The diagonals of a rhombus ABCD meet at M. Angle DAB is 60 degress. P is the midpoint of AD. G is the intersection of PC and MD. AP = 8.

Find MD MC MG CG and GP

so far i got

Md - 13.86

MC - 8

MG 4.62

CG - 9.24

im not sure if im right tho

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- May 6th 2007, 02:46 PMstones44Rhombus
The diagonals of a rhombus ABCD meet at M. Angle DAB is 60 degress. P is the midpoint of AD. G is the intersection of PC and MD. AP = 8.

Find MD MC MG CG and GP

so far i got

Md - 13.86

MC - 8

MG 4.62

CG - 9.24

im not sure if im right tho - May 6th 2007, 06:02 PMtopsquark
The line BD cuts the rhombus in half, bisecting the angles CDA and ABC. Since these angles are (1/2)*(360 - 2*60) = 120 degrees, triangle ABD is an equilateral triangle. Thus MD = 16.

Triangle ACD is an isosceles triangle, with the two equal sides (CD and DA) of 16, and vertex angle 120 degrees, so AC will be

AC = sqrt{CD^2 + AD^2 - 2*CD*AD*cos(12)} <-- By the Law of Cosines

AC = 27.7128

Try working it from here.

-Dan - May 7th 2007, 02:55 PMstones44
ooooh

thanks a bunch....i never see those obvious equilateral or isoscles triangles

and i think you mean MD is 8 because it is half of BD (which is 16)

but: that only helps me get MD and MC...I still have a hard time finding the other three - May 7th 2007, 05:24 PMtopsquark
- May 7th 2007, 05:36 PMtopsquark
I'm not going to do the calculations, but it's not that hard if you go step by step.

Do CG first.

We know that CD = 16 and DP = 8 and angle CDP = 120 degrees. Thus

CP^2 = CD^2 + DP^2 - 2*CD*DP*cos(120)

by Law of Cosines. (Triangle CDP)

Now we can find angle PCD:

sin(angle PCD)/8 = sin(angle CDP)/CP

by Law of Sines. (Triangle CDP)

Now we can find angle CGD:

angle CGD = 180 - angle CDG - (1/2)*120<-- Angle CDG is 1/2 of 120 since BD is the diagonal of the rhombus.

since the sum of the interior angles of a triangle is 180 degrees. (Triangle CGD)

Now we can find CG:

sin(angle CGD)/16 = sin(60)/CG

by Law of Sines (Triangle CGD)

Now do MC:

sin(60)/MC = sin(30)/8 <-- MD = 8

by Law of Sines (Triangle CMD)

-Dan - May 7th 2007, 05:43 PMalinailiescu
to find the other length you shoul notice that CP amd DM are medians in triangle ADC, so G is the center of mass of that triangle.

So, we have GM=1/3 MD,

CG=2/3PC

PG=1/3CP

Use Cosine law in triangle CPA to find CP.

Using sine law in this triangle, after you find the sides will give you the angle ACP.

Hope this helps,

Alina - May 7th 2007, 06:03 PMstones44
thanks..this all helped

is there a way to do it without cosine or sine? i believe there is. - May 7th 2007, 06:06 PMtopsquark
- May 8th 2007, 02:06 PMstones44
i know there is.....the questions is in a subject before trig is taught...hmmm if i can find PC without trig, i can get everything else

- May 8th 2007, 02:33 PMstones44
YESSS I DID IT!!!

ok, so using the median theroum i can find that since DM is 8, and MG is 1/3 of DM (because it is a median of triangle ADC) i find mg is 2.666666.......so then i use pythagoras and do MG^2*MC^2 (MC, which i found previously) and this gets me CG....then i know that cg is 2/3 of PC becuase PC is also a median of triangle ADC....so, PG is 1/3 of PC....

NICE!!!!!!!!!!!!!!!!!