Maths Game (1 Viewer)

[studentcheese]

Come on, let's keep this thread running.. because there are people out there who need help with Maths! (e.g. me.) Okay, here's a question I have trouble with. I don't think its Sc work.. but can someone please help!

http://i420.photobucket.com/albums/pp282/studentcheese/Untitled.jpg

*sigh* This question seems so simple, yet I don't know why I have a different answer to the book.

 

[xXmuffin0manXx]

Come on, let's keep this thread running.. because there are people out there who need help with Maths! (e.g. me.) Okay, here's a question I have trouble with. I don't think its Sc work.. but can someone please help!

http://i420.photobucket.com/albums/pp282/studentcheese/Untitled.jpg

*sigh* This question seems so simple, yet I don't know why I have a different answer to the book.

ill guess 15/4

 

[xV1P3R]

Isn't that the theorem,
If two chords intersect inside a circle then the product of the lengths of the segments of one chordhttp://www.mathwarehouse.com/geometry/circle/chord-of-circle.php equals the product of the lengths of the segments of the other chord.
So if two chords AB and CD meet at X
AX*BX=CX*DX
And this also works if they intersect externally.
so AF*CF=DF*BF
8*5=4*(4+DB)
40=4*(4+DB)
4+DB=10
DB=6
I hope that's right.

 

[xXmuffin0manXx]

Isn't that the theorem,
If two chords intersect inside a circle then the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
So if two chords AB and CD meet at X
AX*BX=CX*DX
And this also works if they intersect externally.
so AF*CF=DF*BF
8*5=4*(4+DB)
40=4*(4+DB)
4+DB=10
DB=6
I hope that's right.

yes yes that is right

ill dig out the theorem in a sec

edit: theorem is the product of the intercepts of intersecting chords are equal

 

[studentcheese]

ahh thanks. I get it now. I had the wrong interpretation of the theorem.