Math question

Son of Liberty · 08-04-2005, 02:59 PM

Y'all love solving them... so Imma keep bringin them

This one's from CB Red book pg. 288

One side of a triangle has length 6 and a second side has length 7. Which of the following could be the area of this triangle?

I. 13
II. 21
III. 24

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

(Answer below)

If you could just tell me how to go about solving this kinda question I'd really appreciate it. Thanks

D

tanman · 08-04-2005, 03:12 PM

The max area is when the given sides are the perpendicular sides of a right triangle (because you can never have another height > 7) so max area = (1/2)(6)(7) = 21.

Edit: I just googled the problem and found a solution on Dr. Math

tkm256 · 08-04-2005, 03:18 PM

The largest area possible is when the angle between the two sides is 90, because the third side must be longer but there is a lot of space within the shape. If you imagine the sides moving apart or together, hinged at their intersection, the area shrinks away from a right triangle. So, if that is true, 1/2 b*h would be 1/2(6*7) which is 3*7 which is 21. 21 units squared is the largest possible area, and smaller areas than 13 are possible because if the right angle was between 6 and the last side (root 13) the area is 10 point something, so the answer is D.

theoneo · 08-04-2005, 03:19 PM

You want to find the range of possible area values. Any answer choice(s) that isn't within this range is not a possible area.

First, find the minimum possible area. You know that the third side must be greater than one unit long. If you think about it, the smallest possible length is not an integer. It would be 1.000000000(infinite number of 0's)1. If you try to picture this 7-6-1.000001 triangle in your head, you can figure out that the area would approach 0.

The maximum possible area of any triangle is when it's a right triangle. if you didn't already know this, you can figure it out using the formula A=BH/2. Take a right triangle whose legs' lengths are b and h. The area is bh/2. Now, if you tilt h over a little bit to make the triangle acute, the base stays the same (b) while the height decreases (a fraction of h). Now the area is (a part of x)(b)/2. This area is smaller than bh/2. The same thing happens if you tilt h to make the triangle obtuse.

Now, if the triangle is right, then the legs would have lengths 6 and 7. The area is 6*7/2, or 21.

The range of the area is 1 < A <= 21.

Choices I and II are within this range, but III is not, so the answer is D.

Son of Liberty · 08-04-2005, 03:48 PM

ah i get it. i didnt know about the right triangle thing being the greatest possible area.

one question though theoneo... you said that the range of area is 1<A<= 21... why is the minimum value 1? shouldn't it be 3? because if the smallest possible value for a side is 1.000000000...1, wouldn't the smallest possible area be 1/2(6)(1.000...1)= 3.0000... or something?

optimizerdad · 08-04-2005, 04:14 PM

I don't think the triangle needs to have integer lengths for all of its sides (though two of them obviously are integer). The area of the triangle can be anything from 0.00000000...00001 to 21 .

theoneo · 08-04-2005, 04:22 PM

Whoooops. I meant the range for the area is 0 < A <= 21.

It wouldn't be 3 because 6-7-1.000000001 is not a right triangle.

Why can't you have the sum of the lengths of the two shorter sides of a triangle be equal to the length of the longest side? Because then you would just have two parallel lines on top of each other. But if the sum is SLIGHTLY greater than the length of the longest side, then it would be a triangle with a very very small area.

Anyway, you don't even need the minimum area in this problem. You could've just done the maximum and you'd be fine. But the unreachable limit for the minimum area of a triangle if you're given exactly two sides will always be 0.

theoneo · 08-04-2005, 04:31 PM

6-7-1.1 triangle: graph y = 2, y = .07084x + 2, and y = -.41773x + 4.92413 in the first quadrant. The area is 1.48397.

6-7-1.000001 triangle: graph y = 12, y = .0002182178767x + 12, and y = -.001309307x + 12.00916515 in the first quadrant. The area is .0045815.

Son of Liberty · 08-04-2005, 04:31 PM

ohhh all right. gotcha. thanks much everyone.

il bandito · 08-04-2005, 05:20 PM

How is the area 0?? Half of 1 x 6 or 7 is 3 or 3.5...

EDIT: Nevermind...I was thinking of the other side, not the height.

gcf101 · 08-04-2005, 10:36 PM

RE: post #4.
theneo, really cool dynamic visualisation - helps a lot on geom. questions.

Another (trig.) way:
For fixed values of sides a and b
area A = ab sinC,
sinC<= 1
A<= ab and
A<max> = ab when sinC=1, C=90deg.

gcf101 · 08-05-2005, 01:54 AM

Should slow down:

it's
A = 1/2 ab sinC.
A[max] = 1/2 ab.

Sorry for misleading!

il bandito · 08-05-2005, 03:10 PM

And as the angle C approaches 0, the area of the triangle should also approach 0.