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2004 AMC 8 真题及详解

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月4日上午11:12

2004 AMC 8 真题

答案详细解析请参考文末

Problem 1

On a map, a? $12$ -centimeter length represents? $72$ ?kilometers. How many kilometers does a? $17$ -centimeter length represent?

$textbf{(A)} 6qquadtextbf{(B)} 102qquadtextbf{(C)} 204qquadtextbf{(D)} 864qquadtextbf{(E)} 1224$

Problem 2

How many different four-digit numbers can be formed by rearranging the four digits in? $2004$ ?

$textbf{(A)} 4qquadtextbf{(B)} 6qquadtextbf{(C)} 16qquadtextbf{(D)} 24qquadtextbf{(E)} 81$

Problem 3

Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for? $18$ ?people. If they shared, how many meals should they have ordered to have just enough food for the? $12$ ?of them?

$textbf{(A)} 8qquadtextbf{(B)} 9qquadtextbf{(C)} 10qquadtextbf{(D)} 15qquadtextbf{(E)} 18$

Problem 4

Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?

$textbf{(A)} 2qquadtextbf{(B)} 4qquadtextbf{(C)} 6qquadtextbf{(D)} 8qquadtextbf{(E)} 10$

Problem 5

Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?

$textbf{(A)} 4qquadtextbf{(B)} 7qquadtextbf{(C)} 8qquadtextbf{(D)} 15qquadtextbf{(E)} 16$

Problem 6

After Sally takes? $20$ ?shots, she has made? $55%$ ?of her shots. After she takes? $5$ ?more shots, she raises her percentage to? $56%$ . How many of the last? $5$ ?shots did she make?

$textbf{(A)} 1qquadtextbf{(B)} 2qquadtextbf{(C)} 3qquadtextbf{(D)} 4qquadtextbf{(E)} 5$

Problem 7

An athlete's target heart rate, in beats per minute, is? $80%$ ?of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from? $220$ . To the nearest whole number, what is the target heart rate of an athlete who is? $26$ ?years old?

$textbf{(A)} 134qquadtextbf{(B)} 155qquadtextbf{(C)} 176qquadtextbf{(D)} 194qquadtextbf{(E)} 243$

Problem 8

Find the number of two-digit positive integers whose digits total? $7$ .

$textbf{(A)} 6 qquadtextbf{(B)} 7 qquadtextbf{(C)} 8 qquadtextbf{(D)} 9 qquadtextbf{(E)} 10$

Problem 9

The average of the five numbers in a list is? $54$ . The average of the first two numbers is? $48$ . What is the average of the last three numbers?

$textbf{(A)} 55qquadtextbf{(B)} 56qquadtextbf{(C)} 57qquadtextbf{(D)} 58qquadtextbf{(E)} 59$

Problem 10

Handy Aaron helped a neighbor? $1 frac14$ ?hours on Monday,? $50$ ?minutes on Tuesday, from 8:20 to 10:45 on Wednesday morning, and a half-hour on Friday. He is paid? $textdollar 3$ ?per hour. How much did he earn for the week?

$textbf{(A)} textdollar 8 qquad textbf{(B)} textdollar 9 qquad textbf{(C)} textdollar 10 qquad textbf{(D)} textdollar 12 qquad textbf{(E)} textdollar 15$

Problem 11

The numbers? $-2, 4, 6, 9$ ?and? $12$ ?are rearranged according to these rules:

        1. The largest isn’t first, but it is in one of the first three places. 
        2. The smallest isn’t last, but it is in one of the last three places. 
        3. The median isn’t first or last.

What is the average of the first and last numbers?

$textbf{(A)} 3.5 qquad textbf{(B)} 5 qquad textbf{(C)} 6.5 qquad textbf{(D)} 7.5 qquad textbf{(E)} 8$

Problem 12

Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for? $24$ ?hours. If she is using it constantly, the battery will last for only? $3$ ?hours. Since the last recharge, her phone has been on? $9$ ?hours, and during that time she has used it for? $60$ ?minutes. If she doesn’t talk any more but leaves the phone on, how many more hours will the battery last?

