¯¯¯ , bd¯¯¯¯¯ , and cd¯¯¯¯¯ are angle bisectors of the sides of △abc . cf=8 m and cd=17 m. what is de ?

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¯¯¯ , bd¯¯¯¯¯ , and cd¯¯¯¯¯ are angle bisectors of the sides of △abc . cf=8 m and cd=17 m. what is de ?

DE= 15 m

Explanation:

Since, in [tex]\triangle CDF[/tex],

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

Thus, [tex]DF^2=225 \Rightarrow DF=\sqrt{225} \Rightarrow DF=15[/tex] m

Again, in [tex]\triangle BDF[/tex] and [tex]\triangle BDE[/tex],

[tex]\angle BED= \angle BFD[/tex] ( Right angles)

BD=BD ( common edges)

[tex]\angle DBE= \angle DBF[/tex] (BD makes the angle bisector of angle B.)

Thus, according to AAS condition- [tex]\triangle BDF\cong \triangle BDE[/tex]

So, DE=DF (CPCT)

Therefore, DE=DF=15m⇒DE=15 m

The length of DF is 12 m.

Step-by-step explanation:

From the given description we can draw a figure as shown below. The lines AD,BD and CD are bisectors. The sides DE,DF and DG are perpendicular to AB,BC and CD respectively.

Draw a circle inside the triangle ABC centered at D. Since the sides DE,DF and DG are perpendicular to AB,BC and CD respectively, therefore the lines DE,DF and DG are radius of the circle.

Use pythagoras is triangle ADG.

[tex]hypotenuse^2=base^2+perpendicular^2[/tex]

[tex]AD^2=AG^2+DG^2[/tex]

[tex]13^2=5^2+DG^2[/tex]

[tex]169=25+DG^2[/tex]

[tex]144=DG^2[/tex]

[tex]12=DG[/tex]

The length of DG is 12, therefore the radius of the circle is 12.

Since DF is the radius of the circle, therefore the value of DF is 12.

[tex]ad¯¯¯¯¯ , bd¯¯¯¯¯ , and cd¯¯¯¯¯ are angle bisectors of the sides of △abc . ag=5 m and ad=13 m. wha[/tex]

DG=16 m

Step-by-step explanation:

Given: AD, BD and Cd are the angle bisectors of the sides of ΔABC and B=12m and BD=20m.

To find: The value of DG.

Solution: It is given that AD, BD and Cd are the angle bisectors of the sides of ΔABC and B=12m and BD=20m, then from the ΔBED, we have

[tex](BD)^{2}=(BE)^{2}+(ED)^{2}[/tex]

Substituting the given values, we have

[tex](20)^2=(12)^2+(ED)^2[/tex]

[tex]400=144=(ED)^2[/tex]

[tex]400-144=(ED)^2[/tex]

[tex]256=(ED)^2[/tex]

[tex]16 m=ED[/tex]

Thus, the value of ED is 16m.

Now, we know that the distance from the mid points of the sides of the given triangle to the circumcenter D are equal, thus

ED=DG

ED=DG=16

Therefore, the value of DG is 16m.

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

Since triangle BDE is right, we can check it for special rules: BD is hypotenuse = 20, and BE = 12, both divisible by 4: get 5 and 3... so it's a 3-4-5 special right triangle, the 4×4 = 16 m for side DE.

Since AD bisects angle A, I think then side DG should be congruent with DG. So then

DG = 16 m

The correct answer is

DF = 12m

Is there a picture Malik

The answer is 16

And also , can people please stop posting did you get the answer yet, just so you could get the points. They only take two answers, and that's a waste of somebody who could've answered or helped.

This is a bit tough but I believe it is 12 m

I just took this quiz. Its 15