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solid geometry - Circle Related

For COMPETITION
Number of Total Problems: 19.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: None

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: None

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 4

From : NCTM

Type: Understanding

Section:solid geometry

Theme:Length
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 5

From : NCTM

Type: None

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 6

From : NCTM

Type: Complex

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 3

Category Circle Related

Analysis

Solution/Answer

Problem Num : 7

From : NCTM

Type: None

Section:solid geometry

Theme:None
Adjustment# :

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 8

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Problem 19

Three semicircles of radius $1$ are constructed on diameter $overline{AB}$ of a semicircle of radius $2$ . The centers of the small semicircles divide $overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

$extbf{(A) } pi - sqrt{3} qquad extbf{(B) } pi - sqrt{2} qquad extbf{(C) } frac{pi + sqrt{2}}{2} qquad extbf{(D...$
'

Category Circle Related

Analysis

Solution/Answer

By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$ . This gives the area of the white region as $frac{5}{6}pi+frac{2cdotsqrt3}{4}=frac{5}{6}pi+frac{sqrt3}{2}$ . The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2pi-frac{5}{6}pi+frac{sqrt3}{2}=frac{7}{6}pi-frac{sqrt3}{2}$ .
Thus the answer is $oxed{ extbf{(E)} frac{7}{6}pi-frac{sqrt3}{2}}$ .

Answer:

Problem Num : 9

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?
$(mathrm {A}) 200+25pi quad (mathrm {B}) 100+75pi quad (mathrm {C}) 75+100pi quad (mathrm {D}) 100+100pi quad (ma...$
'

Category Circle Related

Analysis

Solution/Answer
The area of the circle is $S_{igcirc}=100pi$ , the area of the square is $S_{square}=100$ .
Exactly $1/4$ of the circle lies inside the square. Thus the total area is $dfrac34 S_{igcirc} + S_{square} = oxed{100+75pi} Longrightarrow mathrm{(B)}$ .

Answer:

Problem Num : 10

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?
$mathrm{(A) } frac{3sqrt{5}}{2} qquad mathrm{(B) } frac{7}{2} qquad mathrm{(C) } sqrt{15} qquad mathrm{(D) ...$
'

Category Circle Related

Analysis

Solution/Answer
This is obviously a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$ .
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$ .
The radius $r$ of the inscribed circle can be computed using the well-known identity $frac{rP}2=S$ , where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=5cdot 12/2=30$ and $P=5+12+13=30$ , thus $r=2$ . As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$ .
The distance of these two points is then $sqrt{ (6-2)^2 + (2.5-2)^2 } = sqrt{16.25} = sqrt{frac{65}4} = oxed{frac{sqrt{65}}2}$ .

Answer:

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