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The surface of a mirror is composed of a rectangular piece that is 9

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ajay2121988
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The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   31 Jul 2016, 01:08
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The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5



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I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   26 Aug 2016, 16:05
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Here is my approach
Let the width be x
here the two semicircles make the entire uniform circle of radius pie*x^2/4
area of rect=> 9*x
Given => 9x/pie*x^2/4 = 9/pie => x=4
SMASH THAT D
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   20 Apr 2019, 13:18
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Aviv29 Don't worry I got confused to and it was a silly mistake on my end. Once I explain, you will have an Aha moment

We've been given the length of rectangle = 9
Width of rectangle, lets make it = x

As it mentions that the width of the rectangle is the same as the diameter of the semicircle, to get the radius, you have to divide x by 2.

Therefore, radius of the semicircle = x/2

As area of a circle = pi*r^2, so now lets square the radius of the semicircle and multiply by pi

Area of the two semicircle = pi * ((x/2)^2) = pi * (x^2)/4

Now you can do the ratio.

Given => 9x /pi*(x^2/4) = 9/pi => x=4


Hope this makes sense.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   31 Jul 2016, 01:34
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ajay2121988 wrote:
The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5

I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


Area of 1 semicircular piece will be equal to 1/2 * pi.(D^2)/4 and not pi.(D^2)/4.

When you will equate 9b/(1/2 * pi.(D^2)/4) = 9/pi, it will give b = 4.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   26 Aug 2016, 20:57
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stonecold wrote:
Donnie84 wrote:
Another approach:

First, observe that the radius of the semi-circle and the combined circle will be half of width of the rectangle. Next, as per the data given in the question and answer choices, the width should be a multiple of 2. Options A, C and E are ruled out.=> This is same as doing the long calculation. And width being a multiple of 2 is known only after the initial step.
Let's check with option B.
If width = 2, area of rectangle = 9*2 = 18
Radius of the circle = 1
Area of the circle = pi

Ratio = 18/pi -> ruled out since we want the ration to be 9/pi.

Answer (D).

Validation:

If width = 4, area of rectangle = 9*4 = 36
Radius of the circle = 2
Area of the circle = 4*pi

Ratio = 9/pi -> good!


I meant to say that the width must be even. Width cannot be odd in this case because that would mean a fraction value for the radius and its square would be messy. The values in the question are elegant, so we can eliminate the options with width = 1 or 3 or 5.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   31 Jul 2016, 01:40
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abhimahna wrote:
ajay2121988 wrote:
The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5

I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


Area of 1 semicircular piece will be equal to 1/2 * pi.(D^2)/4 and not pi.(D^2)/4.

When you will equate 9b/(1/2 * pi.(D^2)/4) = 9/pi, it will give b = 4.


Oh, when will I stop doing these silly mistakes !

Anyways, thanks, K +1.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   31 Jul 2016, 01:44
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ajay2121988 wrote:
abhimahna wrote:
ajay2121988 wrote:
The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5

I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


Area of 1 semicircular piece will be equal to 1/2 * pi.(D^2)/4 and not pi.(D^2)/4.

When you will equate 9b/(1/2 * pi.(D^2)/4) = 9/pi, it will give b = 4.


Oh, when will I stop doing these silly mistakes !

Anyways, thanks, K +1.


Thanks.

Simple strategy to avoid silly mistakes is to get completely immersed in the question you are doing.

Every step taken with caution will lead you closer to your dream score. :)
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   03 Aug 2016, 10:43
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paidlukkha wrote:
L*B = area of rectangle
pir^2/2 = area of semi circle

L*B + 2(pir^2/2) = 9/pi
9*x + 2(pix^2/2) = 9/pi

how do I proceed please?


Its is the ratio between the rectangle and semi circles given and NOT the sum.

