Faculty of Mathematics Centre for Education in

Waterloo, Ontario N2L 3G1 Mathematics and Computing

Grade 7/8 Math CirclesNovember 22 & 23, 2016

Math Jeopardy

Let’s play Math Jeopardy! Today, we will be playing a fun game of jeopardy to review what

you have learned so far.

Round One

Mathematical Puzzles

$100 Find three positive whole numbers that have the same answer when added together or

when multiplied together.

1,2,3

$200 1. A number has 3 digits and is odd

2. Two digits are the same

3. The sum of the digits in the tens and ones places is odd

4. The sum of the digits is 4

What is the number?

121

$300 Matching Socks

Sixteen red socks and sixteen blue socks are mixed up in a dresser drawer. The socks

are all identical except for their colour. Suppose Richard wants two matching socks

but there is a black out so the room is dark and he can?t see. What is the smallest

number of socks that Richard must take out of the drawer to guarantee he has a pair

of socks that match?

1

Suppose the first sock Richard takes out is red. Then, the second sock he takes out is

red or blue. If it is red, then he is done. If it is blue, Richard now has 2 socks (one

red and one blue). The third sock Richard takes out is red or blue. If it is red, then

he has a match with the first sock. If it is blue, he has a match with the second sock.

So, Richard needed 3 socks to get a matching pair. This is the same outcome if the

first sock Richard had chosen was blue.

Thus, 3 socks is the smallest number of socks Richard must take out.

$400 Cutting the Pie!

With one straight cut, you can slice a pie into 2 pieces. With two straight cuts, you

can slice a pie into 4 pieces. With three straight cuts, you can slice a pie into as many

as 7 pieces, as shown below.

2

3

4

5 67

1

What is the largest number of pieces that you can make with six straight cuts?

The first cut creates 2 pieces. The second cut intersects the first cut to create 4 pieces.

The third cut intersects both of the previous cuts to create 7 pieces. Let’s organize

this into a table.

# of cuts # of pieces Difference between # of pieces

0 1

1 2 2− 1 = 1

2 4 4− 2 = 2

3 7 7− 4 = 3

2

Notice the difference in number of pieces increases by 1 each time. By following this

pattern, the fourth straight cut will create 7 + 4 = 11 pieces. The fifth cut will

create 11 + 5 = 16 pieces. And finally, the sixth straight cut will create 16 + 6 =

22 pieces of pie.

$500 Toothpick Geometry

Make the fish swim in the opposite direction by moving exactly 3 toothpicks.

Solutions may vary.

Continued Fractions

$100 Express the following as a single improper fraction.

1

1 +1

1 +1

2

=1

1 +13

2

=1

1 +2

3

=15

3

=3

5

$200 Find the gcd(1428, 546) and simplify546

1428to its lowest terms.

1428 = 2× 546 + 336 gcd(1428, 546) = gcd(546, 336)

546 = 1× 336 + 210 gcd(546, 336) = gcd(336, 210)

336 = 1× 210 + 126 gcd(336, 210) = gcd(210, 126)

210 = 1× 126 + 84 gcd(210, 126) = gcd(126, 84)

126 = 1× 84 + 42 gcd(126, 84) = gcd(84, 42)

84 = 2× 42 + 0 gcd(84, 42) = 42

So, gcd(1428, 546) = 42 ⇒ 546÷ 42

1428÷ 42=

13

34

3

$300 Express[1; 4, 5, 2

]as a single improper fraction.

[1; 4, 5, 2

]= 1 +

1

4 +1

5 +1

2

= 1 +1

4 +111

2

= 1 +1

4 +2

11

= 1 +146

11

= 1 +11

46=

57

46

$400 Express159

43as a continued fraction.

