Please Help to find total length of thick line around overlaid rectangles

Sundari

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Nov 14, 2018
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15
Hi,
I am preparing for a test. Please help me to solve this problem.

The perimeter of the rectangle ABCD is 30cm. Three other rectangles are placed so that their centers are at the points A, B, D as in the figure.
The sum of their perimeters is 20cm. What is the total length of the thick line?

perimeter.PNG

A) 50cm
B) 45cm
C) 40cm
D)35cm
E) This is impossible to determine.

Thank you.
 
Hi,
I am preparing for a test. Please help me to solve this problem.

The perimeter of the rectangle ABCD is 30cm. Three other rectangles are placed so that their centers are at the points A, B, D as in the figure.
The sum of their perimeters is 20cm. What is the total length of the thick line?

View attachment 10686

A) 50cm
B) 45cm
C) 40cm
D) 35cm
E) This is impossible to determine.

Thank you.

Have you read our guidelines for submission? You should know that we want to see your own thoughts, so we can tell where you need help. We don't just give answers; and we can't really help without knowing what help is needed.

I can suggest a start, assuming that is where you need help: First think about what the sum of all the heavy lines would be if the rectangles were not overlapping (that is, the sum of the entire perimeters). Then think about how that sum changes as you move each rectangle into place.

There are many other ways to start; you could assign a variable to each side of each rectangle, and simplify the expression representing the length requested, for example. Or, you could choose some specific dimensions for each, and see what the answer is, then think about whether the answer would be the same for other choices.
 
Hi,
I am preparing for a test. Please help me to solve this problem.

The perimeter of the rectangle ABCD is 30cm. Three other rectangles are placed so that their centers are at the points A, B, D as in the figure.
The sum of their perimeters is 20cm. What is the total length of the thick line?

View attachment 10686

A) 50cm
B) 45cm
C) 40cm
D)35cm
E) This is impossible to determine.

Thank you.
Thinking about this problem algebraically may not be the most intuitive approach, but it was the one that seemed obvious to me. Superficially, you have nine unknowns but have only two (obvious) equations. That analysis, if correct, entails that you cannot find all the unknowns and makes E look like the obvious choice. Upon reflection, however, you should realize that the problem does not ask you to find all the unknowns, but only asks you to find one of them.
 
Sure is a "cute li'l problem"!
I did it assigning 8 unknowns:
large rectangle: u by v
rectangle center A: 2a by 2b
rectangle center B: 2c by 2d
rectangle center D: 2e by 2f

2 equations (from the givens):

u + v = 15 [1]

a + b + c + d + e + f = 5 [2]

YOU do the remaining work!!
 
Thank you for the help.


Yes, I had no clue to start the solution.


Now I did with the help.Please confirm that I doing right, kindly let me know whether I am taking the easy way to solve the problem.


Perimeter of ABCD is 30cm
Sum of the perimeter of three rectangles at corner is 20cm.

following the variables as the same as you assigned
large rectangle: u by v
rectangle center A: 2a by 2b
rectangle center B: 2c by 2d
rectangle center D: 2e by 2f

2 equations (from the givens):

2u+2v=30

2(a + b + c + d + e + f )= 20

now to find the length

a+b+a/2+b/2+(v-b/2-d/2)+c/2+d+c+d/2+(u-c/2)+(v-f/2)+e/2+f+e+f/2+(u-e/2-a/2)

which is (a+b+c+d+e+f)+u+v+u+v = 10+30=40

Thank you everyone.

Thanks a lot for all your time and great learning help.
 
Thank you for the help.

Yes, I had no clue to start the solution.

Now I did with the help.Please confirm that I doing right, kindly let me know whether I am taking the easy way to solve the problem.

Perimeter of ABCD is 30cm
Sum of the perimeter of three rectangles at corner is 20cm.

following the variables as the same as you assigned
large rectangle: u by v
rectangle center A: 2a by 2b
rectangle center B: 2c by 2d
rectangle center D: 2e by 2f

2 equations (from the givens):

2u+2v=30

2(a + b + c + d + e + f )= 20

now to find the length

a+b+a/2+b/2+(v-b/2-d/2)+c/2+d+c+d/2+(u-c/2)+(v-f/2)+e/2+f+e+f/2+(u-e/2-a/2)

which is (a+b+c+d+e+f)+u+v+u+v = 10+30=40

Thank you everyone.

Thanks a lot for all your time and great learning help.

Yes, that is correct.

Here is my solution: The total length of all the lines (thick and thin) is 30 + 20 = 50.

The thin lines at each corner add up to 1/2 of the perimeter of that small rectangle (1/4, twice), so they add up to half of 20, which is 10.

We have to subtract that, so the total length of the thick lines is 50 - 10 = 40.

Good work with your algebraic method.
 
The thin lines at each corner add up to 1/2 of the perimeter of that small rectangle (1/4, twice), so they add up to half of 20, which is 10.

sorry, I do not get your solving, there are three small rectangles so 1/4+1/4+1/4 which is 3/4 of 20 .that is 15.




 
The thin lines at each corner add up to 1/2 of the perimeter of that small rectangle (1/4, twice), so they add up to half of 20, which is 10.

sorry, I do not get your solving, there are three small rectangles so 1/4+1/4+1/4 which is 3/4 of 20 .that is 15.


Suppose the three small rectangles have perimeters of P, Q, and R. Then the thin parts of the rectangles are P/4, Q/4, and R/4; and the parts of the large rectangle that are thin are the same. So the total length of line that has been removed from the total is P/4 + Q/4 + R/4 + P/4 + Q/4 + R/4 = P/2 + Q/2 + R/2 = (P + Q + R)/2. But P + Q + R = 20, so this is 20/2 = 10.

The fractions are each multiplied by a length; each is 1/4 of its own rectangle, not of the whole 20. You can't add them together by themselves.
 
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