It is currently 23 Feb 2024, 05:04 |

Customized

for You

Track

Your Progress

Practice

Pays

- Feb
**15**### Head over heels for INSEAD, HEC Paris, or IMD?

04:00 PM EST

-11:00 PM EDT

You’ll LOVE this deal – 20% off J-Term Application Packages! (Valid through 17 March 2024). Upon initial inquiry (phone or online), reference GMAT Club code GMATClub20, and the discount will automatically be applied. - Feb
**27**### How to Choose the Right MBA Program for You!

08:00 PM EST

-09:00 PM EST

How can you find the best fit? On February 27, join mbaMission’s experts to learn the differences that exist among top MBA programs.

In the figure above, line L is tangent to the circle, which is centere
[#permalink]
09 Dec 2022, 06:11

Expert Reply

1

Bookmarks

Question Stats:

Attachment:

GRE In the figure above, line L is tangent to the circle, which is centered at the origin..jpg [ 21.57 KiB | Viewed 1234 times ]

In the figure above, line L is tangent to the circle, which is centered at the origin. The area of the shaded region is equal to the circumference of a circle with radius between 1 and 2 1/4. Which of the following could be values of x ?

Indicate all such values.

2

−3

4

−5

−8.5

−12

_________________

Re: In the figure above, line L is tangent to the circle, which is centere
[#permalink]
11 Dec 2022, 04:00

1

Expert Reply

OE

Use line L to make a triangle, with points at (x, 0), the 45° angle, and the origin. The angle at (x, 0) must be 45° since the sum of the angles of a triangle is 180°. Since line L is tangential to the circle and forms 45° angles with each axis, a line from the point where line L and circle intersect to the origin will form a right angle with line L. The smaller triangle formed—from (x, 0), to where line L and circle meet, to the origin—will be a 45-45-90 triangle, with two sides equal to the radius of the circle. Find the radius of the circle, and you can find x. The area of the shaded region is equal to the circumference of a circle with radius between 1 and 2 1/4. Circumference = 2πr, so the area of the shaded region is between 2π and 4.5π, which means the area of the circle in the figure is between 8π and 18π. Area = πr^2, so the radius of the circle is between \(\sqrt{8}\) and \(\sqrt{18}\), which is to say between \(2 \sqrt{2}\) and \(3 \sqrt{2}\). 45-45-90 triangles have sides of \(a-a-a \sqrt{2}\), where, in this case, a is between \(2 \sqrt{2}\) and \(3 \sqrt{2}\). So the hypotenuse of the triangle, from the origin to (x, 0), is between \(2 \sqrt{2} \sqrt{2}\) and \(3 \sqrt{2} \sqrt{2}\), and therefore between 4 and 6, so x can range from −4 to −6.

_________________

Use line L to make a triangle, with points at (x, 0), the 45° angle, and the origin. The angle at (x, 0) must be 45° since the sum of the angles of a triangle is 180°. Since line L is tangential to the circle and forms 45° angles with each axis, a line from the point where line L and circle intersect to the origin will form a right angle with line L. The smaller triangle formed—from (x, 0), to where line L and circle meet, to the origin—will be a 45-45-90 triangle, with two sides equal to the radius of the circle. Find the radius of the circle, and you can find x. The area of the shaded region is equal to the circumference of a circle with radius between 1 and 2 1/4. Circumference = 2πr, so the area of the shaded region is between 2π and 4.5π, which means the area of the circle in the figure is between 8π and 18π. Area = πr^2, so the radius of the circle is between \(\sqrt{8}\) and \(\sqrt{18}\), which is to say between \(2 \sqrt{2}\) and \(3 \sqrt{2}\). 45-45-90 triangles have sides of \(a-a-a \sqrt{2}\), where, in this case, a is between \(2 \sqrt{2}\) and \(3 \sqrt{2}\). So the hypotenuse of the triangle, from the origin to (x, 0), is between \(2 \sqrt{2} \sqrt{2}\) and \(3 \sqrt{2} \sqrt{2}\), and therefore between 4 and 6, so x can range from −4 to −6.

_________________

gmatclubot

Moderators:

Multiple-choice Questions — Select One or More Answer Choices |
||

## Hi Guest,Here are updates for you: |