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# Math: Circles

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Manager
Joined: 16 Jan 2011
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26 Jun 2011, 11:55
Hi all!
here it is:
Circle A,centre X. XB is the radius. There is a chord AC which intersects XB. D is the point of intersection between XB and AC. BD=2;AC=12;XDA= 90 degrees. What is the circles area?

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Manager
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01 Jul 2011, 13:07
Galiya wrote:
Hi all!
here it is:
Circle A,centre X. XB is the radius. There is a chord AC which intersects XB. D is the point of intersection between XB and AC. BD=2;AC=12;XDA= 90 degrees. What is the circles area?

By basic property of the circle,
the radius bisects the chord AC . ie CD equals AC/2 ie 6

now see, radius XB=XD+BD ie r=XD+2 ie XD=r-2

now concerning Triangle XCD, angle XDC= XDA =90 so pythagorean theorem is applicable
so, sq(XC)=sq(DC)+sq(XD)

plug in values you get sq(r)=sq(r-2)+sq(6)
which gives r=10

now area=pi*r*r=100*pi

Hope it helps.
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Manager
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23 Jul 2011, 21:40
Thanks Bunuel.. love your explanations... very simple, clear and easy to comprehend.. Kudos
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Regards,
Asher

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13 Nov 2011, 23:30
This is probably my worst area in all of gmat. This helps! thanks!

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20 Nov 2011, 06:43
Wow.... I've just started and U've made me fall in love with Circles all over... Thank you...

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Senior Manager
Joined: 15 May 2011
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16 Aug 2012, 11:42
Bunuel wrote:
chauhan2011 wrote:
• If you know the length of the minor arc and radius, the inscribed angle is: 90L/nr

Please correct me if i am wrong but i think the formula should be : 180L/nr

If you know the length $$L$$ of the minor arc and radius, the inscribed angle is: $$Inscribed \ Angle=\frac{90L}{\pi{r}}$$.

The way to derive the above formula:

LengAngle=\frac{180L}{\pi{r}}[/m] --> $$Inscribed \ Angle=\frac{90L}{\pi{r}}$$.

Hope it helps.

Hello, quick conceptual question

The circle represented by the equation x^ 2 + y^ 2 =1 is centered at the origin and has the radius of r= √1 = 1

What is the correlation between the function and the radius for a circle?
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28 Oct 2012, 11:52
2
KUDOS
A secant\chord (except diameter) to a circle divides circle into two region - minor and major. The area of minor region can be calculate by determining area of minor sector minus triangle. Also the direct formula to calculate minor region area is :

$$A=\frac{1}{2}r^2*(\frac{pi*center angle}{180}-sin(\frac{pi*center angle}{180}))$$
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Thanks and Regards!

P.S. +Kudos Please! in case you like my post.

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Math Expert
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11 Jul 2013, 00:06
Bumping for review*.

*New project from GMAT Club!!! Check HERE

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20 Jul 2013, 23:16
Tangent-Secant

Image

Should one of the lines be tangent to the circle, point A will coincide with point D, and the theorem still applies:

PA*PD=PC*PB=Constant

PA^2=PC*PB=Constant - This becomes the theorem we know as the theorem of secant-tangent theorem.

Does this hold true if CB was the diameter of the circle?

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21 Jul 2013, 01:55
naiduashwini wrote:
Tangent-Secant

Image

Should one of the lines be tangent to the circle, point A will coincide with point D, and the theorem still applies:

PA*PD=PC*PB=Constant

PA^2=PC*PB=Constant - This becomes the theorem we know as the theorem of secant-tangent theorem.

Does this hold true if CB was the diameter of the circle?

____________
Yes, it does.
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01 Dec 2013, 05:04
Can somebody explain the properties of a cyclic quadrilateral. Also, do the same properties hold good for a cyclic quadrilateral inscribed in a semicircle with one of its sides being the diameter of that semicircle.

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11 Dec 2013, 21:00
Thanks so much!!

Quick question - under the "Semicircles" section, could you clarify or show a picture of what this means? I don't understand how it would always be true.

"• The angle inscribed in a semicircle is always 90°."

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12 Dec 2013, 02:52
1
KUDOS
Expert's post
catalysis wrote:
Thanks so much!!

Quick question - under the "Semicircles" section, could you clarify or show a picture of what this means? I don't understand how it would always be true.

"• The angle inscribed in a semicircle is always 90°."

Check the diagram below:

Angle ABC is inscribed in semicircle, thus angle B is 90 degrees.

It's the same as the following property: a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

Hope it's clear.
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12 May 2014, 12:05
Bunuel, first of all thanks for another amazing post.

Wanted to check, there are a few concepts in this thread like secant, chord, point theorem etc. Are they tested in GMAT? And i have same question with your other quant concept threads.

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13 May 2014, 00:57
gaurav1418z wrote:
Bunuel, first of all thanks for another amazing post.

Wanted to check, there are a few concepts in this thread like secant, chord, point theorem etc. Are they tested in GMAT? And i have same question with your other quant concept threads.

Some aspects of this properties definitely could be helpful when solving GMAT questions.
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05 Mar 2015, 14:31
Hi Brnuel ,

Triangle ABC is inscribed in a circle, such that AC is a diameter
of the circle (see figure). If AB has a length of 8 and BChas a
length of 15, what is the circumference of the circle?

Aren't we supposed to - after getting AC= 17 ( which is the diameter) to calculate the circumference as 2 π R .

Is not the r supposed to be 17/2 ?. Thanks.

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17 May 2015, 02:38
Excellent work. All the important things in just one post.
Power of point theorem is vague to me. How do we know that the product of two point of intersections is constant? Could someone explain please?

Thanks
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04 Aug 2015, 09:58
Bunnel - on posts for triangle and circle i found the following- which one is correct ?

For a given perimeter equilateral triangle has the largest area.
A circle is the shape with the largest area for a given length of perimeter

Thanks

Bunuel wrote:
gaurav1418z wrote:
Bunuel, first of all thanks for another amazing post.

Wanted to check, there are a few concepts in this thread like secant, chord, point theorem etc. Are they tested in GMAT? And i have same question with your other quant concept threads.

Some aspects of this properties definitely could be helpful when solving GMAT questions.

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16 Aug 2015, 10:03
Bunnel - on posts for triangle and circle i found the following- which one is correct ?

For a given perimeter equilateral triangle has the largest area.
A circle is the shape with the largest area for a given length of perimeter

Thanks

Bunuel wrote:
gaurav1418z wrote:
Bunuel, first of all thanks for another amazing post.

Wanted to check, there are a few concepts in this thread like secant, chord, point theorem etc. Are they tested in GMAT? And i have same question with your other quant concept threads.

Some aspects of this properties definitely could be helpful when solving GMAT questions.

Both.

The first property is for triangles only. Meaning that for a given perimeter of a triangle, equilateral triangle has the largest area.
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24 Jan 2016, 00:35
the Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

Image

• An inscribed angle is exactly half the corresponding central angle. Hence, all inscribed angles that subtend the same arc are equal. Angles inscribed on the arc are supplementary. In particular, every inscribed angle that subtends a diameter is a right angle (since the central angle is 180 degrees).

in the image there's an angle alpha which is outside the circle which could be found out easily if we are given its supplementary angle,
hey bunuel what i cant understand is the relation between the central angle and the inscribed angle

i cant get the image pasted here :/

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Math: Circles   [#permalink] 24 Jan 2016, 00:35

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