Student Solutions to Congruent Rectangles

Novice/Apprentice Solutions || Practitioner/Expert Solutions

Novice and Apprentice Solutions from Public Submissions: These incorrect solutions were submitted by actual students (whose names have been changed) and are presented here to encourage discussion about how to help them with their mathematical thinking.

James, age 15

The perimeter is 166 units.

I knew that area was length times width. So I knew that 27 times 28
gave me 756. So i added 28 plus 28 plus 27 plus 27 to get my answer
of 166.

Jessula, age 14

The perimeter is 27.5 units.

You must take the square root of 756 units squared and that is the
perimeter.

Dre, age 12

The perimeter of the rectangle is 111.2 units^2.

First I tried to solve it algebraicly.  I used xy=756 but there were
too many answers so I tried finding the area of one of the smaller
squares.  I did this by dividing the area of the big square by seven
because there are seven little triangles in the big one.  756/7 = 108
but that got me no where.

Then I saw the answer.  It takes 3 of the rectangles layed across the
long way to go from one side to the other the long way but it takes 4
rectangles stacked up the tall way to get from one side to the other
the long way.

Hmmm I thought I now have 2 equations and 2 unknowns. They are
3/4x = y and xy = 756.

All I need to do now is solve.  I plugged in 3/4x into y in the seccond
equation.  I now have x(3/4x) = 756.  I then multiplied x*3/4x and
got 3/4x^2 = 756.  Then I divided 756 by 3/4 and got x^2 = 1008.  Next I
took the square root of 1008 and got x = aprrox. 31.8.  To solve for
y you must multiply by 3/4 because 3/4x = y.  Therefore y = approx 23.8.

There are 2 x's and 2 y's for both sides of the rectangle. When you add
those up you get 31.8*2+23.8*2 = 111.2.  The perimeter of the rectangle
is 111.2 units^2.

Matteo, age 15

3024 units

If 756 units2 is the area, and area is calculated by multiplying
width times length, then the width and height most both be 756
units.  Then, if perimeter is calculated by 2L+2W the perimeter must
be 3024 units.

Lissa, age 14

My answer is 230.

I divided 756 by 7, since there are 7 equal triangles. I got 108. I
added 108 and 7 and got 115. I doubled that and got 230, since the
formula for perimeter is (2)l+w.

Chung

The perimeter of the larger rectangle is 57 units.

Let the dimensions of the small rectangles be a, b(length a, width
b), then the area of the large rectangle is:

4b(a+b)=756
From the graph, we can see that 4b=3a, so
b=(3/4)a, then we have
3a(a+(3/4)a)=756
    (7/4)a^2=252
         a^2=144
           a=12 ( -12 rejected)
b=(3/4)a=(3/4)*12=9

The perimeter of the larger rectangle is
 4b+a+b
=a+5b
=12+5*9
=12+45
=57

Practitioner and Expert Solutions from Our Archive: These are examples of correct solutions that were submitted by actual students.

Carson, age 13

The Perimeter of the large rectangle is 114 units.

I thought that it might be easier to work with the little rectangles
rather than trying to tackle the large one. I found the area of the
little ones easily by dividing 756 (the large rectangle) by 7 (the
number of little ones). I did this, because the question said all the
small rectangles are congruent. A congruent shape is is a shape that
has the exact same: dimensions, number of sides, and angles, as
another shape. The question said all of the small rectangles were
congruent, so they'd each contain the same surface measurement. That
is why I decided to divide the large rectangle's area into 7.

756/7= 108

So each little rectangle had an area of 108 units. I then noticed in
the picture it shows that, on the horizantal middle line, it takes 4
little sides to make 3 big sides on the little rectangles. Thus, the
small rectangle's long side multiplied by 3 must be the same as the
small rectangle's short side multiplied by 4. (Lonside x 3)= (Small
Side x 4)I would use this information to try to find a factors of 108
(because all the rectangles are congruent).

With this information I began to guess and check. 27 x 4 equals 108 I
thought. However 27(units)x3(long sides) =81, and 4(units)x4 (short
sides)= 16.  81 is greater 16 so that can't work!