Your equation will be

9*x / 2(pix^2/2) = 9/pi

Now proceed and let me know if you need more help!!
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   26 Aug 2016, 20:32
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Another approach:

First, observe that the radius of the semi-circle and the combined circle will be half of width of the rectangle. Next, as per the data given in the question and answer choices, the width should be a multiple of 2. Options A, C and E are ruled out.

Let's check with option B.
If width = 2, area of rectangle = 9*2 = 18
Radius of the circle = 1
Area of the circle = pi

Ratio = 18/pi -> ruled out since we want the ration to be 9/pi.

Answer (D).

Validation:

If width = 4, area of rectangle = 9*4 = 36
Radius of the circle = 2
Area of the circle = 4*pi

Ratio = 9/pi -> good!
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   26 Aug 2016, 21:16
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Yes, that is my opinion and I'm sticking to it. Intuition told me that I should start with options B and D. We are here to learn from each other and improve. I take your feedback constructively.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   03 Aug 2016, 10:08
L*B = area of rectangle
pir^2/2 = area of semi circle

L*B + 2(pir^2/2) = 9/pi
9*x + 2(pix^2/2) = 9/pi

how do I proceed please?
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   18 Apr 2019, 03:40
stonecold wrote:
Here is my approach
Let the width be x
here the two semicircles make the entire uniform circle of radius pie*x^2/4
area of rect=> 9*x
Given => 9x/pie*x^2/4 = 9/pie => x=4
SMASH THAT D



Could someone please explain why we have to divide by 4? "9x/pie*x^2/4"

Thanks!
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   13 Dec 2020, 21:39
ajay2121988 wrote:

The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5



Show Spoiler
I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


Show Spoiler
Attachment:
image.PNG


1) We know from the question stem the ratio of the areas of the rectangle to the semicircles is \(\frac{9}{pi}\).

2) Find the area formulas of each geometric shape. Let x the width of the rectangle (and diameter of the two semicircular pieces).

A of the two semicircular pieces is \(A = pi (r^2) = pi (\frac{x}{2})^2 = pi \frac{x^2}{4}\)
A of the rectangle is 9*x

3) Make a ratio of the two areas we found against the initial ratio given.

\(\frac{9x}{(pi * x^2/4)} = \frac{9}{pi}\)
\(36 pi (x) = 9 pi (x^2)\)
\(x^2 - 4x = 0\)
x = 0 or 4
We know x can't equal 0 since that would mean the areas would be = 0, so x = 4.

Answer D
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   19 Dec 2020, 15:42
We're given the ratio of the area of the rectangular piece to the total area of two semicircular pieces is \(\frac{9}{π}\).

Let \(x\) = width of rectangle
Let \(\frac{x}{2}\) = radius of semi-circle

Area of two semi-circles combined = \(π(\frac{x}{2})^2\) = \(\frac{πx^2}{4}\)

\(9x/\frac{πx^2}{4}\) = \(\frac{9}{π}\)

Solving for x, we get x = 4.

Answer is D.
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   28 Jun 2021, 04:10
Do I have an error of thought?
If I work backwards at this problem, I get for D=4 the following:
I thought the diameter of the two semicircles is equal to the radii of the whole circle, which is formed by the 2 semi-circles.
The area would be the following: \((4*π)^2\), which is 16π.
The area of the rectangular is 4*9=36. So then the ratio is not right.

But if I split up the two semicircles and add the areas together, I got 4π, which is in line with the original answer choice.

Why I can't use the first approach?
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink] New post   14 Nov 2023, 22:40
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ajay2121988 wrote:

The surface of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the rectangular piece, as shown in figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi , what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5



Show Spoiler
I solved it this way.:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.


Show Spoiler
Attachment:
image.PNG



Let radius of the semi circle be r.
Area of rectangle = 9 * 2r
Area of the two semi circles together = \(\pi *r^2\)

Their ratio is \(\frac{9*2r}{\pi*r^2} = \frac{9}{\pi}\)

So r = 2 and width = 2r = 4

Answer (D)
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Re: The surface of a mirror is composed of a rectangular piece that is 9 [#permalink]
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