It is currently 21 Oct 2017, 11:23

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the greatest possible area of a triangular region

Author Message
Manager
Joined: 17 Dec 2005
Posts: 162

Kudos [?]: 35 [0], given: 0

What is the greatest possible area of a triangular region [#permalink]

### Show Tags

16 Jun 2006, 18:10
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the greatest possible area of a triangular region with one vertex at the center of the circle of a radius of 1 and the other two vertices on the circle?

a. sqrt(3)/4

b. 1/2

c. pi/4

d. 1

e. sqrt(2)

Kudos [?]: 35 [0], given: 0

VP
Joined: 25 Nov 2004
Posts: 1481

Kudos [?]: 126 [0], given: 0

Re: PS: Geometry - Triangles [#permalink]

### Show Tags

16 Jun 2006, 21:19
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.

Kudos [?]: 126 [0], given: 0

Manager
Joined: 17 Dec 2005
Posts: 162

Kudos [?]: 35 [0], given: 0

Re: PS: Geometry - Triangles [#permalink]

### Show Tags

16 Jun 2006, 21:38
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.

so i can assume this is a general rule? how could i prove this?

Kudos [?]: 35 [0], given: 0

VP
Joined: 25 Nov 2004
Posts: 1481

Kudos [?]: 126 [0], given: 0

Re: PS: Geometry - Triangles [#permalink]

### Show Tags

16 Jun 2006, 22:17
mrmikec wrote:
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.

so i can assume this is a general rule? how could i prove this?

i guess yes. it should be therom which we can verify with an example.

suppose a an equilatereal triangle with side s = 6.
area = 9 sqrt(3)

now lets change a side, the base, of the original triangle and make the triangle issoceles.
new base, b = 8
now area = 8 sqrt (5).

now lets decrease onle a side, the base, of original triangle and make the triangle issoceles.
new base, b = 4
now area = 8 sqrt(2)

so from the example, the largest area for a given range of sides of a triangle is of equilateral triagle.

Kudos [?]: 126 [0], given: 0

Manager
Joined: 17 Dec 2005
Posts: 162

Kudos [?]: 35 [0], given: 0

Re: PS: Geometry - Triangles [#permalink]

### Show Tags

18 Jun 2006, 13:32
MA wrote:
mrmikec wrote:
MA wrote:
should be A.

a triangle with equal sides has largest area. so the artea of equalitral triangle is sqrt(3)/4.

so i can assume this is a general rule? how could i prove this?

i guess yes. it should be therom which we can verify with an example.

suppose a an equilatereal triangle with side s = 6.
area = 9 sqrt(3)

now lets change a side, the base, of the original triangle and make the triangle issoceles.
new base, b = 8
now area = 8 sqrt (5).

now lets decrease onle a side, the base, of original triangle and make the triangle issoceles.
new base, b = 4
now area = 8 sqrt(2)

so from the example, the largest area for a given range of sides of a triangle is of equilateral triagle.

The OA is B. Can you please explain??? Thanks.
Attachments

Q3.jpg [ 70.24 KiB | Viewed 1326 times ]

Kudos [?]: 35 [0], given: 0

Intern
Joined: 21 Apr 2006
Posts: 18

Kudos [?]: [0], given: 0

Location: San Francisco, CA

### Show Tags

18 Jun 2006, 16:24
If the triangle is 90-45-45 then the sides would be sqrt2 - 1 - 1. The height for this would be 1/2sqrt(2).

A=1/2hb ==> 1/2*1/2sqrt(2)*sqrt(2) ==> 1/2.

This value is is greater than 1/4sqrt(3). Thus, a 90-45-45 triangle is the greatest in value with respect to the area of it.

Can anyone else confirm this?

-Tru

Kudos [?]: [0], given: 0

Manager
Joined: 29 Sep 2005
Posts: 125

Kudos [?]: 2 [0], given: 0

### Show Tags

19 Jun 2006, 05:09
Area = 1/2*Base*height

So in order to maximize the area we should attempt to choose the highest possible values for base and height.
In a right triangle, Base=1 and Height=1.

Height will be maximum when it is equal to radius. Imagine a rt. traingle with vertex at center o.

You could go with a higher base, as close to the length of the diameter as possible but only lesser however (imagine a line almost perpendicular cutting the circle into half, but not quite perpendicular). In this case the height will be small.

Best case if 1 and 1 so ans is 1/2

Kudos [?]: 2 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 377 [0], given: 0

### Show Tags

19 Jun 2006, 06:47
If we know that the two sides (a and b) are always going to be 1, then we could perhaps use this area formula: S=1/2absinO, where O is the angle formed by the two sides. Then we can see that sin90=1, which gives us the greatest value. In that case S=1/2.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Kudos [?]: 377 [0], given: 0

Current Student
Joined: 29 Jan 2005
Posts: 5206

Kudos [?]: 434 [0], given: 0

### Show Tags

19 Jun 2006, 07:19
It`s even simpler than that. With the center of the circle serving as the vertice, then the largest inscribed triangle must be 90-45-45. The two shorter sides, also the length and width, are given as 1.

A=1/2b*h

A=1/2(1)*(1)

A=1/2

(B)

Kudos [?]: 434 [0], given: 0

Director
Joined: 07 Jun 2006
Posts: 513

Kudos [?]: 145 [0], given: 0

### Show Tags

19 Jun 2006, 09:09
Here is another approach,

Imagine the largest rectangle in a circle - it would be a square. So if we fit in a square to the circle of radius 1, then it's sides would be root(2), and therefore area is 2.

Area of square = 2 (maximizes area), and the triangle in question is exactly (1/4) the area of the square, so the area of triangle is

2/4 = 1/2

Kudos [?]: 145 [0], given: 0

19 Jun 2006, 09:09
Display posts from previous: Sort by