I went through this prosses several times and was trying to find
factors of 108 when I found 12 and 9. 12(units)x 3(long sides)= 36
9 (units) x 4(short sides) = 36. They are both the same, and so that
must be the correct answer.

Now I know that the long sides are 12 and the short sides are 9 on
the rectangle. I go back now to the large rectangle. The top has 4 of
the little triangle's short sides, so I multiply 9 x 4= 36. The sides
have one of the little triangle's short sides and little triangle's
long sides. 12+ 9= 21. Thus the dimentions are 21 x 36. To check this
I do the area (length x width). 21x 36= 756! Thats what the question
said. Now i find the perimeter.
Perimeter=(length x 2) + (width x 2)
P= (36x 2) + (21x2)
P= 114 units.

Michael, age 8

The perimeter of the large rectangle is 114 units.

Because the seven small rectangles are congruent I know that --
The length and the width of each small rectangle are the same.
This means that all the small rectangles have the same area.
The seven small rectangles exactly fill the large rectangle.
So, each of the small rectangles has an area that is one-seventh of
the area of the large rectangle.

Because the area of the large rectangle is 756 square units I know
that --
Each of the small rectangles has an area that is 756/7, or 108
square units.

If x is the width of the small rectangle, and y is the length of the
small rectangle:
(See diagram)
Because the area of a rectangle is equal to the width times the
length I know that --
 xy=108

Because opposite sides of a rectangle are equal; and 4 short sides
of the small rectangle fit on one side of the large rectangle, but 3
long sides of the small rectangle fit on the opposite side of the
large rectangle I know that --
      4x=3y
 (1/4)4x=(1/4)3y
       x=3/4 y

By combining these two equations I know that --
       (3/4 y)y=108
 (4/3)(3/4) y^2= 108(4/3)
    (12/12) y^2=(108/3)(4/1)
            y^2=36(4)
            y^2=9*4*4
            y^2=3*3*4*4
            y^2=(3*4)^2
            y^2=12^2 = ((-1)^2)*(12^2)=(-1*12)^2
So,           y=12 or y=-12
But, y cannot be -12 because it's a measurement and measurements
cannot be negative.

So,
 x=(3/4)12
 x=9

By replacing x and y in the diagram with these values I know that --
The long sides of the large rectangle are 3*12, or 4*9, or 36 units
long and the short sides are 12+9, or 9+12, or 21 units long.

I checked these measurements to see if they could be for a rectangle
that has an area of 756 square units, with
36*21=720+36=756

So, the perimeter of the large rectangle is the sum of the lengths
of all of the sides, or
2(36) + 2(21) = 72 + 42 = 114 units.

Brian, age 14

The perimeter of the original largest rectangle is 114 units, with the
base being 36 and the height 21.

I looked at the seven congruent rectangles, and knew that three of
them lying on the longer side was an equal length to four of them
lying on the short side, so I realized that 3A=4B, where A and B are
side lengths.  I also knew that if the total area was 756 units
squared, then one seventh of that was 108 units squared, the area of
each smaller conruent rectangle.  So the sides, A and B, must have a
product of 108, so, B=108/A.  I then plugged what I knew B was in
terms of A into my first equation, and then solved for A.
So,
     3A=4B
     3A=4(108/A)
     3A=432/A
   3A^2=432
    A^2=144
      A=12

With A being 12, I put it into the equation B=108/A, and solved for B,
which turned out to be 9.  I then checked them in both original
equations, and 12*9 does indeed have a product of 108, and 3A does
equal 4B.  With all my checking complete, I set about to solve the
original problem.  I knew the base was 3A, or 36, so that was one
side. The height wwas A+B, which is 21.  I added all the sides up, and
got a sum of 114.  I wasn't done, as a final check I made sure 21*36
equals 756.  It does.  I knew I was as close to correct as I was going
to get alone, so I sent it to you.


I think my answer is correct because all my life algebra hasn't failed
me, and I changed both the equations to y=mx=b format, and graphed
them both and then used to calculator to calculate the intersection,
which it did so at -12, -9, and +12,+9.  this told me that I was
indeed correct, and I shouldn't argue with all the data.