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solid geometry - Area of a rectangle, a triangle

For COMPETITION
Number of Total Problems: 26.
FOR PRINT ::: (Book)

Problem Num : 11

From : NCTM

Type: Complex

Section:solid geometry

Theme:Applied equation
Adjustment# : 0

Difficulty: 3

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Answer:

Problem Num : 12

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A wire is cut into two pieces, one of length $a$ and the other of length $b$ . The piece of length $a$ is bent to form an equilateral triangle, and the piece of length $b$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $frac{a}{b}$ ?
$extbf{(A)} 1qquad extbf{(B)} frac{sqrt{6}}{2}qquad extbf{(C)} sqrt{3} qquad extbf{(D)} 2qquad extbf{(E)} f...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Using the formulas for area of a regular triangle $(frac{{s}^{2}sqrt{3}}{4})$ and regular hexagon $(frac{3{s}^{2}sqrt{3}}{2})$ and plugging $frac{a}{3}$ and $frac{b}{6}$ into each equation, you find that $frac{{a}^{2}sqrt{3}}{36}=frac{{b}^{2}sqrt{3}}{24}$ . Simplifying this, you get $frac{a}{b}=oxed{ extbf{(B)} frac{sqrt{6}}{2}}$

Answer:

Problem Num : 13

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A unit square is rotated $45^circ$ about its center. What is the area of the region swept out by the interior of the square?

$extbf{(A)} 1 - frac{sqrt2}{2} + frac{pi}{4}qquad extbf{(B)} frac{1}{2} + frac{pi}{4} qquad extbf{(C)} 2 - sq...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
First, we need to see what this looks like. Below is a diagram.

For this square with side length 1, the distance from center to vertex is $r = frac{1}{sqrt{2}}$ , hence the area is composed of a semicircle of radius $r$ , plus $4$ times a parallelogram with height $frac{1}{2}$ and base $frac{sqrt{2}}{2(1+sqrt{2})}$ . That is to say, the total area is $frac{1}{2} pi (1/sqrt{2})^2 + 4 frac{sqrt{2}}{4(1+sqrt{2})} = oxed{ extbf{(C) } frac{pi}{4} + 2 - sqrt{2}}$ .

Answer:

Problem Num : 14

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A regular octagon $ABCDEFGH$ has sides of length two. Find the area of $riangle ADG$ .
$extbf{(A) } 4 + 2sqrt2 qquad extbf{(B) } 6 + sqrt2qquad extbf{(C) } 4 + 3sqrt2 qquad extbf{(D) } 3 + 4sqrt2 q...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer

The area of the triangle $ADG$ can be computed as $frac{DG cdot AP}2$ . We will now find $DG$ and $AP$ .
Clearly, $PFG$ is a right isosceles triangle with hypotenuse of length $2$ , hence $PG=sqrt 2$ . The same holds for triangle $QED$ and its leg $QD$ . The length of $PQ$ is equal to $FE=2$ . Hence $GD = 2 + 2sqrt 2$ , and $AP = PD = 2 + sqrt 2$ .
Then the area of $ADG$ equals $frac{DG cdot AP}2 = frac{(2+2sqrt 2)(2+sqrt 2)}2 = frac{8+6sqrt 2}2 = 4+3sqrt 2 Rightarrow$ $oxed{(C)}$ .

Answer:

Problem Num : 15

From : AMC10

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.

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

Category Area of a rectangle, a triangle

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 : 16

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Equilateral $riangle ABC$ has side length $2$ , $M$ is the midpoint of $overline{AC}$ , and $C$ is the midpoint of $overline{BD}$ . What is the area of $riangle CDM$ ? $extrm{(A)} frac {sqrt {2}}{2}qquad extrm{(B)} frac {3}{4}qquad extrm{(C)} frac {sqrt {3}}{2}qquad extrm{(D...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Solution 1

The area of a circle can be given by $frac12 ab ext{sin} C$ . $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is twice the length of $BC$ . Each angle of an equilateral triangle is $60^circ$ so $angle MCD = 120^circ$ . The area is $frac12 (1)(2) ext{sin} 120^circ = oxed{ extbf{(C)} frac{sqrt{3}}{2}}$ .

Solution 2

In order to calculate the area of $riangle CDM$ , we can use the formula $A=dfrac{1}{2}bh$ , where $overline{CD}$ is the base. We already know that $overline{CD}=2$ , so the formula now becomes $A=h$ . We can drop verticals down from $A$ and $M$ to points $E$ and $F$ , respectively. We can see that $riangle AEC sim riangle MFC$ . Now, we establish the relationship that $dfrac{AE}{MF}=dfrac{AC}{MC}$ . We are given that $overline{AC}=2$ , and $M$ is the midpoint of $overline{AC}$ , so $overline{MC}=1$ . Because $riangle AEB$ is a $30-60-90$ triangle and the ratio of the sides opposite the angles are $1-sqrt{3}-2$ $overline{AE}$ is $sqrt{3}$ . Plugging those numbers in, we have $dfrac{sqrt{3}}{MF}=dfrac{2}{1}$ . Cross-multiplying, we see that $2 imesoverline{MF}=sqrt{3} imes1implies overline{MF}=dfrac{sqrt{3}}{2}$ Since $overline{MF}$ is the height $riangle CDM$ , the area is $oxed{mathrm{(C)} dfrac{sqrt{3}}{2}}$ .

