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Rectangle PQST, with dimensions w*h, is inscribed in a circl

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Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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17 Aug 2010, 03:15
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Rectangle PQST, with dimensions w*h, is inscribed in a circle with a radius of 1. Triangle QRS is isosceles with QR = RS and is inscribed in the circle. If triangle QRS and rectangle PQST have the same area, then what is the length of h? (Note: Figure not drawn to scale.)

A. 1/4
B. 2/5
C. 1/3
D. 1/2
E. 2/3
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Dec 2013, 01:24, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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17 Aug 2010, 04:22
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The height of traingle QRS is (1 - (h/2)).

**Since the rectangle PQST shares the centre with Circle, h/2 falls in upper part and remaining h/2 falls in lower part.**

wh = (1/2) (w) (1 - (h/2)) --> h = 2/5

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17 Aug 2010, 04:38
heyholetsgo wrote:
wh=w/2(1-h)--> h=1/3

Can you explain how did you come with the equation ?

wh=w/2(1-h)--> h=1/3

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17 Aug 2010, 04:47
My equations says that the area of the rectangle wh must equal the area of the triangle.
The height of the triangle is the radius minus the height of the rectangle-->(1-h)
Hence the area of the triangle is w*(1-h)
I don't see why we have to divide h by 2!

Thanks for help
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17 Aug 2010, 04:51
heyholetsgo wrote:
My equations says that the area of the rectangle wh must equal the area of the triangle.
The height of the triangle is the radius minus the height of the rectangle-->(1-h)
Hence the area of the triangle is w*(1-h)
I don't see why we have to divide h by 2!

Thanks for help

Chuck it, I misread it and over looked the same area part and hence was confused that what's the relation between the rectangle and the triangle.

Thanks anyway..!

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17 Aug 2010, 05:00
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Because, The rectangle & Circle share the same centre.
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17 Aug 2010, 06:33
Damn, no I get. I always looked at it and thought the rectangle stands on the center!
THX!
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23 Dec 2013, 19:37
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02 Jan 2014, 06:38
nravi4 wrote:
The height of traingle QRS is (1 - (h/2)).

**Since the rectangle PQST shares the centre with Circle, h/2 falls in upper part and remaining h/2 falls in lower part.**

wh = (1/2) (w) (1 - (h/2)) --> h = 2/5

Cheers!
Ravi

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Consider giving me some Kudos if you like it!!!

So why did they give us that the areas where the same?

Would love to hear some reactions

Cheers!
J
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02 Jan 2014, 06:47
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Expert's post
jlgdr wrote:
nravi4 wrote:
The height of traingle QRS is (1 - (h/2)).

**Since the rectangle PQST shares the centre with Circle, h/2 falls in upper part and remaining h/2 falls in lower part.**

wh = (1/2) (w) (1 - (h/2)) --> h = 2/5

Cheers!
Ravi

---------------------------
Consider giving me some Kudos if you like it!!!

So why did they give us that the areas where the same?

Would love to hear some reactions

Cheers!
J

The equation in red above equates the areas of the rectangle and the triangle.
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28 Mar 2014, 06:51
Bunuel wrote:
jlgdr wrote:
nravi4 wrote:
The height of traingle QRS is (1 - (h/2)).

**Since the rectangle PQST shares the centre with Circle, h/2 falls in upper part and remaining h/2 falls in lower part.**

wh = (1/2) (w) (1 - (h/2)) --> h = 2/5

Cheers!
Ravi

---------------------------
Consider giving me some Kudos if you like it!!!

So why did they give us that the areas where the same?

Would love to hear some reactions

Cheers!
J

The equation in red above equates the areas of the rectangle and the triangle.

How do we get height of triangle=(1-h/2)?
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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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28 Mar 2014, 06:55
In other words how do we know that the centre of the circle and the rectangle are the same?What am I missing here?
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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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28 Mar 2014, 09:25
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AKG1593 wrote:
In other words how do we know that the centre of the circle and the rectangle are the same?What am I missing here?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. A rectangle is the sum of two right triangles, thus the diagonals of a rectangle must lie on the diameter of the circle. Therefore the intersection of the diagonals must be the center of the circle.
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Untitled.png [ 4.13 KiB | Viewed 1791 times ]

Hope it's clear.
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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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28 Mar 2014, 10:00
Bunuel wrote:
AKG1593 wrote:
In other words how do we know that the centre of the circle and the rectangle are the same?What am I missing here?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. A rectangle is the sum of two right triangles, thus the diagonals of a rectangle must lie on the diameter of the circle. Therefore the intersection of the diagonals must be the center of the circle.
Attachment:
Untitled.png

Hope it's clear.

Got it!
A very nice observation!Thanks,Bunuel!
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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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09 Apr 2014, 21:38
I am still confused at how we know the height of the triangle is 1-(h/2). Could someone please explain?
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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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10 Apr 2014, 00:19
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Catalysis..letme try

As the triangle is a isosceles triangle, a perpendicular drawn from point R to QS will pass through the centre if extended further.
So, the the length of the line connecting R to centre is 1 (radius)
To get the height of triangle, we need to remove the height of the area covered by rectangle.
Given it is a rectangle, QS = PT (properties of rectangle)
Now QS can only be equal to PT if they are equidistant from the centre of circle (only equidistant chords can be equal)
So, in other words center lies in the middle of the rectangle and hence height from center to QS is h/2
Hence if we remove this from 1 (radius), we get the height of the triangle = 1 - h/2

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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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22 Apr 2014, 17:16
I think I got it. So given that QRS is an isosceles triangle we have that the diameter = 2, is equal to 2a + h

We also know that wh= aw/ 2

Therefore replacing we have that h=2/5

Or also:

Let's call the height of the isosceles triangle 'A'. So, we have wh = aw/2. 2h=a/. Now, we also know that 2h+h+2h=2 which is the diameter of the circle. Therefore, h=2/5. B is the correct answer

Hope this clarifies
Gimme some freaking Kudos if it helps

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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]

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08 Apr 2017, 01:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl   [#permalink] 08 Apr 2017, 01:06
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