$textbf{(A)} 7 qquad textbf{(B)} 8 qquad textbf{(C)} 11 qquad textbf{(D)} 14 qquad textbf{(E)} 15$

Problem 13

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.

I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

Rank the friends from the oldest to the youngest.

$textbf{(A)} text{Bill, Amy, Celine}qquad textbf{(B)} text{Amy, Bill, Celine}qquad textbf{(C)} text{Celine, Amy, Bill}\ textbf{(D)} text{Celine, Bill, Amy} qquad textbf{(E)} text{Amy, Celine, Bill}$

Problem 14

What is the area enclosed by the geoboard quadrilateral below?

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=2; for(int a=0; a<=10; ++a) for(int b=0; b<=10; ++b) { dot((a,b)); }; draw((4,0)--(0,5)--(3,4)--(10,10)--cycle); [/asy]$

$textbf{(A)} 15qquad textbf{(B)} 18frac{1}{2} qquad textbf{(C)} 22frac{1}{2} qquad textbf{(D)} 27 qquad textbf{(E)} 41$

Problem 15

Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?

$textbf{(A)} 5qquad textbf{(B)} 7qquad textbf{(C)} 11qquad textbf{(D)} 12 qquad textbf{(E)} 18$

Problem 16

Two? $600$ ?mL pitchers contain orange juice. One pitcher is? $1/3$ ?full and the other pitcher is? $2/5$ ?full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?

$textbf{(A)} frac18 qquad textbf{(B)} frac{3}{16} qquad textbf{(C)} frac{11}{30} qquad textbf{(D)} frac{11}{19}qquad textbf{(E)} frac{11}{15}$

Problem 17

Three friends have a total of? $6$ ?identical pencils, and each one has at least one pencil. In how many ways can this happen?

$textbf{(A)} 1qquad textbf{(B)} 3qquad textbf{(C)} 6qquad textbf{(D)} 10 qquad textbf{(E)} 12$

Problem 18

Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers? $1$ ?through? $10$ . Each throw hits the target in a region with a different value. The scores are: Alice? $16$ ?points, Ben? $4$ ?points, Cindy? $7$ ?points, Dave? $11$ ?points, and Ellen? $17$ ?points. Who hits the region worth? $6$ ?points?

$textbf{(A)} text{Alice}qquad textbf{(B)} text{Ben}qquad textbf{(C)} text{Cindy}qquad textbf{(D)} text{Dave} qquad textbf{(E)} text{Ellen}$

Problem 19

A whole number larger than? $2$ ?leaves a remainder of? $2$ ?when divided by each of the numbers? $3, 4, 5,$ ?and? $6$ . The smallest such number lies between which two numbers?

$textbf{(A)} 40 text{and} 49 qquad textbf{(B)} 60 text{ and } 79 qquad textbf{(C)} 100 text{and} 129 qquad textbf{(D)} 210 text{and} 249qquad textbf{(E)} 320 text{and} 369$

Problem 20

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are? $6$ ?empty chairs, how many people are in the room?

$textbf{(A)} 12qquad textbf{(B)} 18qquad textbf{(C)} 24qquad textbf{(D)} 27qquad textbf{(E)} 36$

Problem 21

Spinners? $A$ ?and? $B$ ?are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?

$[asy] pair A=(0,0); pair B=(3,0); draw(Circle(A,1)); draw(Circle(B,1)); draw((-1,0)--(1,0)); draw((0,1)--(0,-1)); draw((3,0)--(3,1)); draw((3+sqrt(3)/2,-.5)--(3,0)); draw((3,0)--(3-sqrt(3)/2,-.5)); label("$A$",(-1,1)); label("$B$",(2,1)); label("$1$",(-.4,.4)); label("$2$",(.4,.4)); label("$3$",(.4,-.4)); label("$4$",(-.4,-.4)); label("$1$",(2.6,.4)); label("$2$",(3.4,.4)); label("$3$",(3,-.5)); [/asy]$

$textbf{(A)} frac14qquad textbf{(B)} frac13qquad textbf{(C)} frac12qquad textbf{(D)} frac23qquad textbf{(E)} frac34$

Problem 22

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is? $frac25$ . What fraction of the people in the room are married men?

