Point \(B\) lies on line segment \(AC\), in between \(A\) and \(C\), so that \(BC=10\times AB.\) Line segment \(DE\) is parallel to \(AC\) so that \(BCDE\) forms a rectangle and \(ABE\) forms a right-angled triangle.
If \(AE=25\) and \(BD=74\), determine the exact value of the length of \(AD.\)
Let \(AB=x.\) Then \(BC=10\times AB=10x.\) Let \(CD=y.\) Then since \(BCDE\) is a rectangle, it follows that \(BE=CD=y\) and \(ED=BC=10x.\) We label our diagram accordingly.
Using the Pythagorean Theorem in \(\triangle ABE\), \[\begin{aligned} AB^2+BE^2&=AE^2\\ x^2+y^2&=25^2\\ x^2+y^2&=625\\ y^2&=625-x^2 \end{aligned}\] Since \(\angle BED=90\degree\), it follows that \(\triangle BED\) is a right-angled triangle. Using the Pythagorean Theorem in \(\triangle BED\), \[\begin{aligned} BE^2+ED^2&=BD^2\\ y^2+(10x)^2&=74^2\\ y^2+100x^2&=5476\\ y^2&=5476-100x^2 \end{aligned}\] Then, since \(y^2=625-x^2\) and \(y^2=5476-100x^2\), it follows that \[\begin{aligned} 625-x^2 &= 5476-100x^2\\ 99x^2 &= 4851\\ x^2 &= 49 \end{aligned}\] Thus, \(x=\sqrt{49}=7\), since \(x>0.\)
Then \(y^2=625-x^2=625-(7)^2=576\). Thus, \(y=\sqrt{576}=24\), since \(y>0.\)
It follows that \(CD=y=24\) and \(AC=AB+BC=x+10x=11x=11(7)=77.\)
Since \(\angle ACD=90\degree\), it follows that \(\triangle ACD\) is a right-angled triangle. Using the Pythagorean Theorem in \(\triangle ACD\), \[\begin{aligned} AC^2+CD^2 &= AD^2\\ 77^2+24^2 &= AD^2\\ 6505 &= AD^2 \end{aligned}\] Thus, since \(AD>0\), we can conclude that \(AD=\sqrt{6505}.\)