Two 3x4 rectangles overlap, as shown below. Find the area of the overlapping region. (Which is shaded)
What I've done so far:
I made The smaller segment of BC be x, and thus making The larger one (or a side of the shaded part) be 4-x. Because the sides are all equal, 3^2+x^2=(4-x)^2, using pythagorean theorem. AFter simplifying, I end up with x^2+9=16-16x+x^2, which further simplifies into x=7/16. I plug in the values, but I keep getting 171/16, which apparently is incorrect. Please tell me what I did wrong, so I can fix my mistake.
+26367 Two 3x4 rectangles overlap, as shown below. Find the area of the overlapping region. (Which is shaded)
\(\begin{array}{|rcll|} \hline 3^2+x^2 &=& (4-x)^2 \\ 3^2+x^2 &=& 16 - 2*4x + x^2 \\ 3^2+x^2 &=& 16 {\color{red}{-8}}x + x^2 \\ 9 &=& 16 -8x \\ 8x &=& 16-9 \\ 8x &=& 7\\\\ x &=& \mathbf{\dfrac{7}{8}} \\ \hline \end{array}\)
area
\(\begin{array}{|rcll|} \hline A &=& 3*(4-x) \\ &=& 12-3x \\\\ &=& 12 -\dfrac{3*7}{8} \\\\ &=& 12 -\dfrac{21}{8} \\\\ &=& \dfrac{8*12-21}{8} \\\\ \mathbf{A }&=& \mathbf{\dfrac{75}{8}} \qquad \text{or} \quad \dfrac{150}{16} \\ \hline \end{array}\)
