1. C is the right angle so AB is the hypotenuse and AD is the altitude to the hypotenuse.
The hypotenuse is
[tex]AB=c= \sqrt{15^2 + 20^2} = \sqrt{5^2(3^2+4^2)} = 25[/tex]
The area calculations must match. Let h=CD
[tex] \frac 1 2 a b = \frac 1 2 c h [/tex]
[tex] h = \dfrac{ab}{c} = (15)(20)/25 = 12 [/tex]
Answer: 12
2. This is the same triangle as above. We seek x=AD and DB=25-x
[tex] 12^2 + x^2 = 15^2 [/tex]
[tex] x = \sqrt{3^2(5^2-4^2)} = 9 [/tex]
AD=9 and DB=25-9=16
Answer: AD=9, DB=16
