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The shape of a flower bed is a combination of semicircle and

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The shape of a flower bed is a combination of semicircle and [#permalink] New post 12 Jan 2005, 08:44
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The shape of a flower bed is a combination of semicircle and rectangle with semicircle QR on top and rectangle PQRS at the bottom. If arc QR is a semicircle and PQRS is a rectangle with QR>RS. What is the perimeter of the flower bed?

(1) The perimeter of rectangle PQRS is 28 feet
(2) Each diagonal of rectangle PQRS is 10 feet long
[Reveal] Spoiler: OA
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 [#permalink] New post 12 Jan 2005, 11:04
It is indeed "C". Using Joe's last combined eqn (L – 8) (L – 6) = 0
meaning QR is either 8 or 6. But we know that QR > RS...so if QR is 8 than we can find that RS is 6 which satisfies QR > RS...if QR is 6 than RS is 8 which doesn't satify QR > RS.....meaning QR has to be 8...Suff
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Re: DS 3 [#permalink] New post 12 Jan 2005, 22:04
Given that QR>RS,
Suppose side QR=Length (L), and side RS = Breath (B)
From I,
2(L+B)=28
L+B=14 ..................... (i)

from II,
L^2+B^2=H^2
L^2+B^2=10^2

add 2LBin both sides

L^2+2LB+B^2=100+2LB
(L+B)^2=100+2LB ........... (ii)

putting valu of L+B in eq ii
14^2=100 +2LB
LB=48
L=48/B .............................(iii)

Putting value of L in eq (i)
48/B+B=14
B^2-14B+48 = 0
B=6 and 8,
A=6 and 8
but from question L>B,
L=8 and B=6.
Hence , OA is C.
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 [#permalink] New post 14 Oct 2005, 10:40
B.

Perimeter of flower bed = perimeter of semi circle + perimeter of rectangle - diameter of semicircle

(1) just knowing perimeter of rectangle we cannot find sides of rectangle (they can be 7,7 or 8,6 etc)
(2) of both diagonals are equal, i guess rectangle is square(?). Anyway, we can find sides using pythagorus theorem and once sides known, we can find perimeter of semi circle as well.
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 [#permalink] New post 14 Oct 2005, 10:47
Assume QR = y and RS = x. So y > x.

From statement (1), we have 2x + 2y = 28 -> x + y = 14. Possible values of x and y can be (2,12), (3,11), (4,10), (5,9) and (6,8). SO we can't compute the perimeter of the semicircle. Statement 1 is not sufficient.

From statement (2), we have x^2 + y^2 = 10. There is only one possible (x,y) pair that will satisfy the equation and also the inequality y > x. This is (1,9). We can not compute the perimeter of the semicircle, and thus the perimeter of the flower bed. Statement 2 is sufficient.

B is the answer.
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 [#permalink] New post 14 Oct 2005, 10:49
B.

1. insuff. L and w of rectangle could be anything as long as QR>RS.

2. suff. the diagonal of 10 means that it forms a 6:8:10 triangle PQ:QR:diagonal. Fomr there you should be able to figure out the rest.
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Edited for clarity [#permalink] New post 14 Oct 2005, 13:12
I am lost. I came up with C. Somebody please help me understand if the answer is B. From above,

1. In a Rectangle, I thought the diagonals are always equal.

2. It is not given that x and y have to be integers. So x^2 + y^2=10 can be satisfied by various values of x and y (with x>y)

3. 6:8:10 triangle PQ:QR:diagonal. This is just a ratio - How do we calculate the actual perimeter?

general ex: x=5, y=5*sqrt(3) and diag = 10.
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 [#permalink] New post 14 Oct 2005, 21:31
OK. I will answer each question. Please, if anyone finds a flaw in my logic, let us know. The last thing I want to do it is to give false logic.

1. You are correct. The diagonals are equal in rectangles. They could have just given us 1 diagonal. Maybe they gave us 2 diagonals to throw us off.