159 = 3× 43 + 30 =⇒ 159

43= 3 +

30

43= 3 +

143

30

43 = 1× 30 + 13 =⇒ 159

43= 3 +

1

1 +13

30

= 3 +1

1 +130

13

30 = 2× 13 + 4 =⇒ 159

43= 3 +

1

1 +1

2 +4

13

= 3 +1

1 +1

2 +113

4

13 = 3× 4 + 1 =⇒ 159

43= 3 +

1

1 +1

2 +1

3 +1

4

$500 Solve for x, y, and z.21

16= x +

1

y +1

z

21 = 1× 16 + 5 =⇒ 21

16= 1 +

5

16= 1 +

116

5

16 = 3× 5 + 1 =⇒ 21

16= 1 +

1

3 +1

5

Thus, x = 1, y = 3 and z = 5.

4

Sequences

$100{

8

3,4

3,2

3,1

3, ...

}is a geometric sequence with a constant ratio of

1

2.

$200 Which term in the following sequence is equal to 102?

{4, 11, 18, 25, ...}

This is an arithmetic sequence with a common difference of d = 7. The first term of

the sequence is t1 = 4. Using the formula for the nth term of the sequence, we have...

tn = 4 + (n− 1)7

To determine what term number is equal to 102 in the sequence, let tn = 102. Plug

this into the formula to get...

tn = 102 = 4 + (n− 1)7

98 = (n− 1)7

14 = n− 1

n = 15

So, 102 is the 15th term of the sequence. (i.e. t15 = 102)

$300 Find the 5th term of the following recursive sequence.

tn = 2tn−1 + 5tn−2; t1 = 4, t2 = 2

t3 = 2t2 + 5t1 = 2× 2 + 5× 4 = 24

t4 = 2t3 + 5t2 = 2× 24 + 5× 2 = 58

t5 = 2t4 + 5t3 = 2× 58 + 5× 24 = 236

Thus, t5 = 236

5

$400 What is the 3rd term in the arithmetic sequence where t15 = 161 and t32 = 348?

We know that the 15th term is 161, the 32nd term is 348, and that there are (32−15) =

17 terms with 348 now being the 18th term in the sequence. Now, we can solve for the

common difference, d.

348 = 161 + (18− 1)× d

187 = 17× d

d = 11

Now, again we know that 161 is the 15th term and there are 15−3 = 12 terms between

the 3rd and 15th term. We count 13 terms if we include the 15th term. Now we can

solve for the 3rd term.

161 = t3 + (13− 1)× 11

161 = t3 + 12× 11

161 = t3 + 132

t3 = 29

$500 Without using a calculator, what is the last digit in the 56th term in the following

sequence? {91, 92, 93, 94, ...

}Notice the pattern of the last digit of every term.

t1 = 91 = 9

t2 = 92 = 81

t3 = 93 = 729

t4 = 94 = 6561...

The last digit of every odd term is 9 and the last digit of every even term is 1. Following

this pattern, the last digit in the 56th term of this sequence is 1.

6

Visual Group Theory

$100 The hexagon below shows a rotation of 60◦ from the initial starting position, then a

rotation of 300◦ which undoes the first action.

Draw the hexagon after the following two rotations: R60◦R180◦

$200 Express this rearrangement mathematically.

(1 2 3 4 5

4 1 5 3 2

)

7

$300 Suppose you have the following five balls.

Draw the final configuration of where the five balls are given the following

rearrangement rule: (1 2 3 4 5

2 5 4 3 1

)

$400 Determine the equivalent action of(1 2 3 4 5

3 1 5 2 4

)(1 2 3 4 5

2 5 4 3 1

)=

(1 2 3 4 5

1 4 2 5 3

)

$500 Create a rearrangement rule to return all the balls to their initial position given the

rearrangement rules below.(1 2 3 4 5

3 1 5 2 4

)(1 2 3 4 5

2 5 4 3 1

)

The initial position of the balls is given below.

The rearrangement rule below returns the balls to their initial position.(1 2 3 4 5

1 3 5 2 4

)

8

Random Questions I

$100 There are three people at the dinner table. Two are mothers, and two are daughters.

How is this possible?

The women at the table are grandmother, mother, and daughter.

$200 Express the following as a single improper fraction.

2 +1

5 +1

1 +1

7

= 2 +1

5 +18

7

= 2 +1

5 +18

7

= 2 +1

5 +7

8

= 2 +147

8

= 2 +8

47=

102

47

$300 Young Tony Stark asked his grandmother how old she was. Knowing Tony to be quite

bright, she replied:

“Tony, I have four children born three years apart between each one and the next. I

was 24 when I had my oldest son. Now my youngest is 31. That’s all I’m telling you!”

How old is Tony’s grandmother?

Tony’s grandmother had 4 children in 4× 3 = 12 years since her children were born 3

years apart. She was 24 when she had her first child, so she was 24 + 12 = 36 years old

when she had her youngest child. In present day, her youngest is now 31. Therefore,

Tony’s grandmother is now 36 + 31 = 67 years old.

$400 There are three gentlemen in a meeting: Mr. Red, Mr. Green, and Mr. Gold. They

are wearing red, green and gold ties.

Mr. Red: “How amazing! Our last names are Red, Green, and Gold, and

one of us is wearing a red tie, another is wearing a green tie, and

another is wearing a gold tie.”

Mr. Green: “And none of our tie colours match our names!”

Mr. Gold: “You are right!”

If Mr. Red’s tie is not gold, what is the tie colour of each person?

We know that no one’s last name matches their tie colour. So, Mr. Red’s tie is not red

and we are told that it is not gold. Thus, Mr. Red must be wearing a green tie. Since

Mr. Red is wearing a green tie, Mr. Gold must be wearing a red tie since he cannot

be wearing gold. Since gold is the only tie colour left, Mr. Green must be wearing a

gold tie.

9

$500 Complete the Kenken!

22

223

33

31

11

1 4

44

4

Gauss Prep

$100 Daniel rode his bicycle at a constant speed. After 40 minutes, he cycled 24 km. How

far did he cycle in 30 minutes?

24 km ÷40

40 min. ÷40=

0.6 km

1 min.

Therefore, Daniel cycled 0.6× 30 = 18 km in 30 minutes.

$200 Joe is reading a 400 page book. On Monday, he reads 40 pages. On each day after

the first, the number of pages that he reads is 20 more than on the previous day. Joe

finishes the book on Friday .

On Monday, Joe reads 40 pages so there are 400− 40 = 360 pages left.

On Tuesday, Joe reads 40 + 20 = 60 pages so there are 360− 60 = 300 pages left.

On Wednesday, he reads 60 + 20 = 80 pages so there are 300− 80 = 220 pages left.

On Thursday, he reads 80 + 20 = 100 pages so there are 220− 100 = 120 pages left.

And on Friday, he reads 100 + 20 = 120 pages so there are 120− 120 = 0 pages left.

Thus, Joe finishes his book on Friday.

10

$300 Which of these values is the largest?

(a)4

2− 1

4

(b)4

2 +1

4

(c)4

2− 1

3

(d)4

2 +1

3

(e)4

2− 1

2

Since all options are fractions, we know that the largest value will have the smallest

denominator. This eliminates (b) and (d). Since we want the denominator to be as

small as possible, we should choose the option with the largest fraction subtracted from

2. So,1

4<

1

3<

1

2, therefore (e) is the correct answer.

$400 In how many ways can 101 be expressed as the sum of two integers, both greater than

zero, with the second integer greater than the first?

Beginning with the positive integer 1 as a number in the first pair, we get the sum

101 = 1+100. From this point we can continue to increase the first number by one

while decreasing the second number by one, keeping the sum equal to 101. The list of

possible sums is:

101 = 1 + 100

101 = 2 + 99

101 = 3 + 98...

101 = 50 + 51

After this point, the first number will no longer be smaller than the second if we

continue to add 1 to the first number and subtract 1 from the second. There are 50

possible sums in all.

11

$500 In the diagram, ABCD is a square with side length 6, and WXY Z is a rectangle with

ZY = 10 and XY = 6. Also, AD and WX are perpendicular. If the shaded area is

equal to half of the area of WXY Z, the length of AP is 1 .

