A Math Forum Project

Geometry Forum - Problem of the Week

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Barb Hurley

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From: Cecil Wenisch 

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


Nick Szmyd

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From: nick szmyd 

        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


Elissa Colter

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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.



Jen Cotter and Fernando Davila

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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



Jay Hunt

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From: Cecil Wenisch

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.



William Cumming Quale

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        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
    |__________|______|

        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


Henri Piccioto's tenth grade class at Urban School of San Francisco


Eva Galperin, Perry Garvin, Casey Gage, and Mesha Kussman

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* They (working separately from each other) found that 5 was a 
common factor of 45 and 25 (and15), concluded that b = c = 5, and 
worked it out from there, finding that x=27.



Jason Kibbey and Jen Dillon

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* They (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

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



Michelle Powers and Jenny Phillips

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* They (working together) noticed that 

25   45
-- = --
15    x

They solved the proportion, and got x=27.

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30 June 1995