Question

In the diagram below of right triangle ACB, altitude CD is drawn to
hypotenuse A
If AB = 36 and AC = 12, what is the length of AD?

Answer 1

B
D

C A

\[\text{BC}=36\text{ Sin}[\text{ArcCos}[12/36]]=24 \sqrt{2} \]\[\text{area} = \frac{1}{2}\text{BC } 12 =\text{ }\frac{1}{2}24 \sqrt{2}\text{ } 12 =144 \sqrt{2} \]\[\text{area} = \frac{1}{2}\text{CD } \text{AB} \]\[144 \sqrt{2} = \frac{1}{2}\text{CD } 36 \]\[\text{CD}=8 \sqrt{2}= 11.3137 \]

Answer 2

AD^2 = AC^2 - DC^2
AD^2 = 12^2 - (8 sqrt(2))^2
AD = 4