Question 568·Hard·Math

The altitude makes two smaller triangles that are right at $D$. Their acute angles match, so the triangles are similar: matching sides have the same ratios. For the altitude, those ratios give $CD^2=AD\cdot DB$. This relationship finds the missing piece of the hypotenuse.

Put $CD=12$ and $AD=8$ into the altitude relationship. Type $12^2\sim8x_1$ in Desmos, where $x_1$ stands for $DB$; the $\sim$ tells Desmos to find the value that fits. Under PARAMETERS, it shows $x_1=18$, so $DB=18$.

Point $D$ divides $AB$ into $AD$ and $DB$, so add the pieces to get the whole hypotenuse: $AB=AD+DB=8+18=26$.

Side $BC$ is a leg, one of the two sides forming the right angle at $C$. It touches $DB$, not $AD$. The similar triangles give leg squared equals the whole hypotenuse times the piece beside that leg: $BC^2=AB\cdot DB$. Don’t use $AD\cdot DB$ here; that product belongs to the altitude.

Type $26x_1$; Desmos prints $468$. Then type $\sqrt{26x_1}$; it prints about $21.633$, the positive length of $BC$. The choices use exact square roots, so substitute $x_1=18$: $BC=\sqrt{26(18)}$. Multiply: $BC=\sqrt{468}$. Factor out the perfect square $36$: $BC=\sqrt{36\cdot13}$. Take its square root: $BC=6\sqrt{13}$. So side $BC$ is $6\sqrt{13}$. Choice B.

Once the altitude relationship gives $DB=18$, look at triangle $CDB$. The altitude makes a right angle at $D$, so $CD$ and $DB$ are its legs and $BC$ is its hypotenuse. The Pythagorean theorem gives $CD^2+DB^2=BC^2$.

Type $\sqrt{12^2+18^2}$ in Desmos; it shows about $21.633$. For the exact length, substitute the two legs: $BC=\sqrt{12^2+18^2}$. Square them: $BC=\sqrt{144+324}$. Add: $BC=\sqrt{468}$. Factor out $36$: $BC=\sqrt{36\cdot13}$. Take its square root: $BC=6\sqrt{13}$. So side $BC$ is $6\sqrt{13}$. Choice B.