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POW Solution, Feb. 14-18
Posted:
Feb 21, 1994 6:06 PM
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*****************************************
Problem of the Week, February 14-18
When last we left our heroes, Jill and Jimmy, they had figured out that
the bridge over the river would need to be 65 meters long. Here's a
picture (without the bridge, courtesy of Gino Perrotte):
|
|
| 16m
|
__________ _______|
10m ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 7m
46m
Now, people who use the river are starting to ask questions. In
particular, there is a company who floats barges up and down the river.
Their biggest barges are 30 meters in width. How high above the water
could the top of one of their barges be and still fit under the bridge?
*****************************************
This week we received a lot of responses. About one third of them were right
on, one third were _real_ close, and the other third of them were correct
answers to the wrong problem - many people wrote to tell me what size barge
would fit under any part of the bridge. The most important part of answering
any question is figuring out exactly what is being asked.
There were also a few different methods of finding the answer to last week's
problem. One, and the one I admit I had in mind, used similar triangles.
Another which was popular was to use trig - knowing the angle of elevation of
the bridge, you could find the proper height for the limiting barge.
With that said, here are the folks who submitted correct answers to last
week's problem. Their solutions can be found below.
Tom Corbett Grade 12, Shaler Area High School
Lindsey Becker 9th grade, Fairfield HS
Other solvers from Fairfield High School:
9th grade 10th grade
Scott Dwyer Elissa Colter
Anna Mata Natalie Painchaud
Kathy Medlin Nghiem Vu
Rebecca Naughton
Jan Wilson
Maureen Kennedy Grade 10, Shaler High School
Mel Monteleone Grade 10, Shaler High School
Amanda Balcer Grade 10, Shaler Area High School
Anna "the artist" Luczak Grade 10, Shaler Area High School
Michelle Halliday Grade 10, Shaler Area High School
Kristy Torchia Grade 10, Shaler area high school
Kate Rusbasan Grade 10, Shaler Area High School
Katie Getsy, grade 9, and Ben Ladik, grade 12, Steel Valley High School
*****************************************
From: Tom Corbett <corbett@one.sasd.k12.pa.us>
The highest that the barge can stick out of the water (rounded to the
nearest tenth) is 6.6 meters.
This answer can be derived using proportionality. The point that the
barge will touch (according to the given diagram) will be the left
hand side. To maximize height, figure out the farthest that the
barge can be from the angle created on the left side. Since the
barge is 30 meters wide, the farthest it can be from the left side is
26 meters. [ 46m (water) - 30m(barge) + 10m (land)]
Now make the left side of the barge x. Using proportionality, we
get:
16 x
- = -
63 26
(63 is the length of the leg of the triangle created by the bridge)
Solving for x, we get x to equal 6.6
Tom Corbett, Grade 12, Shaler Area High School
*****************************************
Solution from Lindsey Becker, 9th grade, Fairfield HS
The bridge forms a right triangle.
Draw the height connecting the bridge and a point 30 meters into the
river from the tallest side.
Height is perpendicular because it is the shortest distance from a
point to a line.
Height of the abuttment and the height drawn connecting the river
to the bridge are parallel because both are perpendicular to the
same line.
Forms two similar triangles because if a parallel line is drawn to any
side of a triangle, it forms two similar triangles.
Now the sides are proportional:
26/63 = x/ 16
26 -> the sum of 10 and 16 where 16 is the width of the river
remaining when a 30 m barge goes under the bridge
(46 - 30)
63 -> the sum of the two parts on land and the width of the river
(10 + 46 + 7)
16 -> the bridge abuttment
Solve for x: x = 6.6 meters.
Therefore, a barge 30 m wide and 6.6 m or less high will fit under
the bridge.
*****************************************
From: maureen kennedy <kennedy@one.sasd.k12.pa.us>
Jill and Jimmy's bridgeover the river was 65 meters long. If the
company's biggest barges were 30 meters in length, they have to have
a height requirement because of the bridge. The river itself was 46
meters in length. The bridge is to be slanted at an angle.
