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Hi all! Help me please with one task concerning inscribed triangles and circles 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?

Hi all! Help me please with one task concerning inscribed triangles and circles 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. _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

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? _________________

How to improve your RC score, pls Kudo if helpful! http://gmatclub.com/forum/how-to-improve-my-rc-accuracy-117195.html Work experience (as of June 2012) 2.5 yrs (Currently employed) - Mckinsey & Co. (US Healthcare Analyst) 2 yrs - Advertising industry (client servicing)

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 :

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.

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°."

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.

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.

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. _________________

Would you please help me with this "easy " questions.

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 .

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?

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...