Amazingly hard MATH question, help!

Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: Amazingly hard MATH question, help!
 By Techieguy (Techieguy) on Friday, October 10, 2003 - 09:06 pm: Edit

25. One side of a triangle has length 6 and a second side of 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) II and III only
e) I, II, and III

 By Techieguy (Techieguy) on Friday, October 10, 2003 - 09:07 pm: Edit

Oh, and it would help if someone explained it also

 By Jimjunior (Jimjunior) on Friday, October 10, 2003 - 09:14 pm: Edit

the correct ANSWER is not one of those 5 answers, so that might be part of the problem. The biggest it could be would be an area of 21, if those two sides met at a right angle. It could be anything less than that, so 13 is a possibility. Answer would be I and II

 By Ziggysux4 (Ziggysux4) on Friday, October 10, 2003 - 09:17 pm: Edit

It appears as though the correct answer is not listed. The best way to look at this is to use the formula A = absinC/2, where a and b are two sides and C is the angle in between them. ab = 42 here, so what you want to figure out is whether or not 42 x (sinC/2), or 21sinC, can equal each of the three numbers. Sine of angles from 0 to 180 degrees ranges from 0 (at 0 and 180) to 1 (at 90), including all the values in between. Therefore, the area cannot equal 24, because that would mean that sinC would have to be greater than one (impossible). 21 is certainly possibly, if you have a 90-degree angle between the two sides, and 13 is possible, where sinC = 13/21. Choices i and ii are correct.

 By Mastadecoy2 (Mastadecoy2) on Friday, October 10, 2003 - 09:19 pm: Edit

Why isn't it E? 24 could work too... since one side is 6 and another side is 7 the third side has to be 1<X<13.

 By Techieguy (Techieguy) on Friday, October 10, 2003 - 09:20 pm: Edit

oh whooops

d) should be I and II only

and u guys are right...

 By Btbam (Btbam) on Friday, October 10, 2003 - 09:24 pm: Edit

because the third side is irrelevant. A = .5bh, lets say 7 is the base, to maximize the area we maximize the height, which can be anything up to 6. so the MAX area is .5x7x6 = 21. The triangle's area can be 21, almost 0, or anything in between.

 By Techieguy (Techieguy) on Friday, October 10, 2003 - 09:25 pm: Edit

Can someone please explain it without using Trig?

"The biggest it could be would be an area of 21, if those two sides met at a right angle."

is the biggest area of a triangle always when it is a right angle?

 By Xiggi (Xiggi) on Friday, October 10, 2003 - 09:25 pm: Edit

This question was posted yesterday.

http://www.collegeconfidential.com/discus/messages/69/30407.html

No, 24 cannot work.

 By Xiggi (Xiggi) on Friday, October 10, 2003 - 09:28 pm: Edit

is the biggest area of a triangle always when it is a right angle?

Yes

 By Mastadecoy2 (Mastadecoy2) on Friday, October 10, 2003 - 09:29 pm: Edit

so a right triangle = most area ?

 By Mastadecoy2 (Mastadecoy2) on Friday, October 10, 2003 - 09:29 pm: Edit

ok thx

 By Jason817 (Jason817) on Friday, October 10, 2003 - 09:31 pm: Edit

"so a right triangle = most area ? "

yes

 By Whodunit (Whodunit) on Friday, October 10, 2003 - 10:55 pm: Edit

i dont get it, what if its not a right triangle and the sides are 6,7,12 ..how would you figure that area to be less than 21

 By Xiggi (Xiggi) on Friday, October 10, 2003 - 11:55 pm: Edit

Try it

If your base is 12, your height will be a lot less than 2.

If your base is 7, and the two sides are 6 and 12, you will have an obtuse anhle and the height will be less than 6 and the area will be smaller than 21.