The area of rectangle WXY Z is 10× 6 = 60. Since the shaded area is half of the total

area of WXY Z, its area is 12× 60 = 30. Since AD and WX are perpendicular, then

the shaded area has four right angles, so is a rectangle. Since square ABCD has a

side length of 6, then DC = 6. Since the shaded area is 30, then PD × DC = 30 or

PD × 6 = 30 or PD = 5. Since AD = 6 and PD = 5, then AP = 1.

Round Two

The Matrix

$200 Add the following matrices: 4 −2

3 12

−8 7

+

9 11

2 12

−2 8

=

13 9

5 1

−10 15

$400 Evaluate.

2

7 2 21

−4 5 2

15 3 −8

− 7 6 15

−10 10 3

22 8 4

=

7 −2 27

2 0 1

8 −2 −20

$600 Evaluate.[

1 −3 413

5 7

]−12 6

2 −5

1 7

=

[−14 49

13 26

]

12

$800 Scott, Hugh, and Peter are doing extra chores to earn some extra allowance this

weekend.

• On Friday, they all worked for 2 hours.

• On Saturday, Scott and Hugh worked for 4 hours, and Peter worked for 5 hours.

• On Sunday, Scott worked for 4 hours, and Hugh worked for 5 hours, and Peter

worked for 3 hours.

If Scott earns $5 per hour, Hugh earns $6 per hour, and Peter works $7 per hour, on

which day did they earn the most money altogether?2 2 2

4 4 5

4 5 3

5

6

7

=

36

79

71

Altogether, Scott, Hugh, and Peter earned the most money on Saturday.

$1000 Hill Cipher

Given the key matrix below, encrypt the following message: CIRCLE5 2 6

1 3 7

9 5 4

First, write CIRCLE in matrix form and convert each letter into its corresponding

number. (i.e. A = 0, B = 1, C = 2, ...)

CIRCLE⇒

C C

I L

R E

⇒ 2 2

8 11

17 4

Then multiply the key matrix by the message matrix. Subtract 26 from each element

of the resulting matrix such that all elements are between and including 0 and 25.5 2 6

1 3 7

9 5 4

2 2

8 11

17 4

=

128 56

145 63

126 89

⇒24 4

15 11

22 11

⇒Y E

P L

W L

The encrypted message is YPWELL.

13

Combinatorial Counting

$200 Teodora is getting a new phone! She is deciding between 4 different smartphones and a

red, blue, yellow, or green phone case. How many smartphone and case combinations

can she choose from?

There are 4× 4 = 16 combinations that Teodora can choose from.

$400 Calculate the following:

12!

14!=

/12!

14× 13× /12!=

1

14× 13=

1

182

$600 There’s a crisis on Earth! Given a team has 4 members, how many different team

arrangements can Batman send from the Justice League if there are 35 available su-

perheroes?

There are

(35

4

)= 52360 different teams Batman can send.

$800 Calculate the following:

9C4

8P3

=

(9

4

)÷ 8!

(8− 3)!=

9!

4!(9− 4)!× 5!

8!=

9× /8!

4!/5!×

/5!

/8!=

9

24=

3

8

$1000 Tommy, Chuckie, and three other friends are going to the movies to watch Fantastic

Beasts and Where To Find Them. But Tommy and Chuckie had a fight and refuse to

sit next to each other. How many different ways can Tommy, Chuckie, and friends sit

in the movie theatre?

We can solve this by subtracting the number of ways Tommy and Chuckie can sit

together from the total number of ways the 5 friends can sit in the theatre. There is a

total of 5! = 120 ways they can all sit in the theatre.

Next, we will count the number of ways Tommy and Chuckie can sit together. We

treat Tommy and Chuckie as one person (or a subgroup). Now we are counting the

seating arrangement for 4 friends. So there are 4! = 24 ways to seats the 4 friends.

