Largest Possible Area of Triangle

Expert replies

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Fri May 08, 2009 6:32 am

The sides a,b,c of a triangle satisfy 0<=a<=1<=b<=2<=c<=3. What is the largest possible area of such a triangle?

A. 1

B. 1/2

C. 2/3

D. 2

E. None of the Above

Like this post

Source: Beat The GMAT — Problem Solving | Fri May 08, 2009 6:32 am

vitaly: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Tue May 05, 2009 11:03 am

by vitaly » Fri May 08, 2009 9:03 pm

By the formula of a square, S=w*h/2. Where w is width and h is height.

Let's take c (max of the sides) as a width. Then S=c*h/2.
Which means that for any certain c, S is maximum when h is maximum. And h is maximum when a=b.

If a=b=1 then h = sqrt(1 - (c/2)^2)
=> S = c*sqrt(1 - (c/2)^2)/2 = c*sqrt(4-c^2)/4

Now we need to define c which will give max for the S. Let's find derivative:

S'=sqrt(4-c^2)-c^2/sqrt(4-c^2)=(4-2c^2)/sqrt(4-c^2) => S'=0 when c=sqrt(2)

That's not possible since c>=2.

=> There're no triangles in that set which would have maximum square from all the triangles of that set, answer E.

Like this post

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: Largest Possible Area of Triangle

by Vemuri » Fri May 08, 2009 11:06 pm

IMO E.

When we are given the sides of a triangle & asked to finds its area, we can use a formula to do so. The formula is:

Area = sqrt(s(s-a)(s-b)(s-c))

where s = (a+b+c)/2 & a,b,c are the sides of the triangle.

The question stem tells us that:
a=0,0.1,0.2,......1
b=1,1.1,1.2,......2
c=2,2.1,2.2,......3

If you consider the highest values of a,b,c (i.e. 1,2,3 respectively), then the area of the triangle will be 0, because s=3 & (s-c) = 0

As Vitaly mentioned, if the area of the triangle has to be maximum, the base & height of the triangle should be maximum. So, if we consider c=3 & the other 2 sides equal (so that we get the maximum height), i.e. a=1 & b=1, then the area as per the formula will again be equal to 0 (because s=2 & b=2, so (s-b)=0)

Like this post

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: Largest Possible Area of Triangle

by dtweah » Sat May 09, 2009 1:30 am

dtweah wrote:The sides a,b,c of a triangle satisfy 0<=a<=1<=b<=2<=c<=3. What is the largest possible area of such a triangle?

A. 1

B. 1/2

C. 2/3

D. 2

E. None of the Above

OA is A

Draw 1 and 2 as sides of right triangle whose hypotenuse is 5^.5 which is btw 2 and 3
Largest Area is 1.

Like this post

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: Largest Possible Area of Triangle

by Vemuri » Sat May 09, 2009 1:41 am

dtweah wrote:
OA is A

Draw 1 and 2 as sides of right triangle whose hypotenuse is 5^.5 which is btw 2 and 3
Largest Area is 1.

Oh my God....how in the crazy world are we supposed to determine the third side to be sqrt(5)?

You are right, when the sides are 1,2 & sqrt(5) the area is 1. But, how do we know for sure that this the largest area?

Like this post

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: Largest Possible Area of Triangle

by dtweah » Sat May 09, 2009 2:38 am

Vemuri wrote:
dtweah wrote:
OA is A

Draw 1 and 2 as sides of right triangle whose hypotenuse is 5^.5 which is btw 2 and 3
Largest Area is 1.
Oh my God....how in the crazy world are we supposed to determine the third side to be sqrt(5)?

You are right, when the sides are 1,2 & sqrt(5) the area is 1. But, how do we know for sure that this the largest area?

Because it is the only triangle that can be formed within the region.
Suppose Not. Let's make 1 2 and 3 sides of a triangle as you have been trying to do. 3, since it is the longest side must be less than the sum of the other two sides. But 3 is not less than 3. So no such triangle can be drawn. Since these are the max integer values within each region, we know we can't draw any other triangle there. Hence only one triangle is possible within region.

Like this post

Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member; Thanked: 32 times

Re: Largest Possible Area of Triangle

by Vemuri » Sat May 09, 2009 3:20 am

[quote="dtweah]
Because it is the only triangle that can be formed within the region.
Suppose Not. Let's make 1 2 and 3 sides of a triangle as you have been trying to do. 3, since it is the longest side must be less than the sum of the other two sides. But 3 is not less than 3. So no such triangle can be drawn. Since these are the max integer values within each region, we know we can't draw any other triangle there. Hence only one triangle is possible within region.[/quote]

Sure, but what I am trying to understand is ... how did you determine the 3rd side to be sqrt(5)? I am sure you did not randomly pick some numbers

Like this post

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: Largest Possible Area of Triangle

by dtweah » Sat May 09, 2009 3:28 am

Vemuri wrote:[quote="dtweah]
Because it is the only triangle that can be formed within the region.
Suppose Not. Let's make 1 2 and 3 sides of a triangle as you have been trying to do. 3, since it is the longest side must be less than the sum of the other two sides. But 3 is not less than 3. So no such triangle can be drawn. Since these are the max integer values within each region, we know we can't draw any other triangle there. Hence only one triangle is possible within region.

Sure, but what I am trying to understand is ... how did you determine the 3rd side to be sqrt(5)? I am sure you did not randomly pick some numbers

[/quote]

1^2+ 2^2= c^2

So c = 5^.5

Make 1 and 2 the legs of a right triangle.

Like this post

vitaly: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Tue May 05, 2009 11:03 am

by vitaly » Sat May 09, 2009 5:57 am

>Because it is the only triangle that can be formed within the region.

What do u mean by within this region? There can be infinite number of triangles whose sides match conditions from the original problem.

Triangle 1, 2, sqrt(5) has to be proved to be the largest somehow... It's not certain, that triangle with a 90 degrees will have max area.

Like this post

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Sat May 09, 2009 6:06 am

vitaly wrote:>Because it is the only triangle that can be formed within the region.

What do u mean by within this region? There can be infinite number of triangles whose sides match conditions from the original problem.

Triangle 1, 2, sqrt(5) has to be proved to be the largest somehow... It's not certain, that triangle with a 90 degrees will have max area.

But I proved above that it is the only triangle that can be formed within the restricted range. You are not solving the problem over the set of positive real numbers. Only for restricted values of a b and c.

NO OTHER TRIANGLE CAN BE FORMED WITHIN THAT REGION AND BY THIS ANY TRIANGLE FORMED IS THE LARGEST IN THE INTERVAL UNDER CONSIDERATION. WHAT OTHER PROOF DO YOU WANT?

To disprove the proof given, provide another triangle whose area is larger than 1 for the given interval.

Like this post

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Sat May 09, 2009 6:44 am

Alternative Proof that 1 is largest area.

Use Heron's formula, since triange is scalene, to to derive area

A=(s(s-a)(s-b)(s-c))^.5

s=a+b+b/2

Choose a=1 b= 2 and c =2.999, since c cannot be 3.

s=2.9995

A= (2.9995 (2.9995-1)(2.9995-2)2.9995-2.999))^.5

A=(.002997704)^.5

A=.0547471529

Or Use
A= ab x (Sin C)/2
where c is the longest side of a triangle, and a and b take on other values of the triangle. IF the angle opposite C is 90, then the formula reduces to ab/2 since sin 90 is 1. But this means we have a right triangle whose sides are a and b and hypotenuse is 5^.5. For C less than 90, c will be smaller and the demonstration above should tell you that the area is less than 1. For C greater than 90, you are out the range. Hope this helps.

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Sat May 09, 2009 8:44 am

Seriously people - if you're resorting to derivatives or wacky formulas, you've missed the boat. This is the GMAT, not an MIT applied math entrance exam!

There's only one basic rule you need to know to solve this question:

Given two set sides, a right triangle will have the biggest area (since a 90 degree angle will maximize the height).

Our limitation is the two smaller sides, which max out at 1 and 2, respectively.

Double checking to make sure we don't violate the length of the 3rd side:

1^2 + 2^2 = c^2

5 = c^2

root5 = c

since 2 <= root 5 <= 3, we have a valid triangle.

Now we have a right triangle with base 2 and height 1 (or vice-versa, if you prefer).

Area = 1/2(2)(1) = 1... choose (A).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Like this post

rsadana1: GMAT Instructor; Posts: 119; Joined: Sat Jan 24, 2009 3:52 pmFollowed by: 9 members; Thanked: 16 times

by rsadana1 » Wed May 20, 2009 5:08 pm

I second the explanation given by Stuart.

I am wondering though what would be the answer to this question if the condition for third side was

2<=c<sqrt(5)

In other words could there be a question such that by placing two shorter sides at right angle we maximize the area but get third side that does not satisfy the condition specified?

Like this post

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu May 21, 2009 3:21 pm

rsadana1 wrote:I second the explanation given by Stuart.

I am wondering though what would be the answer to this question if the condition for third side was

2<=c<sqrt(5)

In other words could there be a question such that by placing two shorter sides at right angle we maximize the area but get third side that does not satisfy the condition specified?

It's very unlikely that you'd ever see that on the GMAT. We did the calculation to be 100% safe, but I can't imagine that we'd end up violating the rule.

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review