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Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]
17 Aug 2010, 02:15

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Difficulty:

75% (hard)

Question Stats:

61% (03:27) correct
39% (02:55) wrong based on 112 sessions

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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.)

I don't get your equation. 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!

I don't get your equation. 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 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]
28 Mar 2014, 08:25

Expert's post

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.

Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]
28 Mar 2014, 09:00

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.

Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]
09 Apr 2014, 23:19

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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

Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl [#permalink]
22 Apr 2014, 16:16

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

Answer is thus B

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

Cheers! J

gmatclubot

Re: Rectangle PQST, with dimensions w*h, is inscribed in a circl
[#permalink]
22 Apr 2014, 16:16

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...