Jessica's Solution

From: Jessica, age 16, class: 11

Answer:

The area of the complete square is 81 units.

Explanation:

If the one part of the square has an area of 35 units2, then it could
be 7 units( the top) by 5 units(the right side).  Then, the part of
the square that is 8 units2, could be 4 units(the left side) by 2
units(the botton).  Therefore, the part of the square under the 35
units2 would be 4 units( the right side) by 7 units(the bottom).  And
the other part would be 5 units(the left side) by 2 units(the top).
Thus, the entire square would be 9 units by 9 units.  The area would
then be 81 units2.

Another way to do this problem would be to let the top of the 35
units2 part be 5 units and the right side to be 7 units.  This would
cause the square to be 11 units by 7 units.  Therefore the area would
be 77 units2.

Lastly, another way to solve this problem would be to allow the 35
units2 square to be 35 units(the top) by 1 unit(the right side).  Or
to allow the 8 units2 square to be 8 units by 1 unit.  Then just
switch the lengths of the sides around to get different areas.



From: [the mentor]

>The area of the complete square is 81 units.
>
>If the one part of the square has an area of 35 units2, then it could
>be 7 units( the top) by 5 units(the right side).  Then, the part of
>the square that is 8 units2, could be 4 units(the left side) by 2
>units(the botton).  Therefore, the part of the square under the 35
>units2 would be 4 units( the right side) by 7 units(the bottom).  And
>the other part would be 5 units(the left side) by 2 units(the top).
>Thus, the entire square would be 9 units by 9 units.  The area would
>then be 81 units2.
>
>Another way to do this problem would be to let the top of the 35
>units2 part be 5 units and the right side to be 7 units.  This would
>cause the square to be 11 units by 7 units.  Therefore the area would
>be 77 units2.
>
>Lastly, another way to solve this problem would be to allow the 35
>units2 square to be 35 units(the top) by 1 unit(the right side).  Or
>to allow the 8 units2 square to be 8 units by 1 unit.  Then just
>switch the lengths of the sides around to get different areas.

Hi Jessica.  How do you KNOW that you picked the right numbers, and how do you
KNOW that you put them in the right places?  I think you're forgetting an
important part of the problem - that the enclosing figure is a square.  See
how that affects your answer.

 -[the mentor]

--
[the mentor], for the Geometry Problem of the Week
http://forum.swarthmore.edu/geopow/



  
From: Jessica

Answer:

The area of the complete square is 81 units.

Explanation:

If the one part of the square has an area of 35 units2, then it could
be 7 units( the top) by 5 units(the right side).  Then, the part of
the square that is 8 units2, could be 4 units(the left side) by 2
units(the botton).  Therefore, the part of the square under the 35
units2 would be 4 units( the right side) by 7 units(the bottom).  And
the other part would be 5 units(the left side) by 2 units(the top).
Thus, the entire square would be 9 units by 9 units.  The area would
then be 81 units2.  I know that I am right because if it is a square,
then the lenght must equal the width.  And placing any other values in
for the lenth and width of the smaller squarers would not make them
equal.




From: [the mentor]

>The area of the complete square is 81 units.
>
>If the one part of the square has an area of 35 units2, then it could
>be 7 units( the top) by 5 units(the right side).  Then, the part of
>the square that is 8 units2, could be 4 units(the left side) by 2
>units(the botton).  Therefore, the part of the square under the 35
>units2 would be 4 units( the right side) by 7 units(the bottom).  And
>the other part would be 5 units(the left side) by 2 units(the top).
>Thus, the entire square would be 9 units by 9 units.  The area would
>then be 81 units2.  I know that I am right because if it is a square,
>then the lenght must equal the width.  And placing any other values in
>for the lenth and width of the smaller squarers would not make them
>equal.
>

Hi Jessica.  That looks better - thanks for revising!

 -[the mentor]

--
[the mentor], for the Geometry Problem of the Week
http://forum.swarthmore.edu/geopow/

Lindsay's Solution

 
From: Lindsey, age 15, class: 9/geometery

Answer:

My final answer is 81.

Explanation:
A square is the same lenth on every edge. That was the first step in solving the problem. I found
the mulitples for 35 to be 7 and 5 breaking down the square to 2 lines. The other square with the
number 8 in it had the multiples 2 and 4. I placed the 2 and 7 on the horazontal lines to equal 9.
Then I placed the 4 and 5 on the vertical lines to equal another 9. This made the square 9x9. To
find the area of a square you multipy bolth sides.Nine times nine equals 81, therfore the area of
the square is 81. 


From: [the mentor]

Hi, Lindsey,
Your answer is right, and your methods used to solve are well thought
out.The only problem is in your explanation, as I note below:

>My final answer is 81.
>
>A square is the same lenth on every edge. That was the first step in solving
>the problem. I found the mulitples for 35 to be 7 and 5

Multiples of 35 would be numbers like 35, 70, 105, 140, ... 7 and 5 are not
multiples of 35.

>breaking down the square to 2 lines. The other square with the number 8 in
>ithad the multiples 2 and 4.

Multiples of 8 would be 8, 16, 24, 32, ... not 2 and 4.

>I placed the 2 and 7 on the horazontal lines to equal 9. Then I
>placed the 4 and 5 on the vertical lines to equal another 9. This made the
>square 9x9. To find the area of a square you multipy bolth sides.Nine times
>nine equals 81, therfore the area of the square is 81.

Would you mind changing your solution to replace the word "multiples" with
the correct terminology and resubmit?(We'd prefer not to post solutions
with the word "multiple" used in this way.)Other that that, you've done
good work.

--
[the mentor], for the Geometry Problem of the Week
http://forum.swarthmore.edu/geopow/




From: Lindsey

Answer:

My final answer is 81.

Explanation:
A square is the same lenth on every edge. That was the first step in solving the problem. I found
the factors for 35 to be 7 and 5 breaking down the square to 2 lines. The other square with the
number 8 in it had the factors 2 and 4. I placed the 2 and 7 on the horazontal lines to equal 9.
Then I placed the 4 and 5 on the vertical lines to equal another 9. This made the square 9x9. To
find the area of a square you multipy bolth sides.Nine times nine equals 81, therfore the area of
the square is 81. 


From: [the mentor]

>My final answer is 81.
>
>A square is the same lenth on every edge. That was the first step in solving
>the problem. I found the factors for 35 to be 7 and 5 breaking down the
square
>to 2 lines. The other square with the number 8 in ithad the factors 2 and
4.
>I placed the 2 and 7 on the horazontal lines to equal 9. Then I placed the 4
>and 5 on the vertical lines to equal another 9. This made the square 9x9. To
>find the area of a square you multipy bolth sides.Nine times nine equals 81,
>therfore the area of the square is 81.

Hi Lindsey,
Thank you for submitting your revision to your solution.Your revised
solution is exactly right.(Factors was the word we were looking for.)Your
technique of factoring the areas and trying different combinations until you
found a square was just what we were looking for.Keep up the good work.

--
[the mentor], for the Geometry Problem of the Week
http://forum.swarthmore.edu/geopow/