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Topic: POW Solution, Dec. 13-17
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Problem of the Week

Posts: 292
Registered: 12/3/04
POW Solution, Dec. 13-17
Posted: Dec 20, 1993 10:40 AM
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******************************

Problem of the Week for December 13-17:

____________________________
| | |
| | |
A rectangle is divided into four | 45 | 25 |
rectangles with areas 45, 25, 15, | | |
and x. Find x. |__________________|_________|
| | |
| X | 15 |
|__________________|_________|
******************************

Solutions were received from :

Jen Cotter and Fernando Davila Grade 7, School of the Holy Child, PA
Eva Galperin Grade 10, Urban School of San Francisco
Perry Garvin Grade 10, Urban School of San Francisco
Casey Gage Grade 10, Urban School of San Francisco
Mesha Kussman Grade 10, Urban School of San Francisco
Jason Kibbey and Jen Dillon Grade 10, Urban School of San Francisco
Duskin Drum Grade 10, Urban School of San Francisco
Michelle Powers and Jenny Phillips Grade 10, Urban School of San Francisco
Barb Hurley Grade 10, Steel Valley High School, PA
Nick Szmyd Grade 10, Shaler Area High School, PA
Elissa Colter sophomore, Fairfield High School, CT
Jay Hunt Grade 9, Steel Valley High School, PA
William Cumming Quale Virginia
Howard L. Hansen Western Illinois University
David Radcliffe University of Wisconsin
Leung Wang Yip Tom Hong Kong

A number of the solutions are included below.

========================
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Barb Hurley, grade 10, Steel Valley High School
(Pennsylvania)


To find the value of x, first I made a ratio out of the two given
numbers and reduced it.
Bottom / Top = 15/25 = 3/5


Next I set up an unreduced ratio with x and 45.
Bottom / Top = x/45


Then cross multiply the reduced ratio by the unreduced ratio.

Bottom / Top 3/5 = x/45 45(3) = 5x


Last of all, solve for x. 45(3) = 5x
135 = 5x
27 = x
=========================
From: nick szmyd <szmyd@one.sasd.k12.pa.us>

To find the area of X, you have to start and find the area of
one section of the rectangle. Let's start with 25 since the only way
to get 25 from multiplication is from 25x1 and 5x5. You know that
each one of the sides has to be a multiple of 15 and/or 45. With that
information, the only reasonable answer is 5x5. Now, since each of
the sides is equivalent to 5 units, you must find out the other
side's units of both rectangle 45 and rectangle15. What times 5
gives you 45. The answer is 9. What times 5 gives you 15. The answer
is 3. So, now all the sides are known of each rectangle. The
rectangle X shares the long side of rectangle 45 and the short side
of rectangle 15. Their lengths are 9 and 3, in that order. Now all
you do is multiply them to find the area of rectangle X. The final
answer is 27 units.

Nick Szmyd - Shaler Area High School
============================
From: PDALEY@fair1.fairfield.edu

Problem of the Week for December 13-17
Submitted by Elissa Colter, sophomore, Fairfield High School

By using several corollaries, definitions, and basic multiplication, the
value of x was determined to be 27.

First, 5 is a common factor of both 25 and 15 so the dimensions of the
upper right rectangle are 5 X 5 and the dimensions of the lower rectangle
are 5 X 3. Using the corollary, "the opposite sides of a parallelogram
are congruent", the shorter side of the upper left rectangle is 5, giving
the longer side 9, to yield a product of 45. The lower left rectangle
uses the opposite sides corollary again, giving the shorter side the
measure of 3 and the longer side the measure of 9. Therefore, the area
of the lower left rectangle is 27 square units.

=================================
From: caroline@forum.swarthmore.edu (Caroline Brennan)

Name: Jen Cotter - grade 7 - School of the Holy Child
Fernando Davila - grade 7 - School of the Holy Child

Solution: To find what x equals, we first recalled the formula for area
(A=LxW). We said that the only two multiples for 25 are 5 and 5. So the
shorter side of the '45' rectangle would also have to be 5. 45/5 = 9. So
the longer side would be 9. If the base of the '15' rectangle was 5, the
height would be three. Using 3 and 9, we decided that x = 27.


Teachers: Neil Smart and Caroline Brennan

==================================
From: Cecil Wenisch <wenischc@one.svsd.k12.pa.us>

Submitted by Jay Hunt, grade 9, Steel Valley High School,Pennsylvania

Common Multiples of 25,45: 1,5
Common Multiples of 25,15: 1,5

So, A B
25 could be 5*5 or 1*2
15 could be 5*3 or 1*15
45 could be 5*9 or 1*45

Values in column B are impossible.

So. using values in column A, x = 3*9 = 27.

====================================
From: William Cumming Quale <wquale@cvgs.schools.virginia.edu>

Looking at this, the first thing that hits me is that
the box with area 25 has sides either five and five or one and
twenty-five. As twenty-five is not a factor of either 45 or
15, the sides of the area 25 box must be five and five.
If one of the sides of the 15 box is five, then the
other must be three; the same logic dictates that the other
side of the 45 area box is nine. Now the diagram is as
follows:

9 5
_____________________
| | |
5| 45 | 25 |5
|____________|______|
| x | 15 |3
|____________|______|
5

It can now easily be seen that the two sides of the
rectangle with area x are nine and three, so its area is 27.

Will

=========================================
From: hot@SOE.Berkeley.Edu (Henri Picciotto)

My tenth graders at the Urban School of San Francisco worked on this in class,
individually or in small groups. A number of students came up with a solution
fairly quickly. I asked them to try and show that their answer was the only
one possible. A few succeeded in doing that.

We do not have email at the school, so I am typing the seven answers that were
turned in. I am summarizing, in order to keep the typing manageable. They each
used their own variables, but I will use the ones shown in the figure above.

* Eva Galperin, Perry Garvin, Casey Gage, and Mesha Kussman (working
separately from each other) found that 5 was a common factor of 45 and 25 (and
15), concluded that b=c=5, and worked it out from there, finding that x=27.

* Jason Kibbey and Jen Dillon (working together) figured out that a=x/d,
b=15/d, and c=45d/x. Then they wrote that (15/d)(45d/x)=25. Cancelling the
d's, they got that 675/x=25. They solved for x and got x=27.

* Duskin Drum wrote that x = ad = (45/c)(15/b) = 675/bc = 675/25 = 27

* Michelle Powers and Jenny Phillips (working together) noticed that

25 45
-- = --
15 x

They solved the proportion, and got x=27.





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