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Jet001 [13]
3 years ago
7

AD¯¯¯¯¯ , BD¯¯¯¯¯ , and CD¯¯¯¯¯ are angle bisectors of the sides of △ABC . CF=8 m and CD=17 m.

Mathematics
2 answers:
lara [203]3 years ago
6 0

Answer: DE= 15 m

Explanation:

Since, in \triangle CDF,

After applying, Pythagoras theorem, DF^2=CD^2-CF^2=(17)^2-(8)^2=289-64=225

Thus, DF^2=225 \Rightarrow DF=\sqrt{225} \Rightarrow DF=15 m

Again, in \triangle BDF and \triangle BDE,

\angle BED= \angle BFD ( Right angles)

BD=BD ( common edges)

\angle DBE= \angle DBF (BD makes the angle bisector of angle B.)

Thus, according to AAS condition- \triangle BDF\cong \triangle BDE

So, DE=DF (CPCT)

Therefore, DE=DF=15m⇒DE=15 m


irga5000 [103]3 years ago
3 0
To solve this you have to use the Pythagorean Theorem. 
Which is a²+b²=c²
CF,CD,and DF makes a triangle 
Substitute it in
a²+8²=17² Add
a²+64=289 Subtract 64 to get a alone
a²=225 Square root it 
a=15


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