
It is currently Wed Jun 26, 2019 11:06 am

View unanswered posts  View active topics

Page 1 of 1

[ 10 posts ] 

Author 
Message 
questioner

Post subject: GMAT Geometry Posted: Wed Apr 17, 2013 9:19 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

In the figure above, a circle with center O is inscribed in the square WXYZ. The segment XZ has a length of 3√2. What is the radius of the circle?
A. 1 B. 1.5 C. 2 D. 2.5 E. 3
(B) Triangle XYZ is an isosceles right triangle. Its hypotenuse is 3√2, so the length of each leg is 3. Since the diameter of the circle is equal to the height of the square, the diameter is equal to 3. So the radius is equal to 1.5 inches.
Alternatively, since the sides XY and ZY are equal, set them equal to x. Then 2x² = (3√2)² by the Pythagorean Theorem. Solving for x: x² = 9 and so x = 3. Since x represents the side of the square or the diameter of the circle, we divide by 2 to get the radius. So, the radius must be 1.5. The correct answer is (B).
 I don't understand what 3√2 means, how has this been simplified? How do you extract that each side is 3 from this length?


Top 


Gennadiy

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:20 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

Because we are working with a square, the triangle XZY is a 454590 isosceles right triangle. And, we know from our study of classical right triangles, the hypotenuse of such a triangle is equal to √2 times the length of one of the legs. The length of the leg happens to be the length of the side of the square. So the length of each of the sides of the square is equal to 3.
Now notice that the diameter of the circle is equal to the length of one of the sides of the square. Convince yourself by drawing a vertical like through the circle that passes its center. So, the diameter is equal to 3. The radius is equal to half the diameter, so the radius is equal to 1.5.


Top 


questioner

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:21 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

I am confused how right away in the answer we get the sides of the square to be three.


Top 


Gennadiy

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:21 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

We know that triangle ZXY is isosceles and right. It is a property of an isosceles right triangle that its legs are equal and the hypotenuse is √2 times greater. The alternative method in the explanation proves it.
Therefore, since we know that hypotenuse is (√2 × 3), we clearly see that the leg is 3.


Top 


questioner

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:21 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

I fail to understand how the leap from 3√2 to 3 is achieved. To my mind the symbol 3√2 means the square root of 2 (1.41) multiplied by 3 to give 4.24.


Top 


Gennadiy

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:21 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498


Top 


questioner

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:22 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

The explanation states that each leg is equal to 3 inches but the question states that leg XZ is 3√2. Since it's a square, all legs are 3√2. How can each leg of the triangle (or side of the square) be 3 inches if the problem states that it's 3√2?


Top 


Gennadiy

Post subject: Re: GMAT Geometry Posted: Wed Apr 17, 2013 9:22 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498


Top 


questioner

Post subject: Re: GMAT Geometry Posted: Thu Aug 15, 2013 5:29 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

I didn't understand the last part.


Top 


Gennadiy

Post subject: Re: GMAT Geometry Posted: Thu Aug 15, 2013 5:42 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498


Top 



Page 1 of 1

[ 10 posts ] 

Who is online 
Users browsing this forum: No registered users and 1 guest 

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum

