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Math Forum - Problem of the Week
Submissions for Boxing Up Harry's Broom
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10/16/2007
The smallest width of the box could be about 2.65 feet.
First, I drew a diagram of a rectangle with a 3 foot length and a 4
foot diagonal length going across it, which would make a triangle. I
then used the Pythagorean Theorm (a squared+b squared=c squared).
Since I knew what side b and c were, I inserted what i knew into the
formula, a squared+9=16. Then, I found the difference of 16 and 9
which is 7. Since 7 represents a squared, I found the square root of 7
(about 2.65) which is the shortest width of the box.
10/17/2007
The smallest width is 31.75 inches and the shortest height is 20.78
inches.
Because the box is only 36 inches long, the broom, which is 48
inches, can only be in the box if it is in a diagonal position.
This creates a triangle on the bottom of the box.
If make a the width of the box and one of the legs of the triangle
and b the length of the box which equals 36 and c the hypotenuse of
the triangle which is the broom's length of 48 inches.
You can then use the pythagorean theorem a^2 + b^2 = c^2 to find
a^2 + 36^2 = 48^2
a^2 + 1296 = 2304
a^2 = 1008
a = 31.75 inches (the width of the box)
Extra:
The new box has a length of 24 inches and a width of 36 inches.
So, if you use the pythagorean theorem, you can find out that the
diagonal on the bottom of the box is
24^2 + 36^2 = c^2
1872 = c^2
c = 43.27 inches.
This new value becomes the leg of a triangle that stands upright in
the box. The hypotenuse of this triangle is the broom or 48 inches,
and the height of the triangle is the height of the box.
Leg 1 = 43.27
Leg 2 = a (the height)
Leg 3 = 48 (broom)
a^2 + Leg1^2 = Leg3^2
a^2 + 1872.29 = 2304
a^2 = 431.71
a = 20.78 inches (the height)
10/18/2007
The box would have to be at least 48 inches wide for Harry to fit it in The broom did not fit in the 36 inch long length, yet it fit
widthwise, so the width would have to be 48 inches long to accommodate
laying flat.
the 4 foot long broom, also 48 inches.
10/19/2007
The smallest width of the box that Harry could fit his 4' broom into
flat on the bottom would have to be 31.739" or 2.646'. Extra: The
shortest height of the box would be 20.78" or 1.732'.
Since the length on the box was only 36", or 3', it would be
impossible to lay it straight at the bottom of the box. The only other
way would be to put it inside the box in an angle. If you draw a line
showing the length of the broom, it would make a triangle, the length
of the broom, 48", the hypotenuse. So, just do 362 +
X2 = 482. Subtract 482 by
362, and get the square root of that, and we get the width
that the box would have to be.
Extra
For the extra, I first got the hypotenuse of a triangle with sides 24"
and 36" to get 43.27. Since we know that Harry's broom is 48", we can
do 482 - 43.272 and get the square root of that
to get the height of the box Harry put his broom into.
10/20/2007
The width of the box was 32 inches. Extra = the height of the box
was 21 inches.
one of the two the longest line that can be laid in a rectangle is
the line between one corner on the top right, and one corner on the
bottom left. If we cut the rectangle into two right triangles, this
line becomes the hypothenuse of the triangles. Since Harry's broom
is 48 inches long, and the length is 36 inches, I used the
pythagorean theorim and substituted the variables to the given
numbers.
(36)^ + b^ = (48)^ =
1296 + b^ = 2304.
So b, which is the width of the box, was figured out by the
following steps.
2304 - 1296 = b^
1008 = b^
So, when I solved the root, the closest number that I came up with
was 32. So, the shortest width of the box possible is 32 inches.
Extra :
In a cube, the longest line that could fit in it is between the top,
right-hand corner to bottom left-and corner.
This should be the place Harry put his broom in.
So, first, I got the length of the hypothenuse of the right
triangles that the top square can be divided into, since that line
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would be on of the sides of the triangle that Harry's broom will lie
as the hypothenuse.
24^ + 36^ = c^
576 + 1296 = c^
1872 = c^
So, the c which is the hypothenuse of the top square's triangle is
the square root of 1872. This is one side of the triangle that the
broom is going to be laid on.
So, again, when you substitute the variables of the pythegorean
theorem, it becomes :
1872 + b^ = 2304
I just plainly used 1872 instead of the square root of 1872 squared
because the root and the power cancels each other out, and 2304 is
the square of 48, the length of Harry's broom, or the hypothenuse.
