If two sides of a triangle are 12 and 8, which of the

Question

If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?

I. 35
II. 48
III. 56

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Bunuel · Accepted Answer

Given two sides of a triangle, the greatest area is obtained when there is 90 degrees between them. Thus the area cannot be more than 12*8*1/2=48. Look at the diagram below: Triangles.png If we consider 12 as a base, then you can see that the greatest height is when it coincides with the second side. Now, if we rotate the second side we can make the height close to zero, thus we can have any height from 0 to 8, which implies that we can have any area more than 0 but not more than 48 (12*8*1/2=48). Answer: B. As for other properties. Check Triangles chapter of Math Book: math-triangles-87197.html Hope it helpsl

Jp27 · Answer

Bunuel - I have few other questions related to this. Please correct me if Im wrong with any of the statements below.

1/ Given 2 sides of the triangle max area can be go by forming 90* between them?

2/ Given the hypotenuse of a right angle triangle I can find out the max area by forming isosceles traingle (I find the other 2 sides using the relationship x:x:x root 2)

3/For a given perimeter equilateral triangle has the largest area.
Perimeter = 3+4+5 = 12. 
The area of an equilateral triangle = (s²/4)√3. (s = 12/3)

4/ For a given area equilateral triangle has the smallest perimeter.
Given area = 18 then smallest perimeter possible = 1/2 a^2 = 18
so a = 6 then perimeter 6+6+6 =18....

5/ sum of any 2 sides should be > than the 3side and diff should be < than 3rd side...

Are there any other relationship should I be aware of?

healthjunkie · Answer

Can you explain how we could get an area of 35? Base*Height would have to be equal to 70 right? How could we get B*H to be equal to 70 if the two sides are 12-8 which neither go into 70? Can we assume that the third side doesn't have to be an integer?

EgmatQuantExpert · Answer

Hi healthjunkie , Given two sides, the greatest area will be obtained when the sides are at right angle to each other i.e. they form a right angled triangle. For any other type of triangle, the two sides will not form the base and the height, thus \frac{1}{2} * multiplication of the sides will not give you the area of the triangle. So for example, for a scalene triangle, one of the sides could be the base and then the height to that base from the opposite vertex would need to be calculated which may not be an integer. Thus \frac{1}{2} * base ( either 12 or 8) * height can be equal to 35 Hope this helps

Regards Harsh

EMPOWERgmatRichC · Answer

Hi healthjunkie,

This question is an off-shoot of a Triangle Inequality Theorem question - in those questions, you're given 2 of the sides of the triangle and you're asked for what the third side COULD be. The lengths of the 2 sides dictate the minimum and maximum possible values for the third side and that same logic can be used here.

To emphasize the logic, you should sketch out some quick pictures....

First, the greatest area will be formed when the two sides form a RIGHT angle. With sides of 8 and 12, you have an area of (1/2)(8)(12) = 48.

Next, draw a picture in which the "8 side" does NOT form a right angle with the "12 side"....put the "8 side" at an angle so that it's almost "on top of" the "12 side".... This area is clearly SMALLER than 48 (in fact, it's just a little greater than 0).

If you draw a picture with the "8 side" almost "in line, but going in the other direction" with the "12 side" (so that you form a really long, thin triangle), THAT area is also really small (just a little greater than 0).

These examples serve as proof that any area from "almost 0" up to 48, inclusive is possible. In this question, that means that 35 and 48 are possible areas.

GMAT assassins aren't born, they're made,
Rich

ravindra88 · Answer

I know trignometry is not expected in GMAT, but this particular problem becomes very easy if you know just 2 little concepts of trignometry.

I Concept:  Area of a triangle =  (1/2)a.b.sin(\alpha ), 
where  a  and  b  are any two sides of the triangle and  \alpha  is the angle contained by these two sides.

II Concept:   sin(\alpha ) varies between  0  and  1  (becomes maximum at  \alpha   =90∘ )
So, in this case area will vary between  0  and  48 .

In other words, all the values between 0 and 48 are possible.

There are few problems of area maximization in a triangle which are difficult to be visualized but can be dealt quickly by this method.

GMATinsight · Answer

CONCEPT : 
-  Maximum  Area of the Triangle for given two sides will be obtained when the two sides are taken as perpendicular to each other
-  Minimum  Area of the Triangle for given two sides can always be as small as zero

i.e.  Area_{Max} = (1/2)*12*8 = 48 
i.e.  Area_{Min} ≈ 0

Answer: Option B

southamerica2016 · Answer

What about the triangles properties?