$textbf{(A)} frac13qquad textbf{(B)} frac38qquad textbf{(C)} frac25qquad textbf{(D)} frac{5}{12}qquad textbf{(E)} frac35$

Problem 23

Tess runs counterclockwise around rectangular block? $JKLM$ . She lives at corner? $J$ . Which graph could represent her straight-line distance from home?

$[asy] unitsize(5mm); pair J=(-3,2); pair K=(-3,-2); pair L=(3,-2); pair M=(3,2); draw(J--K--L--M--cycle); label("$J$",J,NW); label("$K$",K,SW); label("$L$",L,SE); label("$M$",M,NE); [/asy]$

Problem 24

In the figure,? $ABCD$ ?is a rectangle and? $EFGH$ ?is a parallelogram. Using the measurements given in the figure, what is the length? $d$ ?of the segment that is perpendicular to? $overline{HE}$ ?and? $overline{FG}$ ?

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(10,0), B=(10,8), A=(0,8); pair E=(4,8), F=(10,3), G=(6,0), H=(0,5); draw(A--B--C--D--cycle); draw(E--F--G--H--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,N); label("$F$",(10.8,3)); label("$G$",G,S); label("$H$",H,W); label("$4$",A--E,N); label("$6$",B--E,N); label("$5$",(10.8,5.5)); label("$3$",(10.8,1.5)); label("$4$",G--C,S); label("$6$",G--D,S); label("$5$",D--H,W); label("$3$",A--H,W); draw((3,7.25)--(7.56,1.17)); label("$d$",(3,7.25)--(7.56,1.17), NE); [/asy]$

$textbf{(A)} 6.8qquad textbf{(B)} 7.1qquad textbf{(C)} 7.6qquad textbf{(D)} 7.8qquad textbf{(E)} 8.1$

Problem 25

Two? $4 times 4$ ?squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

$[asy] unitsize(6mm); draw(unitcircle); filldraw((0,1)--(1,2)--(3,0)--(1,-2)--(0,-1)--(-1,-2)--(-3,0)--(-1,2)--cycle,lightgray,black); filldraw(unitcircle,white,black); [/asy]$

$textbf{(A)} 16-4piqquad textbf{(B)} 16-2pi qquad textbf{(C)} 28-4pi qquad textbf{(D)} 28-2pi qquad textbf{(E)} 32-2pi$

