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GMAT Prep Geometry Question

by ssy » Mon Oct 22, 2007 6:54 am
In the circle in the image attached, PQ is parallel to diameter OR. OR has length 18. What is the length of minor arc PQ?

2pi
(9pi)/4
(7pi)/2
(9p)/2
3pi

Ans 2pi[/img]
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by ri2007 » Mon Oct 22, 2007 7:04 am
Diameter is 18 so circumference of semi circle = 18 pi /2 = 9 pi

PQ ll OR so angle R = angle P = 35

so arc OP = arc QR = 35 * 2 = 70

so arc PQ = 180 - ( arc OP + arc QR) = 40 deg

Since 180 deg corresponds to 9 pi

40 corresponds to 2 pi ( cross multiplication)
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by ssy » Mon Oct 22, 2007 7:13 am
Thank you ri2007 but I am still confused.

Isn't the circumference of the semicircle (pi)(radius)+2pi, so the circumference in this case is [(pi)(9)+18]

Also, can you please explain the following steps

"so arc OP = arc QR = 35 * 2 = 70

so arc PQ = 180 - ( arc OP + arc QR) = 40 deg"

Thanks very much.
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by ri2007 » Mon Oct 22, 2007 7:34 am
Point 1) Circumference is 2 pi r or 2 pi d where d is the diameter,

Point 2) since PQ ll OR and angle R & angle P are the internal angles formed by the transversal - internal angles are equal.

Point 3) semicircle = 180 degrees
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by Suyog » Mon Oct 22, 2007 1:55 pm
I'm also lil confused!
if angle R = 35 then how arc OP = 70 ??
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by ri2007 » Mon Oct 22, 2007 2:21 pm
Suyog wrote:I'm also lil confused!
if angle R = 35 then how arc OP = 70 ??
Its a geometry rule of inscribed angle
minor arc is 2* inscribed angle
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by ssy » Tue Oct 23, 2007 4:40 am
Sorry I am still confused.

Point 1) Do you mean the circumference of a semi circle because isn't the circumference of the semi circle (pi)(radius)+diameter?

https://sg.answers.yahoo.com/question/in ... n6D&show=7

Point 3) How can you subtract the length of the two arcs from 180 degrees?
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by ri2007 » Tue Oct 23, 2007 6:03 am
ssy wrote:Sorry I am still confused.

Point 1) Do you mean the circumference of a semi circle because isn't the circumference of the semi circle (pi)(radius)+diameter?

https://sg.answers.yahoo.com/question/in ... n6D&show=7

Point 3) How can you subtract the length of the two arcs from 180 degrees?
I have made one typing error earlier in explaining the formula for circumference of circle it is 2 pi r or pi d no 2 pi d as typed earlier by mistake. This is the right formula

THis questions is either from OG or Kaplan. They have also give an explaination. You can check it
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by manasi_sh » Tue Oct 23, 2007 7:41 am
Also there is a formula for Length of Arc:
degree measure(here 40)/360* 2 PI R(circumference formula)
so over here Arc PQ = 40/360 *(2PI*9) = 2PI(ANS)
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by 800GMAT » Fri Nov 02, 2007 2:49 pm
Some relevant properties:
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by xcise_science » Tue Nov 27, 2007 11:00 am
can someone explain this part to me:

Since 180 deg corresponds to 9 pi
40 corresponds to 2 pi ( cross multiplication)

I'm not clear on how/why 180 corresponds to 9pi.

thanks
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by gkumar » Mon Oct 19, 2009 11:42 am
I'm still unclear why the degree measures of minor arcs PAO and QAR are equal. I understand why PAO has a measure 70 degrees, but not why QAR has a measure of 70 degrees. Can you please explain via a proof or some other explanation on how angles QAR = PAO?

Here's what I know so far (see attachment)
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by Stuart@KaplanGMAT » Mon Oct 19, 2009 12:24 pm
gkumar wrote:I'm still unclear why the degree measures of minor arcs PAO and QAR are equal. I understand why PAO has a measure 70 degrees, but not why QAR has a measure of 70 degrees. Can you please explain via a proof or some other explanation on how angles QAR = PAO?

Here's what I know so far (see attachment)
Hi!

We know that angle APQ is 70 degrees (35+35). We also know that triangle APQ is isosceles, since both AP and AQ are radii of the circle.

Therefore, angle AQP is also 70 degrees, leaving 40 degrees for central angle PAQ.

So, PAO is 70 and PAQ is 40. 180 - 110 = 70 degrees left over for angle QAR.
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by gkumar » Mon Oct 19, 2009 12:35 pm
Thanks Stuart! I forgot about the isoceles triangle part. All of these angles are hard to keep track and manage under 2 minutes. Is it possible to use Symmetry to assume that QAR is automatically equal to PAO?
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by Stuart@KaplanGMAT » Mon Oct 19, 2009 12:39 pm
gkumar wrote:Thanks Stuart! I forgot about the isoceles triangle part. All of these angles are hard to keep track and manage under 2 minutes. Is it possible to use Symmetry to assume that QAR is automatically equal to PAO?
We need to careful not to assume that everything will be symmetrical, so it's not a safe bet.

However, the fact that all triangles radiating out from the centre of a circle are isosceles is commonly tested, so be on the lookout for such situations.
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