QR questions

[113zami]

1) Triangle JKL is equilateral and triangle JKM is isosceles. If KL=2 what's the distance from L to M? ansr: 0.73

I understand their ansr, but I did it with a different method which I think is equally valid but don't know why I keep getting the wrong ansr. What I did:
Since angle MKJ is 45 degrees and angle LKM is 60 degrees therefore angle LKM is 15 degrees.
Since triangle MKJ is 45-45-90 therefore side MK=sqrt2and side LK=2, now I have 2 sides and an angle so I can use the law of cosines to find length of side ML

So, a=b^2+c^2-(2bc)cosA
a=ML, b=2, c=sqrt2, cosA=cos15
=4+2-(2(2)(sqrt2))cos15
=6-4sqrt2(.96)=.57



2) what's the distance from the midpoint of segment EF to the midpoint of segment GH? Ansr: 5.59

I keep getting 6.5 this is how I did it:
Midpoint of EF=(-5/2, -1/2)
Midpoint of GH=(3,1/2)

Distance between them=sqrt((-5/2-3)^(2)+(-1/2-1/2)^(2))
=sqrt((-11/2)^(2)+(-1)^(2))
=sqrt((121/4)+(1))=(11/2)+(1)=13/2=6.5

3) The figure shows a regular pentagon with all of its vertices on the sides of a rectangle. If BC=1/4, what is the perimeter of the pentagon?

I just don't understand the 2nd step of the correct answer, 1st you use the measure of the interior angle of a regular polygon formula and you find that each angle measures 108, then they say the measure of each angle on either side of vertex E is 36 but how do they know that they should equal eachother, why not 35 degrees and 37degrees, I know the total of these two angles should be 72 degrees, but how do I know they should equal each other
(don't worry about the rest of the problem, I just need this part settled)

thanks in advance

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[jntruong2003]

QUESTION #1

you CAN do it by that method, but i think it would be rather difficult to try to figure what cos15 is. regardless though, you made the mistake in the equation you utilized. the law of cosine is b^2=a^2+c^2-(2ac)cosb.

you put b=a^2+c^2-(2ac)cosb.

if you plug in the math again. you will get the right answer.

 

[jntruong2003]

Question #2

you made the mistake at this part =

=sqrt((121/4)+(1))=(11/2)+(1)=13/2=6.5

you cannot find the sqrt of (121/4) and (1) separately and add them.

sqrt of 121/4 + sqrt of 1 DOES NOT = sqrt (121/4 +1)

just add 121/4 + 1 = 121/4 + 4/4= 125/4

sqrt 125/4 = (5sqrt5)/2= 5.59

hope this helps

 

[Streetwolf]

3) The figure shows a regular pentagon with all of its vertices on the sides of a rectangle. If BC=1/4, what is the perimeter of the pentagon?

I just don't understand the 2nd step of the correct answer, 1st you use the measure of the interior angle of a regular polygon formula and you find that each angle measures 108, then they say the measure of each angle on either side of vertex E is 36 but how do they know that they should equal eachother, why not 35 degrees and 37degrees, I know the total of these two angles should be 72 degrees, but how do I know they should equal each other
(don't worry about the rest of the problem, I just need this part settled)

Okay H = G because it's a regular pentagon. Each are 108 degrees. I think you will agree with that.

A and D are both 90 degrees because you have a rectangle.

So you have angles GFD and HIA are the same because the other 2 angles in both triangles are same between triangles (A = D and H = G).

That means since angle EIH = EFG, that angle EIB = angle EFC (why?).

And finally since B = C, you have FEC = IEB.

Okay now actually plugging in values...

You have HGF = 108 so DGF = 72. With D = 90 you get DFG = 18. Since GFE = 108 you get EFC = 54. Finally since C = 90 you get FEC = 36.

You can either repeat that for the other side or realize that since FEI = 108 you get BEI = 36 also.

 

[113zami]

thank you both

 

[rOOt_cAnAl]

Here's another way to solve question 1:

Triangle JKM is an iso: KM = JM and JKM=KJM=45

Triangle JLK is an equilateral: JL=KL=JK=2 and all angles of triangle JLK = 60

Now, if u draw a line perpendicular JK, it passes through M and L and bisects JK. Let's say it intersects JK at A so JA=KA=2/2 = 1.

So, in triangle AKM, tan 45 = 1 = AM/AK
1 = AM/1
AM = 1

In triangle AKL, sin 60 = (sqrt3)/2 = AL/KL
AL = sqrt3 (KL = 2)

Thus, LM = sqrt3 - 1 = 1.73-1 = .73

I hope this helps!!