What is the length of the altitude drawn to the hypotenuse? the figure is not drawn to scale. a) b) 70 c) d) 19

  we know that
in triangle ABC
cos (alfa)=H/(14+5)-------->   cos (alfa)=H/(19)        equation 1

in triangle BDC
cos (alfa)=14/H   equation 2
1=2
H/(19)=14/H-------> H²=19*14---------> H²=266------> H=√266
H=16.31 units
cos (alfa)=14/16.31----------> cos (alfa)= 0.8584
alfa=arccos(0.8584)=30.86°

we know 
triangle BDC
sin(alfa)=h/H---------> h=sin(alfa)*H
h=sin(30.86)*16.31----------> H=8.37 units---------> √70 units

the answer is the option A) √70

3units/ 5 feet = 60 units/ 100 feet3 units/ 5 feet = 18 units/ 30 feetthe dimensions of the scale drawing are 60 units by 18 units.hope this !

Ithink the answer is: it is likely.