Anyone know this math problem?

Pythagorean theorem is applicable to right triangles only.

Unless we assume bisects here means to divide equally into two, there are infinite number of solutions to this problem.

 

Yeah, Pythagorean only works if the triangle is right. So you can't use that here.

 

Also, DB and AC would have to be perpendicular for the two bisected pieces to be equal to eachother.

 

2+2= 22
see i'm good at that too

 

My clinical eye tells me there's no way that AE= 6 and EC can be 10.7...That's like almost 5 above the AC and that is not proven through visualization .
Still think it's 6 if you maths dudes can't find any EC value under 8
I've even measured in my CTR monitor

 

Web master doing Maths, Amazing stuff man. Me too dont Like Maths wokyoh

 

Look at DA and DC, they almost look the same yet DA=8 and DC=12, the picture is not to scale. The answer is 10.77......

 

IMO, the answer is nine....

My Logic-
That ray becomes an angle bisector
Therefore the angle bisector property comes into play.

 

Let me break it down for those of you that are mathematically challenged:

Since DB bisects AC that means both DEA and DEC form right angles so you have 2 right angle triangles and we all know, or should know that A²+B²=C² thus:

In the triangle DEA side A=DE, side B=AE and side C=DA thus:
DE²+AE²=DA² thus:
DE²+6²=8² thus:
DE²+36=64 thus:
DE²=28 thus:
DE²=square root of 28 thus:
DE=5.2915

In the triangle DEC side A=DE, side B=EC and side C=DC thus:
DE²+AC²=DC² thus:
5.2915²+DC²=12² thus:
28+DC²=144 thus:
DC²=116 thus:
DC²=square root of 116 thus:
DC=10.770329614269008062501420983081

 

Roman, AFAIK bisecting just means dividing into two equal parts. It need not necessarily mean right angles.

 

I hated math as well. But I'm curious to what the answer is
now. 6 seems to be a popular answer, is it right?

 

If it divides in two equal parts and the AC is a 180º angle, then it divides that in two, therefore EC=6...

 

Assuming here bisecting means equally dividing, I get ce=9

ade = cde (equal arcs inscribe equal angles)

dac = 1.5* dca (in ratio of their opposite sides.)

Similarly

de/ae = dac/ade...1
de/ce = dca/cde...2

dividing eqn 1 by eqn 2

ce/ae = dac/dca = 1.5

ce = 1.5*ae

ce = 9

 

Ray DB bisects arc AC thus cutting the arc in half between points A and C which makes Ray AC perpendicular to Ray DB creating 2 right angle triangles.

 

If it divides the arc does that mean it divides the AC in half?

 

Look at what I had said in short in post no.28 of this thread

 

You cannot assume the shape being bisected is a uniform plane.

Twan, go do your own homework
http://www.rfcafe.com/references/mathematical/geometry_arc_circles.htm

 

Hmmm, yes it would and since AC is a straigth line and you have 2 right triangles with 2 sides the same in each triangle then the third must be the same in each triangle, but it isn't which makes the whole problem impossible, my brain hurts.

If AE = 6 and EC = 6 and DE is common to both then DA and DC must be equal but they are not stated as such.

 

See...

We're all right!
This math problem sucks, bring one that really makes sense!

 

In the figure that Twan has provided us, the ray bisects the arc, not the line. When a line bisects another line/ arc/ any other geometric figure, it means that it divides it into two. Only if the line was a perpendicular bisector to AC, would the right angles and equal measures come up.

Therefore in this case, m arc ( axb) = m arc ( bxc)
That means that angle adb = angle cdb

m angle adb = m angle cdb = 1/2 of the measure of the arc.

That makes the ray an angle bisector of the angle D
DA/ AE = DC/CE ( the actual property states that DA/ DC = AE/ CE- )

8/6 = 12/ CE
CE is 9