2004AMC8详细解析

We set up the proportion? $frac{12 text{cm}}{72 text{km}}=frac{17 text{cm}}{x text{km}}$ . Thus? $x=102 Rightarrow boxed{textbf{(B)} 102}$ Since we know? $12$ ?centimeters on the map is equal to? $72$ ?kilometers,? $1$ ?centimeter on the map is simply? $72/12$ , which is? $6$ . The problem is asking how many kilometers a? $17$ -centimeter length represents, so we multiply? $6$ ?by? $17$ ?which is simply 102, your answer.
We can solve this problem easily, just by calculating how many choices there are for each of the four digits. First off, we know there are only? $2$ ?choices for the first digit, because? $0$ ?isn't a valid choice, or the number would a 3-digit number, which is not what we want. We have? $3$ ?choices for the second digit, since we already used up one of the digits, and? $2$ ?choices for the third, and finally just? $1$ ?choices for the fourth and final one.? $2*3*2*1$ ?is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get? $boxed{textbf{(B)} 6}$ .Note that the four-digit number must start with either a? $2$ ?or a? $4$ . The four-digit numbers that start with? $2$ ?are? $2400, 2040$ , and? $2004$ . The four-digit numbers that start with? $4$ ?are? $4200, 4020$ , and? $4002$ ?which gives us a total of? $boxed{textbf{(B)} 6}$ .
Set up the proportion? $frac{12 text{meals}}{18 text{people}}=frac{x text{meals}}{12 text{people}}$ . Solving for? $x$ ?gives us? $x= boxed{textbf{(A)} 8}$ .
There are? $binom{4}{3}$ ?ways to choose three starters. Thus the answer is? $boxed{textbf{(B)} 4}$ .
The remaining team will be the only undefeated one. The other? $boxed{textbf{(D)} 15}$ ?teams must have lost a game before getting out, thus fifteen games yielding fifteen losers.There will be? $8$ ?games the first round,? $4$ ?games the second round,? $2$ ?games the third round, and? $1$ ?game in the final round, giving us a total of? $8+4+2+1=15$ ?games.? $boxed{textbf{(D)} 15}$ .
Sally made? $0.55*20=11$ ?shots originally. Letting? $x$ ?be the number of shots she made, we have? $frac{11+x}{25}=0.56$ . Solving for? $x$ ?gives us? $x=boxed{textbf{(C)} 3}$
The maximum heart rate is? $220-26=194$ ?beats per minute. The target heart rate is then? $0.8*194 approx boxed{textbf{(B)} 155}$ ?beats per minute.
The numbers are? $16, 25, 34, 43, 52, 61, 70$ ?which gives us a total of? $boxed{textbf{(B)} 7}$ .
Let the? $5$ ?numbers be? $a, b, c, d$ , and? $e$ . Thus? $frac{a+b+c+d+e}{5}=54$ ?and? $a+b+c+d+e=270$ . Since? $frac{a+b}{2}=48$ ,? $a+b=96$ . Substituting back into our original equation, we have? $96+c+d+e=270$ ?and? $c+d+e=174$ . Dividing by? $3$ ?gives the average of? $boxed{textbf{(D)} 58}$ .
Convert everything to minutes and add them together. On Monday he worked for? $frac54 cdot 60 = 75$ ?minutes. On Tuesday he worked? $50$ ?minutes. On Wednesday he worked for? $2$ ?hours? $25$ ?minutes, or? $2(60)+25=145$ ?minutes. On Friday he worked? $30$ ?minutes. In total he worked for? $75+50+145+30=300$ ?minutes, or? $300/60=5$ ?hours and? $5cdot 3 = boxed{textbf{(E)} textdollar 15}$ .
From rule 1, the largest number,? $12$ , can be second or third. From rule 2, because there are five places, the smallest number? $-2$ ?can either be third or fourth. The median,? $6$ ?can be second, third, or fourth. Because we know the middle three numbers, the first and last numbers are? $4$ ?and? $9$ , disregarding their order. Their average is? $(4+9)/2 = boxed{textbf{(C)} 6.5}$