2. You are partially correct. x and y could be almost anything with a diagonal of 10 if the triangle wasn't a right triangle. However, in this case the diagonal (hypotenuse) forms a right triangle with the length and wide of the rectangle serving as the legs of the triangle. If there is a hypotenuse of 10, then the 2 sides have to be 6 and 8. The stem says that QR>RS.

3. Yes this is just a ratio, but we are told that the absolute length of the diagonal is 10 feet. We can use this from the reasoning in question 2 to get the absolute length and width of the rectangle. We know that the longer side serves as the diameter of the semicircle. All these vaules can be used to figure out the perimeter.

I hope I explained that well enough. Please let me know if I need to clarify anything.
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Re: help [#permalink] New post 14 Oct 2005, 22:46
chriswil2005 wrote:
2. You are partially correct. x and y could be almost anything with a diagonal of 10 if the triangle wasn't a right triangle. However, in this case the diagonal (hypotenuse) forms a right triangle with the length and wide of the rectangle serving as the legs of the triangle. If there is a hypotenuse of 10, then the 2 sides have to be 6 and 8. The stem says that QR>RS.

I hope I explained that well enough. Please let me know if I need to clarify anything.


Thanks for your explanation Chris. I did not get the point# 2. Pythagorean theorem says a^2 + b^2 = c^2, where c is hypotenuse.

If we take a = sqrt(80) and b = sqrt(60), then we would still get c = 10 which is diagonal.

Is it a MUST to have other 2 sides to be 6 & 8 and not others?
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 [#permalink] New post 15 Oct 2005, 05:54
I simply can not understand what your reasonings are. For me the only correct answer is C. Let me explain the following:

Just take two sets of numbers for wich the sum of the squares of theirs elements is 100.

x=8 and y=6
x=9 and y=sqrt(19) . Please, note that 1^2+9^2 is different of 10 and of 10^2.

So, there are two different values for the perimeter.

For x=8 and y=6 the perimeter will be P=pi*x/2 + x +2y=32.56

For x=9 and y=sqrt(19) the perimeter will be P=pi*x/2 + x +2y=45.99

This prove that (2) is simply insufficient. (1) and (2) together are sufficient.

Please, post OA. I can not uderstand what people is arguing!!

Last edited by automan on 30 Oct 2005, 05:54, edited 1 time in total.
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Re: Edited for clarity [#permalink] New post 15 Oct 2005, 13:42
gsr wrote:
general ex: x=5, y=5*sqrt(3) and diag = 10.


My question is the same as sudhagar's and automan's -

QR>RS and diagonal is 10
QR=5*sqrt(3)
RS=5
Diag=10

The ratio thing will work out only if we know 2 sides.

IMHO the answer is C
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 [#permalink] New post 15 Oct 2005, 14:17
OK, I can be way wrong here, so I am hoping someone will jump in and correct me. I am 98% sure that B is the answer. I understand there are multiple values for x^2+y^2=100. Lets start by saying that the sides are in fact 6(x), 8(y) and 10 (hypo). We know that 10 is constant becasue it is stated in (2), so we cannot adjust this number, but we can try and adjust 6 and 8 so along as x^2+y^2=100.

Try to imagine a right triangle on a coordinate system with sides 6, 8, 10. Let side 6 run from (0,0) to (0,6), side 8 run from (0,0) to (8,0), and side 10 connects them. Remember 10 has to stay constant. Let's increase the distance of the 6 side to, say, 7. So the new coordinate's of the side are (0,-1) to (0,6). The 8 side will change length, but the triangle is no longer a right triangle and the x^2+y^2=100 theorem no longer applies. I guess what I am trying to prove is that a right triangle with a hypotenuse of 10, the other 2 sides have to be 6 and 8. Am in wrong in this thinking?

From there, we are told which side of the rectagle is the longest and that it represents the diameter of the semicircle.

