GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 Sep 2018, 18:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# What is the greatest possible area of a triangular region with one

Author Message
TAGS:

### Hide Tags

Intern
Joined: 24 Feb 2016
Posts: 27
Location: India
Schools: LBS (A)
GMAT 1: 650 Q45 V34
GPA: 4
What is the greatest possible area of a triangular region with one  [#permalink]

### Show Tags

Updated on: 02 Oct 2017, 00:56
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:52) correct 20% (02:29) wrong based on 20 sessions

### HideShow timer Statistics

What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

Originally posted by arabella on 02 Oct 2017, 00:50.
Last edited by Bunuel on 02 Oct 2017, 00:56, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Manager
Joined: 30 Mar 2017
Posts: 69
Re: What is the greatest possible area of a triangular region with one  [#permalink]

### Show Tags

02 Oct 2017, 01:20
B 36 . The area will be maximum when 12 us the base and radius is the height

Sent from my Moto G (5) Plus using GMAT Club Forum mobile app
Senior SC Moderator
Joined: 22 May 2016
Posts: 1980
What is the greatest possible area of a triangular region with one  [#permalink]

### Show Tags

02 Oct 2017, 07:17
arabella wrote:
What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24
(B) 36
(C) 40
(D) 48
(E) 72

One side of this triangle, its base, IS the diameter. Hence it is a right triangle inscribed in a semicircle, no matter where the third vertex is placed on the circle.

For any right triangle, isoceles right triangle has the maximum area. Maximum area is achieved with maximum height. Maximum height, as becomes clear from inscribing the right triangle in a circle, bisects and is perpendicular to the diameter. That maximum height is the radius.

Imagine two right triangles in a semicircle. The first has a vertex on the circle that, on a face clock, would be at 12 o'clock.

The second triangle has a vertex on the circle that, on a face clock, would be at 10 o'clock.

Drop altitudes from both vertices. The first has a height of radius = 6.

The second has a height that is not the radius, and is < 6. Base for both = 12

Area = $$\frac{b*h}{2}$$

First ∆: $$\frac{(12)(6)}{2}$$ = 36

Second ∆: $$\frac{(12)(<6)}{2}$$ < 36

The first triangle has the greater area, with height = radius. No height of any inscribed right triangle can be greater than the radius. Stated differently, the altitude that is the radius yields the maximum height, which yields the maximum area.

First triangle has the maximum area every time. (And it's an isosceles right triangle.)

From above, that triangle's area, where base = diameter = 12, and h = radius = 6, is 36.

_________________

In the depths of winter, I finally learned
that within me there lay an invincible summer.

What is the greatest possible area of a triangular region with one &nbs [#permalink] 02 Oct 2017, 07:17
Display posts from previous: Sort by

# What is the greatest possible area of a triangular region with one

## Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.