The two abuttments on either end are 10 and 7 meters, left to right.
On the 7 meter side, there is a 16 meter high wall that is the
original wall. Jill and Jimmy have to figure how high the company's
largest barges must be to satisfy both the public and the company.
To do this problem, proportions will need to be used in order to find
the exact point at which the barges' height is passable. First,
find the total length of the river and both abuttments. This totals
63 meters. Next, find how much of the river is left with the barge
in it, or 46 - 30 = 16m, plus the 10m abuttment, w3hich is 26m.
Using the 16m wall, we find the proportion:
63/16 = 26/x
X = 6.603 meters, which is the maximum height of the barge
Maureen kennedy, grade 10, shaler
*****************************************
From: Carl Detzel <detzel@one.sasd.k12.pa.us>
Problem of the week by: Mel Monteleone grade 10
It is given that the largest barges are 30 meters in width
leaving 16 meters of the river, because the river is originally 46
meters. The question is, "How high above the water could the top of
one of their barges be and still fit under the bridge?" The way to
find out would be using the idea of similar triangles. One would
have to take the height of land to the bridge which is given to be
16, and the width of the whole land area which is given to be 63
meters. The other triangle's height has to be found so it is given
the name "x", which would be over 26 because of the ten meters inland
and the left over 16 meters. The two should be set equal to each
other, which would look like this:
16/63 = x/26
Next, you should cross multiply, and afterwards, it looks
like this:
63x = 416
After cross- multiplying, you would need to divide, which
gives the answer:
x = 6.6031746
The barge must be less than 6.6 meters in height.
/|
/ | original:
/ |
/ |
/ | 16
/ |
/ |
/ |
/_________|^^^^^^^^^^^^^^^^^|_____|
10 m. 46 m. 7 m.
/|
/ |
/ |
/ |
/| / | 16
/ | / |
/ | x / II |
/ I | / |
/________________| / __________|^^^^^^^^^^^^^^^|____|
16 63
*****************************************
From: kate rusbasan <rusbasan@one.sasd.k12.pa.us>
To do the problem of the week I used similar triangles.
Pretend that there is a bridge connecting the angle.
|
|
| |
| x | 16m
|----------|-------------------------| |
| 16m | 30m | |
--------^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^-----|
10m 46m 7m
On the picture I sectioned off the 30 meters because it is the length
of the longest barge.
I took these two triangles and made an equation of x /26 = 16/63
|
|
|
| | 16
| x |
| |
------------------| ------------------------------------|
26 63
After I got the equation, I crossed multiplied getting:
63x = 416 63x/63 = 416/63
By doing this I got x= 6.603
By: Amanda Balcer
*****************************************
From: m1 <m1@one.sasd.k12.pa.us>
Assuming the triangle is still a right triangle. Intoducing
AB which is perpendicular to the river, and touches the river 28
meters inland from the 7 meter mark. X= the height. 16 is the
distance from the left side of the barge to the 10 meter mark. 63
meters is the river + the inlands. 26 meters equals the new triangle
formed of the river = 10 meters. Using cross multiplication:
26/63=X/16
416=63X
X=6.6
In conclution, the tallest barge that can fit under the
bridge is 6.6.
|
X |
| |
16 | 30 | 16
------^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^----|
10 46 7
Anna "the artist" Luczak
Grade 10
Shaler Area High Scool
*****************************************
From: m4 <m4@one.sasd.k12.pa.us>
|
|
|
| | |
x| 30M |
-------|^^^^^^|^^^^^^^^^^^^^^^^^^^^^^-----|
10M | 16M | 46M | 7M
In this picture the water equals 46M. By the definition of
similar triangles I concluded that the highest barge could be 6.6m,
since the bridge equals 65m. I took measure of the bottom of the new
little triangle which equals 26m, and put that over the measure of X.
I then cross multiplied that with 63, which is the measure of the
entire total on the bottom, and I put that over 16m which is the
measure of abuttment. When I cross multiplied I got the solution of
6.6m.