However, we need to consider that there are 2! = 2 ways of arranging Tommy and

Chuckie’s seating arrangement. (i.e. Tommy & Chuckie and Chuckie & Tommy are

2 different seating arrangements.) There are 2! × 4! = 2 × 24 = 48 different seating

arrangements such that Tommy and Chuckie are sitting next to each other.

Therefore, there are 120− 48 = 72 different ways they can sit in the movie theatre.

14

Areas of Triangles

$200 Find the area of 4ABC.

A

B C

10 cm

12 cm

8 cm

4ABC is an isosceles triangle so the height of our triangle intersects the base, BC at

the midpoint of BC. Half of the base is 6 cm. Now, we can use Pythagorean Theorem

to find the height of 4ABC.

62 + h2 = 102 ⇒ h =√

100− 36 =√

64 = 8cm

We now have the base and height of 4ABC, so using the basic area formula, we get...

A4ABC =12× 8

2= 48 cm2

$400 Find the area of 4DEF .

D

F

E17 mm

25 mm26 mm

15

Use Heron’s formula!

s =1

2(17 + 26 + 25) = 34

A4DEF =√

34(34− 17)(34− 26)(34− 25)

=√

34(17)(8)(9)

=√

41616

A4DEF = 204 mm2

$600 Given the coordinates below, find the area of 4PQR.

P (3, 13), Q(12, 1), R(28, 17)

Use Shoelace Theorem!

A4PQR =1

2

∣∣∣∣∣∣∣∣∣∣3 13

12 1

28 17

3 13

∣∣∣∣∣∣∣∣∣∣=

1

2|(12× 13 + 28× 1 + 3× 17)− (3× 1 + 12× 17 + 28× 13)|

=1

2|(156 + 28 + 51)− (3 + 204 + 364)|

=1

2|235− 571|

=1

2|−336|

=1

2(336)

A4PQR = 168 units2

16

$800 The perimeter of 4XY Z is 72 cm. If XY = XZ and Y Z = 20 cm, what is the area

of 4XY Z?

X

Y Z

26 cm 26 cm

20 cm

4XY Z is isosceles since XY = XZ. So, XY =72− 20

2= 26 cm. Now we can use

Heron’s formula:

s =1

2(72) = 36 cm

A4XY Z =√

36(36− 26)(36− 26)(36− 20)

=√

36(10)(10)(16)

=√

57600

A4XY Z = 240 cm2

$1000 Given that the perimeter of 4DEF is 42, what is the area of hexagon ABCDEF?

A

B

CD

E

F

25 30

6

13

5 14

15

17

Notice that ∠ABC = ∠BCA, this means 4BCA is an isosceles triangle and that

AB = AC = 25. Use Heron’s Formula to find the area of 4ABC:

s4ABC =1

2(25 + 25 + 30) = 40

A4ABC =√

40(40− 25)(40− 25)(40− 30) =√

40(15)(15)(10) =√

90000 = 300

Next, since AC = 25, DF = 25 − 5 − 6 = 14. We also know that the perimeter of

4DEF is 42, so DE = 42− 13− 14 = 15. Use Heron’s formula again to find the area

of 4DEF :

s4DEF =1

2(13 + 14 + 15) = 21

A4DEF =√

21(21− 13)(21− 14)(21− 15) =√

21(8)(7)(6) =√

7056 = 84

The area of hexagon ABCDEF is the area of 4ABC minus the area of 4DEF so:

AABCDEF = A4ABC− A4DEF = 300− 84 = 216 units2

Logic

$200 James made some cookies. He ate one cookie and gave half of the rest to Guntaas.

Then he ate another cookie and gave half of the rest to Daniel. James now has 5

cookies. How many cookies did James start with?

Work backwards! Right now, James has 5 cookies. Before this, he gave half his cookies

to Daniel so there were 5 × 2 = 10 cookies. Then before James ate a cookies, there

were 10 + 1 = 11 cookies. Next, James gave half his cookies to Guntaas so then there

were 11 × 2 = 22 cookies. Then, James ate another cookie. Therefore, James started

off with 22 + 1 = 23 cookies.