When not being used, the cell phone uses up? $1/24$ ?of its battery per hour. When being used, the cell phone uses up? $1/3$ ?of its battery per hour. Since Niki's phone has been on for? $9$ ?hours, of those? $8$ ?simply on and? $1$ ?being used to talk,? $8(1/24) + 1(1/3) = 2/3$ ?of its battery has been used up. To drain the remaining? $1/3$ ?the phone can last for? $frac{1/3}{1/24}=boxed{textbf{(B)} 8}$ ?more hours without being used.
If Bill is the oldest, then Amy is not the oldest, and both statements I and II are true, so statement I is not the true one.If Amy is not the oldest, and we know Bill cannot be the oldest, then Celine is the oldest. This would mean she is not the youngest, and both statements II and III are true, so statement II is not the true one.Therefore, statement III is the true statement, and both I and II are false. From this, Amy is the oldest, Celine is in the middle, and lastly Bill is the youngest. This order is? $boxed{textbf{(E)} text{Amy, Celine, Bill}}$ .Bashing through/ eliminating the optionsThe first statement tells us that bill is the oldest. That is only what we need to know to solve now. In the answer choices, you can easily see that the first one is the only option where Bill is the oldest. Thus, the answer is? $boxed{textbf{(E)} text{Bill, Aline , Celine, Bill}}$ .
Assign points to each of the four vertices and use the shoelace theorem to find the area. Letting the bottom left corner be? $(0,0)$ , counting the boxes, the points would be? $(4,0),(0,5),(3,4),$ ?and? $(10,10)$ . Applying the?Shoelace Theorem, $[text{Area} = frac12 begin{vmatrix} 4 & 0 \ 0 & 5 \ 3 & 4 \ 10 & 10 end{vmatrix} = frac12 |(20+30)-(15+40+40)| = frac12 |50-95| = boxed{textbf{(C)} 22frac12}]$ Apply?Pick's Theorem?on the figure, and you will get $[boxed{textbf{(C)} 22frac12}]$
The first ring around the middle tile has? $6$ ?tiles, and the second has? $12$ . From this pattern, the third ring has? $18$ ?tiles. Of these,? $6+18=24$ ?are white and? $1+12=13$ ?are black, with a difference of? $24-13 = boxed{textbf{(C)} 11}$ .
The first pitcher contains? $600 cdot frac13 = 200$ ?mL of orange juice. The second pitcher contains? $600 cdot frac25 = 240$ ?mL of orange juice. In the large pitcher, there is a total of? $200+240=440$ ?mL of orange juice and? $600+600=1200$ ?mL of fluids, giving a fraction of? $frac{440}{1200} = boxed{textbf{(C)} frac{11}{30}}$ .
For each person to have at least one pencil, assign one of the pencil to each of the three friends so that you have? $3$ ?left. In partitioning the remaining? $3$ ?pencils into? $3$ ?distinct groups, use?Ball-and-urn?to find the number of possibilities is? $binom{3+3-1}{3} = binom{5}{3} = boxed{textbf{(D)} 10}$ .Like in solution 1, for each person to have at least one pencil, assign one of the pencil to each of the three friends so that you have? $3$ ?left. In partitioning the remaining? $3$ ?pencils into? $3$ ?distinct groups, use number of non-negetive integral soutions. Let the three friends be? $a$ ,? $b$ ,? $c$ ?repectively. $a$ ?+? $b$ ?+? $c$ ?=? $3$ ?The total being 3 and 2 plus signs, which implies? $binom{3+2}{3} = binom{5}{3} = boxed{textbf{(D)} 10}$ .Solution by?phoenixfireLike in solution 1 and solution 2, for each person to have at least one pencil, assign one of the pencil to each of the three friends so that you have? $3$ ?left. In partitioning the remaining? $3$ ?pencils into? $3$ ?distinct groups use casework. Let the three friends be? $a$ ,? $b$ ,? $c$ ?repectively.
Case? $1$ ? $a$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$a$ ?=0,

