pradeepss wrote:
If two sides of a triangle are 6 and 12, respectively, which of the following could NOT be the area of this triangle?
A. 1
B. 6
C. 17
D. 29
E. 38
Bunuel could you help!
Official answer from
veritas:
Solution: E.
If you know two sides of a triangle, the maximum area of the triangle can be obtained by setting these two sides as the base and height of a right triangle. For our triangle this gives us a maximum area of (6 * 12) / 2, or 36, so anything higher than this is invalid and the answer is (E). An alternative approach is to consider the minimum possible value of the triangle. Say we violated our third side rule and set the triangle's sides as 6, 12, and 18, respectively. Our "maximum" area here would be the set the base as 18, but since the triangle has no height -- the three sides only "fit together" as two parallel lines -- the height is 0 and the area is 0. But were we to extend one of the other two sides by an infinitesimally small amount, our area would be infinitesimally greater than 0, so presumably our minimum value is merely above zero. This means the greatest answer choice will be the number that violates the rule, and since no answer violates our minimum, (E) must again be the correct choice.
This is a very poor question. What is the source ?
Consider a line of 12 inch. Consider two line of 6 inches each, kept exactly above the first line and covering it completely.
Can you form a triangle by moving the two 6 inches lines away from the base line of 12.
No you cannot. The moment you move the any of the 6 inch line there would be a gap. The end points of one or more lines will not meet. If you keep the end points of the two 6 inches lines fixed with the end point of the 12 inches line then the other free end of the 6 inch lines will never meet.
This is the basis of triangle Triangle inequality Rule :-
THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD.
Therefore even if we assume the unknown side to be second smallest, then this unknown side must be must be greater than 6 . What can be the maximum length of the unknown side. Less than 18
6<unknown side<18
In the earlier case when two lines of 6 inches were lying exactly above the 12 inches line, the end points of all lines were collinear. Such a collection of three collinear points is called a degenerate triangle. (although it just looks like a line having three points marked over its length).
The area of an degenerate triangle is zero.
Any other combination of sides of a triangle (which follows the triangle inequality rule) will have an area of greater than zero.
Now the area of a triangle is given by \(\frac{1}{2} *b*h\)
and we know that the height is greatest when the angle is 90 degree
Keeping the smallest side of 6 as the base, the height can be 6.1 or 12 or 17.99 (there can be infinite values for height. I am just taking the two limiting cases at the either ends 6.1/17.99 and 12 is already given as the second side in question stem )
AND THE AREA CAN BE \(\frac{1}{2} *6 *(6.1\) OR \(12\) OR \(17.99\)) = approx 18.3, 36 or approx 72
SO TECHNICALLY SPEAKING ALL OPTIONS IN YOUR QUESTION ARE WRONG.Bunuel can you reconfirm my line of reasoning ?
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