In triangle ABC, CA = 4, and CB = 6. Which of the following could be t

Two sides given and we need to find range of area possible

Max area = 1/2* 6*4 = 12

Min area > 0

So III is not possible

D - I and II

1) We have to figure out the range for the third side of the triangle: 2 < AB < 10

2) We should test extreme cases to understand the range of the area by calculating the area of triangle through 3 sides:

S=sqrt{p*(p-a)*(p-b)*(p-c)}
p = (a + b + c)/2

The area will be smallest with the smallest side. Let's take AB = 2.005
p = (2.005 + 4 + 6)/2 = 6.0025
S = sqrt {6.0025 * 3.9975 * 2.0025 * 0.0025} ~ sqrt { 6*4*2 / 400} - it's definitely less than 1. So, the area of the triangle could be 1 (actually, AB = 2.04135 if area is 1)

Out of all triangles, the equilateral triange has the biggest area. In our case it's impossible, but it gives an idea that the triangle will have the biggest area if AB is within 5-8 range. It's easy to go and test 6,7,8 and get that the maximum area will be ~12. It means that area larger than 12 is impossible, and 11 is possible.

Thus, it leaves us with D: areas 1 and 11 possible.

Again, please provide any easier/faster solutions, I would greatly appreciate it.

Call the 3rd unknown side X


Draw a straight line = 6 units


Then draw a line directly on top of it that is 4 units ——— and another line that is 2 units to completely cover the original line.


Because of the triangle inequality theorem, the side lengths of 2, 4, and 6 can not work. We can not “lift” 2 of the 3 sides up such that we can connect them at vertices and create a triangle. They can only be placed directly on top of the line of length 6.


As we lift the line of length 4 above the line of length 6, we can imagine side X increasing just a little bit to reach the vertex and create a valid triangle.


We could pick up side 4 ever so slightly and imagine side X = a tiny minuscule length above 2. This would give us a triangle with barely any Area. Theoretically, a triangle with Area of 1 could be made.


As we pick up the side of length 4 and move it in a counterclockwise direction away from the line of length 6, we can create various triangles with different areas.

Side X will compensate with the appropriate length to connect side 4 to line 6 and create a triangle.

You can draw a semi-circle of sorts that includes various triangles with known side lengths 4 and 6 and unknown side length X.

The Area of this triangle will be Maximized when the side of length 4 is exactly perpendicular to the side of length 6.

Again, this can be seen if you create the visual drawing as described above.

Therefore, the MAX Area must be = (1/2) (4) (6) = 12

Therefore, the range of possible Areas will be:

0 < [Area of triangle with known side lengths 4 and 6] </= 12


Thus, only I and II are possible.

III, an area of 14, is NOT possible.


Takeaway Concept:

Given 2 known, CONSTANT side lengths of a triangle, the Area is MAXIMIZED when those 2 known, constant side lengths are placed perpendicular to each other.

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area of a triangle can is 1/2 ab sin(c) where c is the internal angle between the sides a,b
0 <= sin(c) <= 1 , and is maximized at c = 90 or a right triangle
so the area of the triangle can take on values from 0 to 12
I, II