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solid geometry - Circle Related

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

Problem Num : 11

From : AMC12

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$ext{(A)} pi qquad ext{(B)} 1.5pi qquad ext{(C)} 2pi qquad ext{(D)} 3pi qquad ext{(E)} 3.5pi$
''>''
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$ext{(A)} pi qquad ext{(B)} 1.5pi qquad ext{(C)} 2pi qquad ext{(D)} 3pi qquad ext{(E)} 3.5pi$
''

Category Circle Related

Analysis

Solution/Answer
The outer circle has radius $1+1+1=3$ , and thus area $9pi$ . The little circles have area $pi$ each; since there are 7, their total area is $7pi$ . Thus, our answer is $9pi-7pi=oxed{2piRightarrow ext{(C)}}$ .

Answer:

Problem Num : 12

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
An $8$ -foot by $10$ -foot floor is tiles with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

$mathrm{(A)} 80-20pi qquadmathrm{(B)} 60-10pi qquadmathrm{(C)} 80-10pi qquadmathrm{(D)} 60+10pi ...$
'

Category Circle Related

Analysis

Solution/Answer
There are 80 tiles. Each tile has $[mbox{square} - 4 cdot (mbox{quarter circle})]$ shaded. Thus:
$egin{align*}mbox{shaded area} &= 80 left( 1 - 4 cdot dfrac{1}{4} cdot pi cdot left(dfrac{1}{2} ight)^2 ight)...$

Answer:

Problem Num : 13

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?
$mathrm{(A) } frac{3sqrt{2}}{pi}qquad mathrm{(B) } frac{3sqrt{3}}{pi}qquad mathrm{(C) } sqrt{3}qquad math...$
'

Category Circle Related

Analysis

Solution/Answer
Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle.
In a circle with a radius $r$ the side of an inscribed equilateral triangle is $rsqrt{3}$ .
So $s=rsqrt{3}$ .
The perimeter of the triangle is $3s=3rsqrt{3}$
The area of the circle is $pi r^{2}$
So: $pi r^{2} = 3rsqrt{3}$
$pi r=3sqrt{3}$
$r=frac{3sqrt{3}}{pi} Rightarrow B$

Answer:

Problem Num : 14

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$ , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

$mathrm{(A) } frac{1}{6}pi-frac{sqrt{3}}{4}qquad mathrm{(B) } frac{sqrt{3}}{4}-frac{1}{12}piqquad mathrm{(C) ...$
'

Category Circle Related

Analysis

Solution/Answer

Let $[X]$ denote the area of region $X$ in the figure above.
The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$ .
The area of the smaller semicircle is $[A+B] = frac{1}{2}picdot(frac{1}{2})^{2}=frac{1}{8}pi$ .
Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures $60^circ$ .
The area of the $60^circ$ sector of the larger semicircle is $[B+C] = frac{60}{360}picdot(frac{2}{2})^{2}=frac{1}{6}pi$ .
The area of the triangle is $[C] = frac{1^{2}sqrt{3}}{4}=frac{sqrt{3}}{4}$ .
So the shaded area is $[A] = [A+B]-[B+C]+[C] = left(frac{1}{8}pi ight)-left(frac{1}{6}pi ight)+left(frac{sqrt{3}}{4} ight)=oxed{mathr...$ .

Answer:

Problem Num : 15

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Circles $A$ , $B$ , and $C$ are externally tangent to each other and internally tangent to circle $D$ . Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$ . What is the radius of circle $B$ ?

$mathrm{(A) } frac{2}{3} qquad mathrm{(B) } frac{sqrt{3}}{2} qquad mathrm{(C) } frac{7}{8} qquad mathrm{(D) ...$
'

Category Circle Related

Analysis

Solution/Answer
Let $O_{i}$ be the center of circle $i$ for all $i in {A,B,C,D}$ and let $E$ be the tangent point of $B,C$ . Since the radius of $D$ is the diameter of $A$ , the radius of $D$ is $2$ . Let the radius of $B,C$ be $r$ and let $O_{D}E = x$ . If we connect $O_{A},O_{B},O_{C}$ , we get an isosceles triangle with lengths $1 + r, 2r$ . Then right triangle $O_{D}O_{B}E$ has legs $r, x$ and hypotenuse $2-r$ . Solving for $x$ , we get $x^2 = (2-r)^2 - r^2 Longrightarrow x = sqrt{4-4r}$ .
Also, right triangle $O_{A}O_{B}E$ has legs $r, 1+x$ , and hypotenuse $1+r$ . Solving,
$egin{eqnarray*}r^2 + (1+sqrt{4-4r})^2 &=& (1+r)^2\1+4-4r+2sqrt{4-4r}&=& 2r + 1\1-r &=& left(f...$
So the answer is $mathrm{(D)}$ .