$400 There are three Dalmatian puppies: Spot, Socks, and Patches. Spot has fewer spots

than Socks, but more spots than Patches. Which puppy is the spottiest?

Spot has fewer spots than Socks and has more spots than Patches, so Spot is in the

middle in terms of number of spots. Since Spot has fewer spots than Socks, then Socks

must be the spottiest puppy.

$600 How can you use only five 5s and only addition to make 565?

555 + 5 + 5 = 565

18

$800 Fill in the blanks!

$1000 Let’s play Clue! Miss Scarlet, Colonel Mustard, Mr. Green, Mrs. Peacock and Mrs.

White were involved in a theft. One of the five stole a credit card from one of the other

four. The following facts are known:

1. A man and woman were eating at McDonalds at the time of the theft

2. The thief and victim were together at the bank at the time of the theft

3. Colonel Mustard was not with a married woman at the time of the theft

4. Mr. Green was not with Mrs. White at the time of the theft

5. One of the married women was alone at the time of the theft

6. One of the married women was at the bank at the time of the theft

7. The victim is a man

Who is the thief and who is the victim?

By statements 5 and 6 , Miss Scarlet must be the woman eating at McDonalds. By

3 , Colonel Mustard was not with a married woman at the time of the theft so he must

be the man eating at McDonalds with Miss Scarlet which satisfies 1 . This means Mr.

Green is the victim. Then by 4 and 6 , Mrs. Peacock must the the married woman at

the bank since Mr. Green and Mrs. White cannot be together. This means Mrs. White

was alone at the time of the theft. 6 and 7 implies that a married woman is a thief so

Mrs. Peacock is the thief. Therefore, the thief is Mrs. Peacock and the victim is Mr. Green.

19

Random Questions II

$200 What are the missing values for x, y, and z?[15 4

23 17

]− 3

[5 3

6 2

]=

[0 −5

5 11

]

$400 Which one is the better deal? 10 chocolates for $15 or 5 chocolates for $8?

10 chocolates for $15 is $1.50 a piece and 5 chocolates for $8 is $1.60 a piece. Thus,

the better deal is 10 chocolates for $15.

$600 What is the measure of angle x?

A

B

C

D

E

F

G H

I

60◦

130◦20◦ 80◦80◦

80◦

20◦

70◦

$800 If the area of a square is 64 m2, what is the perimeter?

The square has a side length of√

64 = 8 m. Thus, the perimeter of the square is

4× 8 = 32 m.

$1000 There are 52 cards in a standard deck of cards. How many ways can Jiin select 4 cards

from the deck if 2 of the cards are spades? (Hint: There are 13 cards per suit.

Of the 4 cards, Jiin selects from the deck, 2 of them must be spades. There are 13

spades from the deck of 52 cards so there are

(13

2

)= 78 ways to choose 2 spades.

Now, Jiin needs to choose 2 cards from the remaining 52 − 13 = 39 cards. There

are

(39

2

)= 741 ways to choose the remaining 2 cards. Therefore, in total, there are(

13

2

)(39

2

)= 78 × 741 = 57 798 ways that Jiin can select 4 cards from the deck if 2

of the cards are spades.

20

Gauss Prep

$200 If 10x − 10 = 9990, then x is equal to 4

Since 10x − 10 = 9990, then 10x = 9990 + 10 = 10000. If 10x = 10000, then x = 4

since 10 000 ends in 4 zeroes.

$400 In a class of 40 students, 18 said they liked apple pie, 15 said they liked chocolate cake

and 12 said they did not like either. How many students in the class liked both?

Of the 40 students, 12 did not like either dessert. Therefore, 40 − 12 = 28 students

liked at least one of the desserts. But 18 students said they liked apple pie, 15 said

they liked chocolate cake, and 18 + 15 = 33, so 33− 28 = 5 students must have liked

both of the desserts.

