2},-1)$ and the right point is $(2,-\frac{4}{3})$.
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