So, it all comes down to this.
2304 - 1872 = b^
432 = b^
b, the height of the triangle can be estimated to about 21 in order
to be able to fit the broom.
Reflection.
Well, the problem was about something that I knew a lot about, and I
think it should be considered easy because a algebra student could
solve it without any help!
10/21/2007
The smallest width of the box could be 9.165 inches or the square
root of 7 feet. The answer to the extra is 1.732 or the square root of
3.
3^2+x^2=4^2
x^2=7
x= 9.165 or sqrt(7)
Extra:
3^2+2^2+x^2=4^2
Pythagorean theorem
H=Height L=Length W=Width B=Broom Y=Hypotenuse of
the triangle on the base of the box
Y^2=L^2+W^2
Pythagorean Theorem
B^2=Y^2+H^2
Pythagorean Theorem
B^2=L^2+W^2+H^2
Substitution
x^2=3
x= 1.732 or sqrt(3)
10/21/2007
Part 1: Square root of 7 feet. Part 2: Square root of 3 feet
Part 1:
Step 1: First I read the problem and it told me that the broom was
48 inches long. The box was 36 inches long.
Step 2: Since the length, width, and the broom form a right
triangle, i could use the Pathagorean Theorum. For C squared i put
in 4 squared, length of broom, for A squared i put in 3 squared and
left B squared as the width.
Step 3:
I solved it and got the square root of 7 feet.
Part 2:
Step 1: I used the formula h(height) squared plus w (width) squared
plus l (length) squared equals 4 squared.
Step 2: I left h squared alone and put 3 squared in for w, and 2
squared in for l.
Step 3:
I then solved and got the square root of 3 feet.
I used the pythagorean theorem to discover what the other side COULD
be if the box was 36 inches long. By squaring 36 and adding that
number to itself, I discovered the hypotenuse would be 50 inches, 2
inchses longer than Harry's broom. Using 35 squared, the hypotenuse
was 35 inches. Using 34 squared, it was roughly 48 inches, the
perfect width!
10/21/2007
34 inches
10/21/2007
The width of the first box must be at least 2.65 feet. The height of the To solve the first problem, I used the pythagorean theorem. To fit properly,
would have had to lie diagnally across the bottom of the box, creating two ri
second box must be at least 1.72 feet.
with the broom's length as the hypotenuse.
Since a^2+b^2=c^2, and C (the hypotenuse) was the length of harry's broom, c^
The legs of the triangle (a and b) were 36 inches (or 3 feet) and "b" because
unknown.
I then plugged in 16 for c^2 and 9 for a^2 since 3^2=9.
I subtracted 9 from 16 to get the value of b^2 (16-9=7), and then took the pr
square root of 7 to get the width of the box, which is approximately 2.65 fee
To solve the extra, I also used pythagorean theorem.
The broom must have fit diagonally from the top corner of the box to the bott
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the opposite side, which also forms a right triangle with the length of the b
being the hypotenuse (c^2=16).
To get the legs of the triangle you must perform another calculation using th
find the distance across the bottom of the box, diagonally. A=2 and B=3 since
the given width and lengths of the box, so 3^2+2^2=13, and the square root of
3.61, making that the distance in feet across the bottom of the box, and one
the triangle containing Harry's broom.
The legs of this triangle are 3.61 and "b" since the height is unknown.
3.61^2=13.021, and 4^2=16, so to get the squared value of "b", 13.021 is subt
from 16 to yeild an answer of 2.9679.
The square root of this number is 1.72 feet making this the minimum height of
10/21/2007
the minimum with of the box was 31.8 inches.
a=length
b=width
c=hypotenuse
9+x^2=16
x^2=7
x=2.65 feet
x=31.8 inches
10/22/2007
It was 48 inches long.
12 times 4=48
10/24/2007
The max width is 9 in for the broom.
9+9+9+=36
10/24/2007
7 ft. is the smallest wotdh the box could be.
36 in. = 3 ft.
4 ft. + 3 ft. = 7 ft.
7 ft.
10/26/2007
The smallest width the box could be is the square root of seven.
EXTRA: THe shortest height the box can be is the square root of
three.
I started with the formula a squared + b squared = c squared.
a squared + b squared = c squared.
