What is the maximum possible area of triangle?

Bunuel, can you please comment on this? Is it not a GMAT type question? If that is the case, then there is no point discussing it.

To maximize the area of a triangle and if you know the length of two sides make them perpendicular to maximize the area. Hence A is sufficient.
However, I do have my doubts about this answer as well..
where are the experts?

Hi Smita04,

This is very much a GMAT query.

Evaluating statement 1 only:
Here, we know that the length of the two sides are 7 cms and 14 cms respectively. Now just picture this. Let us try to a triangle with base = 14 cm and then try to put the 7 cm side such that ther area is the maximum.




Let the side AB = 7 cm and BC = 14 cm. The figure shows 3 possibilities for AB that would result in the maximum possible area for triangle ABC.
Now, we k now that area
= 1/2 * base * altitude = 1/2 * BC * altitude.
Now, the area will be maximum for the maximum value of the altitude. This is only possible with AB as the altitude, as in the other two cases the length of the altutide goes down.
Hence, ABC is right angled and the maximum area = 1/2 * 14 * 7
We can eliminate options B, C and E.


Evaluating statement 2 only:
Let us picture this.



Let side AB = 7 cm. Now depending on the size of the circle, the area of the triangle can keep increasing. Hence, statement 2 alone is insufficient.
Hence D is eliminated.
Answer is A.

Hope this helps.

Regards,

Shouvik.

For the greatest area, the triangle should be a right angles triangle.
Also, Area = 1/2 * base * height.

If you make 14 the greatest side (hypotenuse), and 7 as base, the height would be smaller than 14.

To maximise area, consider 14 as height, in that case Area = 1/2 * 7 * 14.

A is sufficient.

B is clearly not sufficient, because we have to know atleast one more side, or the radius of the triangle.
Answer - A.

Hope it helps.

For any triangle ABC its area = 1/2*b*c*sinA=1/2*c*b*sinC = 1/2*c*a*sinB.

since two sides are known the area would be 1/2*7*14*sin(angle bw sides 7 & 14)

maximum value of sine of angle is 1 when angle is 90 degrees..
so maximum area is 49.

So statement A is itself sufficient.

Since triangle is inscribed as of statement B the side 7 can be a chord or diameter.So not much data.

So answer is A.

Hope its clear to you all.

Good question. I feel this can be perfect GMAT question.

Q ->What is the maximum possible area of triangle ?
1) Let the third side is x which ranges from 7<x<21
where x can take any value within this range.

Now the three sides of the triangle are 7, 14, 7<x<21. Using these 3 sides numerous triangles can be made but only 1 triangle will have the maximum area , which we can find out (but it is not required to find the max area)
Sufficient

2) Outright insufficient

Answer A

(1) You are absolutely right. Since only two sides of the triangle are given, the area varies depending on the third side. Since we can make the area as small as we wish and the third side must be between 7 and 21, the area must be finite for every one of these possible triangles. So, there must be a maximum, and because this is a DS question, we are not supposed to find that maximum. Trigonometry is out of question on the GMAT, but even without it, we can figure out when the area is maximum.

Since the area of a triangle is the same regardless which side we take as a base, we can consider 14 (denoted by BC in the attached drawing) as a constant base, and look at the various triangles that can be formed. Angle ABC varies between 0 and 180, with the side AC of constant length 7. Maximum height corresponding to BC is obtained when AB is perpendicular to BC, and in fact we now know that the maximum area will be 14*7/2 = 49.
Indeed (1) is sufficient.

(2) Any triangle can be inscribed in a circle and through two given points (apart at a distance of 7) there are infinitely many circles passing through.
Obviously not sufficient.

Hence answer A.

Despite what some comments claim above, this is most definitely not a GMAT question. It makes no logical sense in a DS question to ask for a maximum value of something, so you could never see a problem like this on the real test.

Here is an image to complement the classic answer


The horizontal black line segment is the initial fixed side.

The length of the second side is the radius of the red circle. All possible triangles with sides of these respective lengths can be generated by these radii; three examples are given in the picture: one in blue, one in green, and one in black. In each case, the base ×
height formula has the same base, and one sees that the height is greatest when the two line segments are perpendicular to one another.

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