|By Thunder77 (Thunder77) on Sunday, June 20, 2004 - 08:05 pm: Edit|
If you have time, can you please provide an answer and explanation
1) The length of two adjacent sides of a parallelogram are 6 and 15. If the degree measure of the included angle is 60, what is the length of the shorter diagonal of the parallelogram?
2) in right triangle ABC, altitude CD is drawn to hypotenuse AB. If the coordinates of points A, B, and D are (3,-1), (3,34), and (3,6), respectively, what are possible coordinates of point C? (There is only two)
That's it. If you have time and know how to do it, please help!
|By Aim78 (Aim78) on Sunday, June 20, 2004 - 08:24 pm: Edit|
Is this from an SAT book?
1.) You make a triangle with one side = 6, one side = 15, the other side is x (the small diagonal), and the angle opposite that is 60. Use law of cosines.
x^2 = 36+225-(2)(6)(15)(cos60)
x^2 = 171
x = square root of 171 ~ 13.08
2 is kind of complicated but if you get the basic graph, you'll see that you need to make the slope of the line going through A and the slope of the line going through B perpendicular. Doesn't seem like an SAT problem so I'm not going to do it.
|By Jenesaispas (Jenesaispas) on Sunday, June 20, 2004 - 08:28 pm: Edit|
You could do it that way for 1, or, if you want the simpler yet more time-consuming method...which, in essence, is the same thing:
If you set it up, with 15 on the long side and 6 on the shorter side, and a 60 degree angle inside, then you can just draw an altitude from the 120 angle to the side of length 15.
Now you have a 30-60-90 triangle. You know the hypotneuse... it's 6. So you find the base, it's 3, and then the other leg, which is 3-sqrts(3). And now you have another triange. It has a base of 3sqrts3, and a longer leg of 12.
So, it's just 12^2 plus (3sqrts3)^2=shorter diagonal^2. I got 3sqrts19. And that's approximately 13.08.
|By Thunder77 (Thunder77) on Sunday, June 20, 2004 - 08:43 pm: Edit|
What about #2? Can it be done without graphing it?
|By Jenesaispas (Jenesaispas) on Sunday, June 20, 2004 - 09:49 pm: Edit|
Well, I'm not so sure about this answer, but I'll try. I think that if you just roughly sketch it... you get two trainges on either side of the line x=3. I think that if you draw an altitude from the 90-degree angle to the shared hypotneuse, the result is two similar (well, three) triangles. Isn't that some property of triangles?
So, since we're tyring to find the length of the altitude for the coordinate, we assign x to its length. So, 31/x = x/3 (Since the hypo's length is 34..etc.)? that would mean x^2=93. So you just add and subtract that from 3, and you have your x-coordinates.
Does this sound right?
|By Optimizerdad (Optimizerdad) on Sunday, June 20, 2004 - 10:11 pm: Edit|
We know that AD=7 and DB=28. If x is the altitude of the original triangle ABC, then CDA and CDB are also right angled triangles; therefore, from these latter triangles,
AC^2 = AD^2 + x^2 = 49 + x^2 and
CB^2 = DB^2 + x^2 = 28^2 + x^2
From the original right-angled triangle ABC, we know that AB^2 = CB^2 + AC^2
(28+7)^2 = 28^2 + x^2 + 49 + x^2
1225 - 784 - 49 = 2x^2
392 = 2x^2
196 = x^2
x = +/- 14
So C can have coordinates of (3+14, 6) or (3-14,6)
|By Benzinspeicher (Benzinspeicher) on Sunday, June 20, 2004 - 11:47 pm: Edit|
optimizerdad, i did it the exact same way as you did and got the same answer; there's some verification
|By Thunder77 (Thunder77) on Monday, June 21, 2004 - 12:28 am: Edit|
Just two more please. One of them I don't get what they mean by "gutter" and the other one I dont understand the problem as well
1) a rectangular sheet of aluminum 25ft long and 12 inches wide is to be made into a rain gutter by foldig up the two longer parallel sides the same number of inches at right angles to the sheet. how many inches on each side should be folded upso that the gutter will have its greatest capacity?
2) Student editors are designing the yearbook so that each rectangular page will have a permimter of 40 inches The printer advises that each page must have a margen of 1.5 inches on the bottom and 1 inch on the other three side. What must be the dimensions of the page if the printed area is to be as great as possible?
A good explanation would be appreciated(like Optimizer Dad's, for ex) !!
Thanks to ALL who helped!
|By Foreignboy (Foreignboy) on Monday, June 21, 2004 - 12:46 am: Edit|
-Converting everything to inches, the sheet is 300" times 12"
-Assume the folded up width to be x. The volume (V) of the gutter is therefore (300)(x)(12-2x). Imagine it in 3D.
V = (300)(x)(12-2x)
= 3600x - 600x^2
dV/dx = 3600 - 1200x
as dV/dx = 0, x = 3"
|By Foreignboy (Foreignboy) on Monday, June 21, 2004 - 12:54 am: Edit|
-Assume the width to be x.
-As P = 40", the length = (40-2x)/2 = 20 - x
-Width of printed area = x - 2
-Length of printed area = 20 - x - 2.5 = 17.5 - x
-Printed Area, A, = (x - 2)(17.5 - x) = -x^2 + 19.5x - 35
-dA/dx = -2x + 19.5 = 0
Therefore x = 9.75
Width = 9.75, length = 10.25
Hope that helps. These are definitely not SAT questions.
|By Thunder77 (Thunder77) on Monday, June 21, 2004 - 11:54 am: Edit|
Thanks! Nice Explanation
But for the Gutter problem, the Ans says 0.3" . ???
One more , its a rate problem. Im just going over problems for the final and this one I dont really understand
teacher drove 280 miles to attend a math conference and arrivd 1 hour late. The teacher figured that, had she increased her average speed by 5 miles per hour, she would have arrived at the time for which the conference was scheduled. What was her average rate of speed?
|By Needhelp06 (Needhelp06) on Monday, June 21, 2004 - 12:12 pm: Edit|
let "a" be the average speed of the car.
let "j" be the time it takes to get to the math conference.
280mi * 1hr/a mi = j + 1 hrs
280 = aj + a
280mi * 1hr/(a+5)mi = j hrs
280 = aj + 5j
setting the two equations equal to each other gives:
aj+a = aj+5j
a = 5j
substitute back into original equation:
280 * 1/5j = j+1
5j^2 + 5j - 280 = 0
j^2 + j -56 = 0
because timeis positive, j = 7 hrs.
wefound that a = 5j, so a = 35mi/hr.
|By Thunder77 (Thunder77) on Monday, June 21, 2004 - 12:21 pm: Edit|
BTW, I just got a new calculator, how do you solve an equation with the TI-83 plus?
|By Thunder77 (Thunder77) on Monday, June 21, 2004 - 03:49 pm: Edit|
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