Again, I can be way wrong with this. Let me know your thoughts
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 [#permalink] New post 15 Oct 2005, 17:16
Let me try to put forward our (gsr, automan and mine) concern in simple terms:

Lets take a book and a pen. The angles of 2 sides of the book (rectangular one for easier imagination) is almost perpendicular. Lets keep one of end of pen touching the edge of longer side and other end touching the edge of shorter side. Imaging pen as the hypotenuse side of the right triangle.

Now move the pen back and forth with its end still touching the edges. You will notice, that the length of the pen does not change and the right angle also does not change. The angles that change are the angles that are formed between the pen and the edge of the book. But those angles does not govern the pythagorean theorem, because we have the right angle intact.

Hope I made our point clear. Please post back if I have made any wrong assumption or any of the above lacks clarity.
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 [#permalink] New post 15 Oct 2005, 17:34
Hi people.

just pick numbers...

6^2+8^=10^2
9^2+(sqrt(19))^2=10^2
(sqrt(49))^2+(sqrt(51))^2=10^2
.............. And so on.....

Actually, there are millions of rectangular triangles having 10 as hipothenuse.
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 [#permalink] New post 15 Oct 2005, 17:47
OK, I am totally wrong. :oops: . I looked much closer and inspected some books, I assume too much on the 10:8:6 triangle. Thanks for the heads up. Please be aware when listen to my answers and explainations. Sorry for the in covenience :lol:
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 [#permalink] New post 15 Oct 2005, 18:33
chriswil2005 wrote:
OK, I am totally wrong. :oops: . I looked much closer and inspected some books, I assume too much on the 10:8:6 triangle. Thanks for the heads up. Please be aware when listen to my answers and explainations. Sorry for the in covenience :lol:


Chris: There is nothing to feel sorry about. We are sharing our knowledge and most of us makes mistakes. The good part is to correct it before the test day :-).

With this discussion most of us would never forget the 3:4:5 ratio for atleast years and hope we do not fall for this on test day.
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 [#permalink] New post 15 Oct 2005, 22:08
We can not assume this is a 6:8:10 triangle. There are infinite numbers of triangles with a hypotenuse of 10. B is insufficient.

A and B together confirms that it must be a 6:8:10. Longer side 8 is the diameter of the semicircle, which we can use to solve the length of the arc QR.

Answer must be C.
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 [#permalink] New post 16 Oct 2005, 18:31
Excellent, guys...it's time for OA. We cannot assume that 6:8:10 is the only ratio in statement 2.

OA - C
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Re: The shape of a flower bed is a combination of semicircle and [#permalink] New post 18 Mar 2014, 12:08
Option C.
S1:l+b=14.
Insufficient as miltiple values are possible for length and breadth.
S2:l^2+b^2=100
Again insuff

Combining the two
We can find out l*b=48
And l+b=14
To get single value of l & b.After that we can find out the perimeter of the whole figure.

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Re: DS 3 [#permalink] New post 02 Jun 2014, 05:13
MA wrote:
Given that QR>RS,
Suppose side QR=Length (L), and side RS = Breath (B)
From I,
2(L+B)=28
L+B=14 ..................... (i)

from II,
L^2+B^2=H^2
L^2+B^2=10^2

add 2LBin both sides

L^2+2LB+B^2=100+2LB
(L+B)^2=100+2LB ........... (ii)

putting valu of L+B in eq ii
14^2=100 +2LB
LB=48
L=48/B .............................(iii)

Putting value of L in eq (i)
48/B+B=14
B^2-14B+48 = 0
B=6 and 8,
A=6 and 8
but from question L>B,
L=8 and B=6.
Hence , OA is C.


Almost correct. But last steps are unnecessary. See:

If you have x+y = 14
and xy = 48
Then obviously one will be 6 and the other 8.
No need to play hero doing quadratics

Therefore, we know that QR>RS, then QR = 8

This is enough to solve the problem

Hope it helps
Cheers

J :)
Re: DS 3   [#permalink] 02 Jun 2014, 05:13
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