3 squared + b squared = 4 squared.
If you want to find b squared you can do the equation:
b squared = 4 squared - 3 squared.
That means b squared = 7.
Now since square roots are the oposite of squared the answer is:
b = the square root of 7.
EXTRA: For this problem I used the same stratagy.
a squared + b squared = c squared.
the square root of 13 squared + b squared = 4 squared.
b squared = 4 squared - the square root of 13 squared.
b squared = 16 - 13.
b squared = 3.
This means:
b = the square root of 3.
10/27/2007
The answer to both problems is 31.75 inches.
First, I drew a picture of the situation.It formed a triangle with
the broom as the hypotenuse, 3 feet as one side, and x for the
other. I used the Pythagorean theorem to solve:
36^2 + x^2 = 48^2
1296 + x^2 = 2304
I subtracted to get:
x^2 = 1008
x = 31.749 or x = 31.75
The extra problem had the same situation but 3 dimensional, but the
answer was the same.
To get the first answer set up the pythagoriam theorm to asy that if the widt
squared + 36 squared= 48 squared. You then multiply to get x squared+ 1296=23
you subtract to get x squared=1008. You then take the square root of both sid
x=31.7490. For the Extra if the diaganol of the base of the box is y and the
can set up the equations 24 squared + 36 squared= y squared. You can also get
x squared + y squared= 48 squared. You then can take the fist equation and si
squared= 1872. Next you substitute this into the second equation to get 1872+
squared=48 squared. This simplifies to x squared= 432, you then take the squa
both sides to get the x=20.7846
10/28/2007
The first box could be 31.7490 inches wide approximated to four
decimal points. The answer for the extra is the box could be 20.7846
inches high approximated to four decimal points.
10/28/2007
Hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm. I don't know any thing you know so i gess. you people are
soooooooooooooooooooooooooooooooooo compla-cated. I guessed and
I'm not sure but I think it is:9852983957876-28758277@#%$&^&
guessed again because you know i am very
stupid.dddddddddddddddddddddddddddduuuuuuuuuuuuuuuuuuuuuuuuuuuhhhhhhh
%!$%62357634526.
hhhhhhhhhhhhhhhhhh
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Short Answer
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10/28/2007
The smallest possible width of the first box was 31.74901... inches
long. The shortest possible height of the second box was 20.7846...
inches long.
The broom laying flat in the box would make the hypotenuse of a
triangle. That means that 36'' ^2 + x^2 = 48'' ^2 or 48^2-36^2=x^2.
Solving that, x being the width, is about 31.75''.
In the second box, the broom lays on a diagonal along the bottom, so
the diagonal is about 43 inches, and then it creates another triangle
with the height being x. X turns out to be about 20.78 inches.
10/28/2007
The smallest width that the bok could be is 31.75 inches.
I got this answer using the pathagorean theorem. 36 inches became a,
the unknown was b, ansd 48 inches, the diagnal became c. I then
solved the problem using the pathagorean theorem and got my answer.
10/28/2007
The smallest width the box could be is 31.749 inches. For the extra: To help me find my answer I first drew a box. I then noticed that
the diagonal of the box would be where the broom sits so it would be
The smallest hieght the box could be is 1.732 feet or 20.784 inches. 48 inches. The diagonal made a triangle in the box so I used the
Pythagorean Theorem to find the width. a^2 + b^2 = c^2, or c^2-b^2 =
a^2. I then filled in the numbers I knew. 48^2-36^2=a^2. So 23041296=31.749^2. So 31.75 inches is the smallest height possible.
The Extra:
Once again I drew a picture to help me. But this time I drew a
cube. I saw that if I found the hypotenuse of the bottom square on
the cube I would be able to find the height. So a^2 + b^2 = c^2,
2^2+3^2=the square root of 13. I then noticed that this number
would be the base of the triangle that included both the brooms
height and the height of the box. So once again I used the same
theorem c^2-b^2 = a^2, 4^2-13=the square root of 3. The square root
of 3 is 1.732 feet, so the smallest height the box could be is 1.732
feet or 20.784 inches.
