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Based on one of the properties of a centroid. Segment GD is 1/3 the length of AD.

Sorry my method is pretty complicated so hang on.

Triangles ABD and ABC are similar. Since D is the midpoint, and they share AB as a side, their ratio is 1:2

We can find the area of ABC using Heron's formula. After one million years, we have a simplified radical of \(2\sqrt{4466}\).

Since their ratio is 1:2, we know that the area of ABD is now \(\sqrt{4466}\).

Side BD is 18 / 2 = 9

We know sides AB = 15 and BD = 9. So we can work backwards from Heron's formula to find the length of AD.

Working backwards from heron's formula: 

we calculate s. Since the length of side C is unknown, we can call it X.

We find that S  is 12 + 0.5x. We know the area is \(\sqrt{4466}\). So we can make the following equation:

\((0.5x+12)(0.5x+12-15)(0.5x+12-9)(0.5x+12-x)=4466\)

(0.5x+12)(0.5x-3)*(0.5x+3)(0.5x+12-x)=4466 = {x=-(2*sqrt(86)), x=2*sqrt(86), x=-(2*sqrt(67)), x=2*sqrt(67)}

So side C is either \(2\sqrt{86}\), or \(2\sqrt{67}\).

That means GD is either: \(\frac{2\sqrt{86}}{3}\), or \(\frac{2\sqrt{67}}{3}\)....

Yup! THis is completely messed up! What a crazy method! I got two possible sides too, uh.. Help?

There are 15 splits if we do not count the order in which the 4 numbers are arranged, This comes from the observation that picking two distinct numbers out of the 6 possible, not counting order, is \({ 6\choose 2}\)=15. Each of the splits can be ordered in 6 different ways. For instance, two ones and two twos can be arranged as

(1, 1, 2, 2)

(1, 2, 1, 2)

(1, 2, 2, 1)

(2, 1, 2, 1)

(2, 2, 1, 1)

(2, 1, 1, 2)

so there are \({6 \times15}\)=90 ordered splits. There are a total of \(6^4\)outcomes when 4 six-sided dies are cast. So the probability of an ordered split is \(\frac {90} {6^4}=\frac{5}{72}\)

Oh.

Got it!

I think it looks like a test or something that is worth marks.

conics and math

idk the points are just staring at me in the face...

Okay, nvm, it's conics...

And this very suspiciously looks like homework...

Yes and it is maths too.

What have you tried so far?

I just drew it and the answer to the first question is staring me in the face.

Sorry to interrupt-

Is this SCIENCE???

Note that all angles in a triangle sum to 180.

Notice \(\Delta ABE\), Angle BEA is 90, angle A is 54, so angle ABE is 180 - 90 - 54 = 36.

Notice \(\Delta BDA\), Angle HDB is 90, angle B is 52, so angle ​DAB is 180 - 90 - 52 = 38

Notice \(\Delta ABH\), Angle ABE is 36, angle DAB is 38, so angle AHB is 180 - 36 - 38 = 106

OS first I drew a visual representation of it to get a better picture of what I'm working with. But from then on I am unsure as to what I should do...

Still no bananas.  This is a combination question.

the figure doesn not show any other angles if u draw it out

am i missing somehting??

Casewrok gives us 5^3 + 5^4 + 5^5 = 3875 possible rolls, so the answer is 3875/7776.

I am not good with conics so I probably will not attempt to help anyway . 

But if I was to help I would want you to spell out what you had tried in order to help yourself first.

That's fine if you can help me later, thank you so much

Looks like you are doing a test right now.

Chris if you want to answer these could you please leave it for a couple of hours. 

Use 6, 7, 8, 9.

Make up 2 number: 96 x 87 =8,352 is slighly bigger than: 97 x 86 =8,342

By similar triangles, [G1G2G3]/[ABC] = 1/6.

Draw FH  through G   such that  FH  is parallel to   DE.....and let F lie on AB and H lie on AC

Then triangles DEA , FHA   and  BCA  are similar

And since G is a median  then AG  is  2/3  of the distance  from A to BC

So...AF  is 2/3  of AB

And AH  is 2/3 of AC

So   DF  = (2/3)AB - (1/2)AB = (1/6)AB

So  DF /  AF   =  [ (1/6)AB]  / [(2/3)AB] =   (1/6) / (2/3)  =   1/4

And  because triangles  DEA  and FHA are similar....then  DF /AF  =   GM / GA  =  1/4

so lazy he didn't even remove latex or converted

Draw   median AE   to bisect  BD

Let the intersection of G1G2  and AD   =  F

Since G1G2 is parallel  to BC, then   triangles  A G1 F   and  AED   are similar

But since   AE is a median.....then  AG1  is 2/3  of AE

So....  AE  = 3/2  of AG1

So....ED  is (3/2)  of    G1F  =  (3/2) (2)  = 3

And, by symmetry, ED  = (1/4) BC  

So BC   =  4ED  =  4 * 3  = 12

So  (1/2) BC  = BD  = 6 

And by the Pythagorean Theorem

AD^2  + BD^2  =  AB^2

8^2 + 6^2  = AB^2

100  =  AB^2

10  = AB  = AC

So....the perimeter  of ABC  =  BC + AB + AC  =  12 + 10 + 10  =  32

The  area is given by

49  =  (1/2)r^2 (theta)   where theta is in radians

Multiply  both sides by 2  and divide by r^2  and we have that

98 / r^2  =  theta      (1)   

The perimeter of the sector =  2r  + r (theta) 

So

28  =  2r   + r (theta)    (2)

Sub (1)  into (2)  and we have that

28  =  2r + r (98 / r^2)      simplify

28  = 2r  +  98/r      divide through by 2

14  = r  + 49/r       multiply through by r

14r = r^2  + 49      rearrange as

r^2 - 14r + 49  = 0      factor

(r - 7)^2  =  0          take the square root

r - 7    = 0

r  = 7 

And theta  =  98 / 7^2   =  98 /49   =   2 

So....the arc length of the sector is

r * (theta)   =  7 (2)   =    14 cm

Everyone is finding the middle finger.

And that is Ronaldo.

The question is asking which sibling, not which finger

You are convincing me to give a middle finger to you, Guest. grrr.....