Question

Circle centred at A and B each have radius's as shown. Point O is midpoint of AB and
0.4-2. Segments OC and OD are tangents to circles centred at A and B respectively and EF
is a common tangent. What is area of shaded region ECODF?
E
F

Answer 1

Answer:

The area we are trying to find is simply $ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD})$. Obviously, $\overline{EF}\parallel\overline{AB}$. Thus, $ABFE$ is a rectangle, and so its area is $b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{2}$.

Since $\overline{OC}$ is tangent to circle $A$, $\triangle{ACO}$ is a right triangle. We know $AO=2\sqrt{2}$ and $AC=2$, so $\triangle{ACO}$ is isosceles, a $45$-$45$ right triangle, and has $\overline{CO}$ with length $2$. The area of $\triangle{ACO}=\frac{1}{2}bh=2$. By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.

$\overarc{AEC}$ (or $\overarc{BFD}$, for that matter) is $\frac{1}{8}$ the area of its circle. Thus $\overarc{AEC}$ and $\overarc{BFD}$ both have an area of $\frac{\pi}{2}$.

Plugging all of these areas back into the original equation yields $8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}$

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Step-by-step explanation: