Problem Solving

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

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.

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)

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.

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?

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.

[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

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.

>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.

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.

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.

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).

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?

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.