26/X =63/16
6.6m
Michelle Halliday
Grade 10
Shaler Area High School
*****************************************
From: Carl Detzel <detzel@one.sasd.k12.pa.us>
Problem of the week: Kristy Torchia grade10 shaler area high
school
It's already given that the biggest barges are thirty meters
in width. This leaves sixteen meters. You divide the big triangle
using similarity so you have a big triangle and a smaller one. The
smallest side of the big one is still 16 because the side is the same
as the original triangle. Next you would take ten and sixteen from
the smaller triangle and add them which would give you 26. The
height of the smaller triangle would be given the name "x".
Therefore, back to the big triangle, the height is still 16, and the
length of the base is 63. Then you have x/26=16/63. You have the
16/63 and x/26 because of similarity. After cross-multiplying you
have 63x=416. You have to divide by 63. The 63's leave you with x
and other side would give you 416/63 or 6.603. Once solved the final
answer is 6.603 meters for the highest point it could be.
.|
. | 16m
. |
. | |
. X| |
. | 30m | |
---------^^^^^^^^^^^^^^^^^^^^^^^^---------|
10m 46m 7m
*****************************************
From: Drew Ludwig <ludwig@one.sasd.k12.pa.us>
Mike Gelik
shaler grade ten
The height of the barge would have to be less than 6.6 meters high.
When the barge is extended from the river bank it would reach out 30
meters out into the water, going straight up perpindicular to the ege
of the barge it would take 6.6 meters to reach the bottom of the
bridge. The bridge extends from the bank at a 14 degree angle so the
barge coul easily fit unger the bridge at the high parts but as the
height of the bridge went down the height of the barge goes down as
well. The heights of the bridge and the barge are proportional. The
barge will have to travel as close to the bank on the abutment side
as possible, it has to like scrape the sides. The statements above
will only be true considering these two things.
1. the water in the river is deep enough for thhe barge to navigate
at all pionts
2. the barge has to be thick enough to hold things, it couldnt be one
milimeter thick.
If these are true the height of the barge would be less than 6.6
meters high in order for it to fit under the bridge, but there is
going to be some extra space at the one side so there could some more
height on that side and it could fit.
*****************************************
From: kate rusbasan <rusbasan@one.sasd.k12.pa.us>
This is my solution for the problem of the week. Using the drawing
of the bridge, I only considered the case of the barge using the left
side of the river. Taking the length of 56, which is the length of
the full river plus the far left land which is under the bridge at
the greatest slope, I then subtracted 30 for the use of the barge.
The resulting length is 26. Using a formula of height/length =
height/length. I then solved for x and the maximum height of the
barge that would allow it under the bridge was 6.6 meters. This will
always be dependant on the height of the river.
x = 16
--- ---
26 63
x will equal 6.6
By: Kate Rusbasan
*****************************************
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>
To: pow@forum.swarthmore.edu
Subject: POW for Feb. 14-18
Solution to problem of the week for Feb.14-18.
Submitted by Katie Getsy, grade 9, and Ben Ladik, grade 12, Steel Valley High
School, Pa.
]
] ] 16 m
y
x degrees
[----------^^^^^^^^^^^^^^^^]^^^^^^^^^^^^^^^^^^^^^^-------]
10 16 30 7
To find the height of the barge we took the 16 m. (height) of the bridge, the
63 m. (width) of the land to the abuttment on the other side, and we figured
out the angle of elevation which would be 14 degrees.
tan x = 16 / 63
tan x = .25
x = 14 degrees
Then we used the fact that the barge would go to the far right so it would
have
the maximum clearance possible, and we used trigonometry, (which we recently
learned), to find the height of the barge. We took the 14 degree angle, the
26m,
length, and used "y" to find the height. We took tangent 14 degrees and set
it equal to "y" over 26. Next, we got .2493 as the tangent of "y" and
multiplied it by 26. We got 6.5 meters.
tan 14 degrees = y / 26
.2493 = y / 26
6.5 = y
In order for the barge to fit , it must be less than 6.5 meters high.
*****************************************
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