10/28/2007
For the first problem, we knew that harry's broom was 4 feet long
The smallest width that the first box could be is radical 7, or 2.65,
and that the length of the box that it fix into lying on the bottom
feet. The shortest height that the second box could be is radical 3, or is 36 inches. Because we know the broom was lying flat on the
bottom and the length of the box was shorter than the lenngth of the
1.73, feet.
broom, harry must have placed the broom diagnol from one corner to
the opposite corner of the box. This would create a right triangle
where the hypotenuse was the length of the brrom, 4 feet, and one
side was the length of the box, 36 inches or 3 feet. I used the
pythagoreary theorem a^2+b^2= c^2 to find the length of the other
side, which is the width of the box. Because I had the values of
c^2 and b^2, I subracted b^2 from c^2 to find c^2 and then took the
square root of b^2. I plugged in (4)^2-(3)^2= 7, and then the
square root of 7, 2.65 feet, would be the smallest width that the
box could be.
For the second question, the problem stated that the broom didn't
fit on the bottom, but it did fit in the box. This meant the broom
must have been positioned diagnally, with one end at one upper
corner of the box, and the other end at the opposite bottom corner.
Because of this, I drew a 3 dimensional box with the broom placed
inside. Then, because I knew that the width of the box was 3 feet,
and the length was 2 feet. With these known, I looked at the top
face of the box. On this face I drew a diagonal line from the top
left corner to the bottom right corner, thus creating a right
triangle. I then knew that one side was 3 feet and the other side
was 2 feet because we already knew the length and width of the box.
Then to find the hypotenuse, I used the pythagorean theorem and
plugged in 3 for a and 2 for b. (3)^2+(2)^2= 13, and then I took
the square root of 13 to get just c. I then realized that the
diagonal line that I had drawn in was on the same plane as the
broom, and that I had formed a new triangle with the diagonal line,
the broom and one vertical side of the box, which was the height. I
did the pythagorean theorem again by plugging in the square root of
13 as one side and the length of the broom, 4 feet, as the
hypotenuse. So, (4)^2-the square root of 13=the square root of 3.
The square root of 3, or 1.73 feet, was the shortest height that the
box could be.
10/28/2007
The smallest width the box could be is 2.65 ft. or rad 7 ft. Extra: The I used the pathagorian theorum to determine the height of the box if
the broom were to lie on a diagonal along the bottom. 9+b^2=16.
shortest height of the box is 1.73 ft. or rad 3 ft.
b=2.65 ft.
Extra: I found the hypotenues of the top of the box by doing
9+4=c^2. c=rad 13. Then imagine the broom layed from one corner of
the box to the opposite corner, and do (rad 13)^2+b^2=16. b=rad 3 or
1.73 ft.
10/28/2007
The smallest the width of the box could be is 31.752 inches and the The box is 36 inches long and the hypotenuse of the box would have
to be 4 feet, or 48 inches, so that the broom could fit in it. From
height could be radical 3 feet.
this, if you use the Pythagorean therom in reverse to find the
missing width, you find that it is 2.646 feet or 31.752 inches.
The second box is 2 feet wide and 3 wide so for the broom to fit it
would have to fit diagonally. From this, you have to find the
hypotenuse of the base of the box, which are radical 13. Then the
hypotenuse from the top corner to the alternate base corner has to
be 4 feet so if you use the Pythagorean therom and subtract the base
side to from the hypotenuse you get 3. Then because this is done
using the Pythagorean therom you have to take the square root and
you get the height to be radical 3 feet.
10/28/2007
4 of 6
The first box must be 31.75in wide. The box in the extra must be
20.78in high.
For the first part I read it and realized immediately that it was
just a pythagorian theorem problem; The broom was the hypotunuse and
I already had the length, so I subtracted the length squared from
the hypotenuse squared (48^2 - 36^2), then square rooted the result
to get 31.75. For the extra, it was a double pythagorian theorem
11/24/10 1:42 PM
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problem; I found the hypotenuse of the bottom of the box. Then I
used that as one of the legs, and, once again having the hypotenuse
of 48in, did the same thing as in the first one, (48^2 - 43.27^2) to
get 20.78in.
10/28/2007
The smallest possible width of the box would be radical 7 (ft.) extra: If the first box he found was 3 feet (36 inches)long, and he was
able to fit his four foot long broom diagonally across the bottom,
the shortest height possible for the box is radical 3 (ft.)
the it's possible to find the minimum width of the box by using
pythagorian theorem. Make the three foot side A, and the four foot
broom would be the hypotenus (C), then solve.
a^2 + b^2 = c^2
3^2 + b^2 = 4^2
9 + b^2 = 16
b^2 = 7
b = rad. 7
the smallest possible width of the box is rad. 7.
