Question

guys I need 6 and 7 problems. can you help me

Answer 1

(6). ABC is an equilateral triangle where AD perpendicular to BC
if you draw diagram then you see two triangle form e.g ABD & ACD and both are right angle triangle .

now,
for ∆ ABD
use Pythagoras theorem
AB^2=BD^2+AD^2
hence
AD^2=AB^2-BD^2 ---------------(1)
in the same way for ∆ACD
AD^2=AC^2-CD^2 -------------(2)

from equation (1) and (2)
2AD^2=(AB^2+AC^2)-(BD^2+CD^2)

but we know traingle is. equilateral
so,
AB=BC=CA
and BD=CD=BC/2

hence ,
AD=√3/2AB=√3/2BC =√3/2CA

(5) ∆ ABC~ ∆DEF
so ,
AB/DE=BC/EF=AC/DF
1/2 =8cm/EF
EF=16cm

Answer 2

use Pythagoras theorem
AB^2=BD^2+AD^2
hence 
AD^2=AB^2-BD^2 ---------------(1)
in the same way for ∆ACD 
AD^2=AC^2-CD^2 -------------(2)

from equation (1) and (2)
2AD^2=(AB^2+AC^2)-(BD^2+CD^2)

but we know traingle is. equilateral 
so, 
AB=BC=CA
and BD=CD=BC/2

hence ,
AD=√3/2AB=√3/2BC =√3/2CA

(5) ∆ ABC~ ∆DEF
so ,
AB/DE=BC/EF=AC/DF
1/2 =8cm/EF
EF=16cm