Answer:

Problem Num : 16

From : NCTM

Type: Complex

Section:solid geometry

Theme:None
Adjustment# : 0

Difficulty: 1

Category Circle Related

Analysis

Solution/Answer

Problem Num : 17

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
An equilateral triangle has side length $6$ . What is the area of the region containing all points that are outside the triangle but not more than $3$ units from a point of the triangle?
$mathrm{(A)} 36+24sqrt{3}qquadmathrm{(B)} 54+9piqquadmathrm{(C)} 54+18sqrt{3}+6piqquadmathrm{(D)} left(2sqrt{...$
'

Category Circle Related

Analysis

Solution/Answer
The region described contains three rectangles of dimensions $3 imes 6$ , and three $120^{circ}$ degree arcs of circles of radius $3$ . Thus the answer is $3(3 imes 6) + 3 left( frac{120^{circ}}{360^{circ}} imes 3^2 pi ight) = 54 + 9pi Longrightarrow mathrm{(B)}.$

Answer:

Problem Num : 18

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A square of side length $1$ and a circle of radius $dfrac{sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?
$extbf{(A)} dfrac{pi}{3}-1 qquad extbf{(B)} dfrac{2pi}{9}-dfrac{sqrt{3}}{3} qquad extbf{(C)} dfrac{pi}{18} ...$
'

Category Circle Related

Analysis

Solution/Answer
The radius of circle is $frac{sqrt{3}}{3} = sqrt{frac{1}{3}}$ . Half the diagonal of the square is $frac{sqrt{1^2+1^2}}{2} = frac{sqrt{2}}{2} = sqrt{frac12}$ . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle Therefore the picture will look something like this:

Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$ . Also note that $OX=frac12$ because it is half the sidelength of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for $a.$
$a^2+left( frac12 ight) ^2 = left( frac{sqrt{3}}{3} ight) ^2$
Solving, $a= frac{sqrt{3}}{6}$ and $2a=frac{sqrt{3}}{3}$ . Since $AB=AO=BO$ , $riangle AOB$ is an equilateral triangle and the central angle is $60^{circ}$ . Therefore the sector has an area $pi left( frac{sqrt{3}}{3} ight) ^2 left( frac{60}{360} ight) = frac{pi}{18}$ .
Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is
$frac{s^2sqrt{3}}{4} = frac{frac13 sqrt{3}}{4} = frac{sqrt{3}}{12}$
Putting it together, we get the answer to be $4 left( frac{pi}{18}-frac{sqrt{3}}{12} ight)= oxed{ extbf{(B)} frac{2pi}{9}-frac{sqrt{3}}{3}}$

Answer:

Problem Num : 19

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

$mathrm{(A)} 3-2sqrt2qquadmathrm{(B)} 2-sqrt2qquadmathrm{(C)} 4(3-2sqrt2)qquadmathrm{(D)} frac12(3-sqrt2)qqua...$

'

Category Circle Related

Analysis

Solution/Answer
Draw some of the radii of the small circles as in the picture below.

Out of symmetry, the quadrilateral in the center must be a square. Its side is obviously $2r$ , and therefore its diagonal is $2rsqrt{2}$ . We can now compute the length of the vertical diameter of the large circle as $2r + 2rsqrt{2}$ . Hence $2R=2r + 2rsqrt{2}$ , and thus $R=r+rsqrt{2}=r(1+sqrt{2})$ .
Then the area of the large circle is $L = pi R^2 = pi r^2 (1+sqrt 2)^2 = pi r^2 (3+2sqrt 2)$ . The area of four small circles is $S = 4pi r^2$ . Hence their ratio is:
$egin{align*}frac SL & = frac{4pi r^2}{pi r^2 (3+2sqrt 2)} \& = frac 4{3+2sqrt 2} \& = frac 4{3+2sqrt...$

Answer:

Array ( [0] => 8794 [1] => 8032 [2] => 7783 [3] => 7786 [4] => 7806 [5] => 3010 [6] => 7876 [7] => 8137 [8] => 7899 ) 191 2