$600 Winifred earns $10/hour and works 8 hours per day for 10 days. She first spends 25%

of her pay on food and clothing, and then pays $350 in rent. How much of her pay

does she have left?

In 10 days, Winifred works 8×10 = 80 hours. So in these 10 days, she earns 80×10 =

$800 . Since 25% =1

4, she spends

1

4× $800 = $200 on food and clothing, leaving her

with $600. If she spends $350 on rent, she will then have $600− 350 = $250 left.

$800 The values of r, s, t, and u are 2, 3, 4, and 5, but not necessarily in that order. What

is the largest possible value of r × s + u× r + t× r?

We first recognize that in the products, r×s, u×r and t×r, r is the only variable that

occurs in all three. Thus, to make r× s + u× r + t× r as large as possible, we choose

r = 5, the largest value possible. Since each of s, u and t is multiplied by r once only,

and the three products are then added, it does not matter which of s, u or t we let equal

2, 3 or 4, as the result will be the same. Therefore, let s = 2, u = 3 and t = 4. Thus,

the largest possible value of r×s+u×r+t×r is 5×2+3×5+4×5 = 10+15+20 = 45.

21

$1000 In the diagram, ABCD is a square with area 25 cm2. If PQCD is a rhombus with

area 20 cm2, the area of the shaded region, in cm2, is 11 cm 2

Since ABCD is a square and has an area of 25 cm2, then the square has a side length

of 5 cm. Since PQCD is a rhombus, then it is a parallelogram, so its area is equal to

the product of its base and its height. Join point P to X on AD so that PX makes

a right angle with AD, and to Y on DC so that PY makes a right angle with DC.

Then the area of the shaded region is the area of rectangle ABZX plus the area of

triangle PXD. Since the area of PQCD is 20 cm2 and its base has length 5 cm, then

its height, PY , must have length 4 cm. Therefore, we can now label DX = 4, DP = 5

(since PQCD is a rhombus), AX = 1, and AB = 5. So ABZX is a 1 by 5 rectangle,

and so has area 5 cm2. Triangle PXD is right-angled at D, and has DP = 5 and

DX = 4, so by Pythagorean Theorem, PX = 3. Therefore, the area of triangle PXD

is1

2× 3× 4 = 6 cm2. So, in total, the area of the shaded region is 11 cm2.

22

Final Jeopardy

A rectangular piece of paper ABCD is folder so the edge CD lies along edge AD, make a

crease DP . It is unfolded, and then folded again so that edge AB lies along the edge AD,

making a second crease AQ. The two creases meet at R, forming triangles PQR and ADR,

as shown. If AB = 5 cm and AD = 8 cm, what is the area of quadrilateral DRQC, in cm2?

To find the area of quadrilateral DRQC, we subtract the area of 4PRQ from the area of

4PDC. First, we calculate the area of 4PDC.

We know that DC = AB = 5 cm and that ∠DCP = 90◦. When the paper is first folded,

PC is parallel to AB and lies across the entire width of the paper, so PC = AB = 5 cm.

Therefore, the area of 4PDC is5× 5

2=

25

2= 12.5 cm2.

Next, we calculate the area of4PRQ. We know that4PDC has PC = 5

cm, ∠PCD = 90◦, and is isosceles with PC = CD. Thus, ∠DPC = 45◦.

Similarly, 4ABQ has AB = BQ = 5 cm and ∠BQA = 45◦. Therefore,

since BC = 8 cm and PB = BC − PC, then PB = 3 cm. Similarly,

AC = 3 cm. Since PQ = BC − BP − QC, then PQ = 2 cm. Also,

∠RPQ = ∠DPC = 45◦ and ∠RQP = ∠BQA = 45◦.

Using four of these triangles, we can create a square of side length

2 cm (and thus an area of 4 cm2). The area of one of these triangles (for

example, 4PRQ) is1

4of the area of the square, or 1 cm2.

So the area of quadrilateral DRQC is therefore 12.5− 1 = 11.5 cm2.

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