EXTRA: possible to solve using a cube. if the length of the cube is
3 feet and the width is 2 feet and it's possible to fit a 4 foot
broom into the box, then the broom must go diagonally across fromm
opposite corners. First, find the diagonal of the top 2 ft. by 3 ft.
side using pythagorian theorm.
4 + 9 = c^2
13 =c^2
rad. 13 = c
the diagonal of the top of the box is rad. 13, and it meets the 4
foot broom in a vertex, so you can form a triangle with them and the
height. If know 2 out of 3 sides of a triangle, then the third can
be found using pythagorian theorem.
(rad. 13)^2 + b^2 = 4^2
13 + b^2 = 16
b^2 = 3
b = rad. 3
rad. 3 is the final side of the triangle and the shortest possible
height for the box.
10/28/2007
The smallest width of Harry's box would be 2.646 feet or 31.749
inches. Extra: The shortest height of the box would be 3.464 feet or
41.569 inches.
First, I drew a picture and labeled the length of the box "3 feet"
and drew the broomstick on a diagnol and labeling it "4 feet." Then
I labeled the height of the box "x" since I was trying to figure
that out. I did the pathagorean theorem and figured out that the
height of the box was the square root of 7 or 2.646 feet. In inches
it would be labeled as 31.749 inches.
For the Extra, I drew a similar box, but labeled the information I
already knew differently. The length was 2 feet, hypotenuse was 4
feet because it was the length of Harry's original broomstick, and I
was again trying to solve for "x" for the height of the box. I used
pythagorean theorem again and solved. My answer came to be the
square root of 12 or 3.464 feet. In inches it would be labeled as
41.569 inches.
10/28/2007
The width of the first box is about 31.75 inches. the answer to the
extra was 20.78.
I used the pathagorean theorem, a squared +b squared = c squared, and
plugged in the c, or hypotenuse, of the trangle. The hypotenuse is
the diaganal of the triangle, while the a and b values are the length
and width. I knew one value, which I plugged in for b, which was the
length of 36 inches. I squared both of these values to find that a
squared + 1296= 2304. I then subtracted 1296 from both sides and came
up with a squared= 1008. I took the square root of 1008 and found the
value of a, or the width, of 31.74901573 which i rounded to 31.75.
Extra:
I used the pathagorean theorem, a squared +b squared = c squared. I
then pl,ugged in the length and width, 24 and 36 inches to find that
the squares of these equals 1872 inches squared. the length of the
broom squared is 2304 inches squred so by subtraction i found that
just the length and width were 432 square inches less than needed. So
the height must make up for it and if 432 is the square of the
height, the square root is 20.78 inches.
10/28/2007
The smallest width the box could be is 31.749 inches, and for the
extra, the shortest the box could be is 15.620 inches.
The broom is 4 feet long, which translates to 48 inches. The length
of the box is 36 inches.
Thus, we have one side of the triangle, as well as the hypotenuse.
A^2+B^2=C^2
C^2-B^2=A^2
48^2-36^2=31.749^2
The length is 31.749, or any number larger.
Onto the extra:
A box is 2 feet by 3 feet. 24 inches by 36 inches.
24^2+36^2 = 43.267^2
Now, that is the diagonal length of the box. The broom will go along
that length, but vertically. So we have that number, as well as the
length of the broom, so we do the pythagorean theorem again.
46^2-43.267^2=15.620^2
And thus, 15.620 is the height necessary (or anything higher)
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10/28/2007
31.749 inches
I drew a picture of what the box would look like and then how the
broom would have to lie in the box. I realized that the diagram made
a triangle. I had values for one side and the hypotenuse so I decided
to solve the problem by using the Pythagorean Theorem.
36^2 + B^2 = 48^2
1296 + B^2 = 2304
B^2 = 1008
B = 31.749
EXTRA:
The shortest height of the box would also be 31.749 inches because
they are actually the same measurements and also form a triangle so
when you do the Pythagorean theorem you get the same answer.
10/28/2007
The width is about 2.7 ft
I looked in a text and decided to ask my mom for help. So, I asked my mom and
use the pythagoream Theorem because I knew the length of the box was smaller
was smaller than the length of the broom. So I knew the broom fit diagonally
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