$b$ ?+? $c$ ?=? $3$ ,

$b$ ?= 0,1,2,3 ,

and respective values of? $c$ ?will be? $c$ ?= 3,2,1,0 ,

Which means? $boxed{textbf 4}$ ?solutions.

Case? $2$ ,

$a$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$a$ ?=1,? $1$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$b$ ?+? $c$ ?=? $2$ ,

$b$ ?= 0,1,2 ,

and respective values of? $c$ ?will be? $c$ ?= 2,1,0 ,

Which means? $boxed{textbf 3}$ ?solutions.

Case? $3$ ?,

$a$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$a$ = 2,

$2$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$b$ ?+? $c$ ?=? $1$ ,

$b$ ?= 0,1 ,

and respective values of? $c$ ?will be? $c$ ?= 1,0 ,

Which means? $boxed{textbf 2}$ ?solutions.

Case? $4$ ,

$a$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$a$ ?= 3,

$3$ ?+? $b$ ?+? $c$ ?=? $3$ ,

$b$ ?+? $c$ ?=? $0$ ,

$b$ ?= 0 ,

and respective value of? $c$ ?will be? $c$ ?= 0 ,

Which means? $boxed{textbf 1}$ ?solution.

Therefore there will be total 10 solutions.? $boxed{textbf{(D)} 10}$ .
The only way to get Ben's score is with? $1+3=4$ . Cindy's score can be made of? $3+4$ ?or? $2+5$ , but since Ben already hit the? $3$ , Cindy hit? $2+5=7$ . Similarly, Dave's darts were in the region? $4+7=11$ . Lastly, because there is no? $7$ ?left,? $boxed{textbf{(A)} text{Alice}}$ ?must have hit the regions? $6+10=16$ ?and Ellen? $8+9=17$ .Taking the 2 largest scores to narrow out the points we getEllen with either? $10,7$ ?or? $9,8$ ?and Alice with either? $10,6$ ?or? $9,7$ , the 2 pairs that work are Alice= $10,6$ ?Ellen= $9,8$ , therefore Alice scored the? $6$ .
The smallest number divisible by? $3,4,5,$ ?and? $6$ , or their least common multiple, can be found to be? $60$ . When? $2$ ?is added to a multiple of number, its remainder when divided by that number is? $2$ . The number we are looking for is therefore? $62$ , and between? $boxed{textbf{(B)} 60 text{and} 79}$ .
Working backwards, if? $3/4$ ?of the chairs are taken and? $6$ ?are empty, then there are three times as many taken chairs as empty chairs, or? $3 cdot 6 = 18$ . If? $x$ ?is the number of people in the room and? $2/3$ ?are seated, then? $frac23 x = 18$ ?and? $x = boxed{(text{D}) 27}$ .
An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from? $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from? $1$ . $[1-frac24 cdot frac23 = 1- frac13 = boxed{textbf{(D)} frac23}]$
Assume arbitrarily (and WLOG) there are? $5$ ?women in the room, of which? $5 cdot frac25 = 2$ ?are single and? $5-2=3$ ?are married. Each married woman came with her husband, so there are? $3$ ?married men in the room as well for a total of? $5+3=8$ ?people. The fraction of the people that are married men is? $boxed{textbf{(B)} frac38}$ .
For her distance to be represented as a constant horizontal line, Tess would have to be running in a circular shape with her home as the center. Since she is running around a rectangle, this is not possible, ruling out? $B$ ?and? $E$ ?with straight lines. Because? $JL$ ?is the diagonal of the rectangle, and? $L$ ?is at the middle distance around the perimeter, her maximum distance should be in the middle of her journey. The maximum in? $A$ ?is at the end, and? $C$ ?has two maximums, ruling both out. Thus the answer is? $boxed{textbf{(D)}}$ .
The area of the parallelogram can be found in two ways. The first is by taking rectangle? $ABCD$ ?and subtracting the areas of the triangles cut out to create parallelogram? $EFGH$ . This is $[(4+6)(5+3) - 2 cdot frac12 cdot 6 cdot 5 - 2 cdot frac12 cdot 3 cdot 4 = 80 - 30 - 12 = 38]$ The second way is by multiplying the base of the parallelogram such as? $overline{FG}$ ?with its altitude? $d$ , which is perpendicular to both bases.? $triangle FGC$ ?is a? $3-4-5$ ?triangle so? $overline{FG} = 5$ . Set these two representations of the area together. $[5d = 38 rightarrow d = boxed{textbf{(C)} 7.6}]$
If the circle was shaded in, the intersection of the two squares would be a smaller square with half the sidelength,? $2$ . The area of this region would be the two larger squares minus the area of the intersection, the smaller square. This is? $4^2 + 4^2 - 2^2 = 28$ .The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle. This value can be found with Pythagorean or a? $45^circ - 45^circ - 90^circ$ ?circle to be? $2sqrt{2}$ . The radius is half the diameter,? $sqrt{2}$ . The area of the circle is? $pi r^2 = pi (sqrt{2})^2 = 2pi$ .The area of the shaded region is the area of the two squares minus the area of the circle which is? $boxed{textbf{(D)} 28-2pi}$ .Find the area of the overlapping squares.? $2 cdot 4^2 - 2^2=28$ . Now, we find the chord of the circle to be 2, so then the radius of the circle would be? $sqrt{2}$ . Now, the area of the circle is? $pi r^2$ , so we put in the values to get? $2pi$ . Now, we can subtract the two values. Now, we can find the answer to be? $boxed{textbf{(D)} 28-2pi}$ .