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# Area of the Region Problem | AMC-10A, 2007 | Problem 24

Try this beautiful problem from Geometry: Area of region from AMC-10A, 2007, Problem-24. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Area of the region

## Problem on Area of the Region – AMC-10A, 2007- Problem 24

Circle centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$?

• $\pi$
• $7\sqrt 3 -\pi$
• $8\sqrt 2 -4-\pi$

### Key Concepts

Geometry

Triangle

similarity

Answer: $8\sqrt 2 -4-\pi$

AMC-10A (2007) Problem 24

Pre College Mathematics

## Try with Hints

We have to find out the area of the region $ECODF$ i.e of gray shaded region.this is not any standard geometrical figure (such as circle,triangle…etc).so we can not find out the value easily.Now if we join $AC$,$AE$,$BD$,$BF$.Then $ABFE$ is a rectangle.then we can find out the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]

Can you find out the required area…..?

Given that Circle centered at $A$ and $B$ each have radius $2$ and Point $O$ is the midpoint of $\overline{AB}$, and $OA = 2\sqrt {2}$

Area of $ABEF$=$2 \times 2 \times 2\sqrt 2$=$8\sqrt 2$

Now $\triangle{ACO}$ is a right triangle. We know $AO=2\sqrt{2}$and $AC=2$, so $\triangle{ACO}$ is isosceles, a $45$-$45$ right triangle.$\overline{CO}$ with length $2$. The area of $\triangle{ACO}=\frac{1}{2} \times base \times height=2$. By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.now the $\angle CAO$ = $\angle DBO$=$45^{\circ}$. therefore $\frac{360}{45}=8$

So the area of arc $AEC$ and arc $BFD$=$\frac{1}{8} \times$ area of the circle=$\frac{\pi 2^2}{8}$=$\frac{\pi}{2}$

can you finish the problem……..

Therefore the required area by [ area of rectangle $ABEF$- (area of arc $AEC$+area of $\triangle ACO$+area of $\triangle BDO$+ area of arc $BFD$)]=$8\sqrt 2-(\frac{\pi}{2}+2+2+\frac{\pi}{2}$)=$8\sqrt 2 -4-\pi$