"any side must be smaller than the sum of the other sides and greater than their difference". I thought it was not possible to have a side way smaller. The properties are not applied in this case?

EMPOWERgmatRichC · Answer

Hi gabriela2015,

That rule applies to ALL triangles, regardless of what the question is asking about. Here though, we're NOT asked for the third side, we're asked for the possible AREAS when we have those two specific side lengths.

With a side of 8 and a side of 12, the third side would fall into the falling range:

4 < third side < 20

The range of the areas will still end up being:

0 < Area <= 48

GMAT assassins aren't born, they're made,
Rich

mohd_int · Answer

Could you tell me please how you got the possible range of the third side (4<third side < 20)
Thanks

EMPOWERgmatRichC · Answer

Hi mohd_int,

There's a Geometry rule called the Triangle Inequality Theorem - and the rule is that the sum of ANY 2 sides of a triangle MUST be greater than the third side. Thus, if we call the 3 sides of a triangle A, B and C, then...

A+B > C
A+C > B
B+C > A

All 3 inequalities MUST exist for a triangle to exist. Here's a simple example to prove the point: Can you have a triangle with sides of 1, 1 and 100? Try drawing it. You would NOT have an actual triangle, you would have a really long line with 2 tiny lines attached at the ends (and those tiny lines would NOT meet).

So if you have one side of a triangle that is 12, then the other two sides MUST sum to a total that is GREATER than 12. If one of those two sides is an 8, then the third side must be GREATER than 4. In that same way, 12+8 = 20, so the third side must be LESS than 20.

GMAT assassins aren't born, they're made,
Rich

beeblebrox · Answer

Hi  EMPOWERgmatRichC

In a triangle: s1+s2>s3 & |s1-s2|<s3.
Which helps us in inferring that with 12 and 8 as two sides, third side will be 4<s3<20. So the sides will possibly be be 5,6,7,......19. I am only considering integers for the sake of simplicity. 
That being said, will the smallest area of a triangle with above constraints not be :  (Sides = 5,12,8; Area = 0.5*5*8)  & the max area be  (Sides=19,12,8 & Area = 0.5*19*12). ?
So will the range of Area not be between 20 and 112?
Hence my answer will be all of the above numbers can be area of the triangle.
What's is wrong with my thought process?

chetan2u , sir can you help?

chetan2u · Answer

In triangle 19-12-8, you are taking the area to be 0.5*12*19. This means 19 and 12 are height and base. 
But if that is the case, then the third side has to be hypotenuse and thus, greater than 19 or exactly equal to  \sqrt{19^2+12^2}=22.5 ….A triangle that does not give you third side as 8.

Thus, you take the third unknown side to be the hypotenuse and the height and base be the given sides. This area = 0.5*12*8=48

EMPOWERgmatRichC · Answer

Hi beeblebrox, When it comes to maximizing the area of a possible triangle when you know two of the sides, there are two facts that you have to remember: 1) Any side of the triangle can be the 'base' of the triangle, but the 'height' is always PERPENDICULAR to that base (re: it forms a 90 degree angle) - which means that the height might actually be outside of the triangle (and you have to be careful not to confuse any diagonal lines with the height). 2) When you know the exact lengths of two of the sides, then the largest possible area will be when those two sides form a 90 degree angle (and a right triangle). In your two examples, neither of the measures that you use for the 'height' is actually the height. GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

KarishmaB · Answer

Not sure if the doubt is why (Sides=19,12,8 & Area = 0.5*19*12) is not possible. But if it is, note that we cannot have a right triangle such that it has legs 19 and 12 (to get the Area = 0.5*19*12) and hypotenuse 8. The hypotenuse will be greater than both legs. We can have a triangle with sides 8, 12 and 19 but it will not be a right triangle and its altitude will be much much smaller and hence its area will be very small too. Similarly, sides 5, 8 and 12 do not give a right triangle (does not satisfy pythagorean theorem) and hence its area is not (0.5 * 5 * 8). In fact this triangle also has a very small altitude and hence a very small area. The area of a triangle with given two sides can go very very small (just a tiny bit more than 0 to make it a triangle). But the maximum area given two sides will be when the two sides form two legs of a right triangle. The figures below show how triangles are formed when we are given two sides of say 5 cm and 4 cm. Note the very small altitudes (and hence very small areas) when the third side is the smallest possible or greatest possible. Screenshot 2023-03-14 at 10.06.58 AM.png

If two sides of a triangle are 12 and 8, which of the

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If two sides of a triangle are 12 and 8, which of the

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