Chords $\overline{AB}$ and $\overline{XY}$ of a circle meet at $T$. If $XT = 4$, $TY = 6$, and $AT = 2TB$, then what is $AB$?

From the intersecting chord theorem, we know that.....

 

XT * TY  =  AT * TB

                                            The problem tells us that  XT = 4,  TY  =  6 ,  and  AT = 2(TB) .

4 * 6  =  2(TB) * TB

24  =  2(TB)²

                                            Divide both sides by  2  .

12  =  (TB)²

                                            Take the positive (since TB is a length) square root of both sides.

√12  =  TB

 

 

AT  =  2(TB)

                                            We know that  TB = √12

AT  =  2√12

 

 

AB  =  AT + TB

                                            Plug in the values we know for  AT  and  TB .

AB  =  2√12 + √12

                                            Combine like terms.

AB  =  3√12

                                            We can simplify  √12  since  √12  =  √(2 * 2 * 3)  =  √(2²) * √3

AB  =  3(2√3)

AB  =  6√3

Tip:

 

The intersection of both chords loosely form the shape of an "x." For me, this fact reminds me that multiplication is involved in the individual segments of the larger chord.