Answer:

Problem Num : 17

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

''
In the overlapping triangles $riangle{ABC}$ and $riangle{ABE}$ sharing common side $AB$ , $angle{EAB}$ and $angle{ABC}$ are right angles, $AB=4$ , $BC=6$ , $AE=8$ , and $overline{AC}$ and $overline{BE}$ intersect at $D$ . What is the difference between the areas of $riangle{ADE}$ and $riangle{BDC}$ ?

$mathrm {(A)} 2 qquad mathrm {(B)} 4 qquad mathrm {(C)} 5 qquad mathrm {(D)} 8 qquad mathrm {(E)} 9 qquad$

'''>'In the overlapping triangles $riangle{ABC}$ and $riangle{ABE}$ sharing common side $AB$ , $angle{EAB}$ and $angle{ABC}$ are right angles, $AB=4$ , $BC=6$ , $AE=8$ , and $overline{AC}$ and $overline{BE}$ intersect at $D$ . What is the difference between the areas of $riangle{ADE}$ and $riangle{BDC}$ ?

$mathrm {(A)} 2 qquad mathrm {(B)} 4 qquad mathrm {(C)} 5 qquad mathrm {(D)} 8 qquad mathrm {(E)} 9 qquad$

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Solution 1

Since $AE perp AB$ and $BC perp AB$ , $AE parallel BC$ . By alternate interior angles and AA~, we find that $riangle ADE sim riangle CDB$ , with side length ratio $frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 cdot frac{4}{7} = frac{16}{7}$ and $h_{CDB} = 3 cdot frac 47 = frac {12}7$ . Subtracting the areas, $frac{1}{2} cdot 8 cdot frac {16}7 - frac 12 cdot 6 cdot frac{12}7 = 4$ $Rightarrow$ $oxed{(B)}$ .

Solution 2

Let $[X]$ represent the area of figure $X$ . Note that $[ riangle BEA]=[ riangle ABD]+[ riangle ADE]$ and $[ riangle BCA]=[ riangle ABD]+[ riangle BDC]$ .
$[ riangle ADE]-[ riangle BDC]=[ riangle BEA]-[ riangle BCA]=frac{1}{2} 8*4-frac{1}{2} 6*4= 16-12=4$ $Rightarrow$ $oxed{(B)}$ .

Answer:

Problem Num : 18

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle? $mathrm{(A) } 20piqquad mathrm{(B) } 25piqquad mathrm{(C) } 30piqquad mathrm{(D) } 40piqquad mathrm{(E) ...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Since the area of the square is $40$ , the length of the side is $sqrt{40}=2sqrt{10}$ . The distance between the center of the semicircle and one of the bottom vertecies of the square is half the length of the side, which is $sqrt{10}$ .
Using the Pythagorean Theorem to find the square of radius, $r^2 = (2sqrt{10})^2 + (sqrt{10})^2 = 50$ . So, the area of the semicircle is $frac{1}{2}cdot pi cdot 50 = 25pi Longrightarrow oxed{ ext{(B)}}$ .

Answer:

Problem Num : 19

From : AMC10

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?
$mathrm{(A) } 120 qquad mathrm{(B) } 180 qquad mathrm{(C) } 240 qquad mathrm{(D) } 360 qquad mathrm{(E) } 4...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base 3 and height 80. $3cdot 80=240Rightarrow oxed{mathrm{(C) }}$

Answer:

Problem Num : 20

From : AMC10B

Type:

Section:solid geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Rhombus $ABCD$ is similar to rhombus $BFDE$ . The area of rhombus $ABCD$ is $24$ and $angle BAD = 60^circ$ . What is the area of rhombus $BFDE$ ?
$mathrm{(A) } 6qquad mathrm{(B) } 4sqrt{3}qquad mathrm{(C) } 8qquad mathrm{(D) } 9qquad mathrm{(E) } 6sqr...$
'

Category Area of a rectangle, a triangle

Analysis

Solution/Answer
Using properties of a rhombus, $angle DAB = angle DCB = 60 ^circ$ and $angle ADC = angle ABC = 120 ^circ$ . It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$ . Let the lengths of the sides of rhombus $ABCD$ be $s$ .
The longer diagonal of rhombus $BFDE$ is $BD$ . Since $BD$ is a side of an equilateral triangle with a side length of $s$ , $BD = s$ . The longer diagonal of rhombus $ABCD$ is $AC$ . Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$ , $AC = 2 cdot frac{ssqrt{3}}{2} = ssqrt{3}$
The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $frac{s}{ssqrt{3}} = frac{sqrt{3}}{3}$ . Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $left( frac{sqrt{3}}{3} ight) ^2 = frac{1}{3}$
Let $x$ be the area of rhombus $BFDE$ . Then $frac{x}{24} = frac{1}{3}$ , so $x = 8 Longrightarrow oxed{mathrm{(C)}}$ .

Answer:

Array ( [0] => 3079 [1] => 8199 [2] => 7958 [3] => 7977 [4] => 7756 [5] => 8038 [6] => 7794 [7] => 8056 [8] => 7802 [9] => 8061 ) 261 2 3